{
  "index": "2024-B-4",
  "type": "COMB",
  "tag": [
    "COMB",
    "ANA",
    "NT"
  ],
  "difficulty": "",
  "question": "Let $n$ be a positive integer. Set $a_{n,0} = 1$. For $k \\geq 0$, choose an integer $m_{n,k}$ uniformly at random from the set $\\{1,\\dots,n\\}$, and let\n\\[\na_{n,k+1} = \\begin{cases} a_{n,k} + 1, & \\mbox{if $m_{n,k} > a_{n,k};$} \\\\\na_{n,k}, & \\mbox{if $m_{n,k} = a_{n,k}$;} \\\\\na_{n,k}-1, & \\mbox{if $m_{n,k} < a_{n,k}$.}\n\\end{cases}\n\\]\nLet $E(n)$ be the expected value of $a_{n,n}$. Determine $\\lim_{n\\to \\infty} E(n)/n$.",
  "solution": "The limit equals $\\frac{1-e^{-2}}{2}$.\n\n\\noindent\n\\textbf{First solution.}\nWe first reformulate the problem as a Markov chain.\nLet $v_k$ be the column vector of length $n$ whose $i$-th entry is the probability that $a_{n,k} = i$, so that $v_0$ is the vector $(1,0,\\dots,0)$.\nThen for all $k \\geq 0$, $v_{k+1} = A v_k$ where $A$ is the $n \\times n$\nmatrix defined by\n\\[\nA_{ij} = \\begin{cases}\n\\frac{1}{n} & \\mbox{if $i = j$} \\\\\n\\frac{j-1}{n} & \\mbox{if $i = j-1$} \\\\\n\\frac{n-j}{n} & \\mbox{if $i = j+1$} \\\\\n0 & \\mbox{otherwise.}\n\\end{cases}\n\\]\nLet $w$ be the row vector $(1, \\dots, n)$; then the expected value of $a_{n,k}$ is the sole entry of the $1 \\times 1$ matrix $w v_k = w A^k v_0$. In particular, $E(n) = w A^n v_0$.\n\nWe compute some left eigenvectors of $A$. First,\n\\[\nw_0 := (1,\\dots,1)\n\\]\nsatisfies $Aw_0 = w_0$. Second,\n\\begin{align*}\nw_1 &:= (n-1, n-3, \\dots, 3-n, 1-n) \\\\\n&= (n-2j+1\\colon j=1,\\dots,n)\n\\end{align*}\nsatisfies $Aw_1 = \\frac{n-2}{n} w_1$: the $j$-th entry of $Aw_i$ equals\n\\begin{align*}\n&\\frac{j-1}{n} (n+3-2j) + \\frac{1}{n} (n+1-2j) + \\frac{n-j}{n} (n-1-2j) \\\\\n&\\quad= \\frac{n-2}{n} (n-2j+1).\n\\end{align*}\nBy the same token, we obtain\n\\[\nw = \\frac{n+1}{2} w_0 - \\frac{1}{2} w_1;\n\\]\nwe then have\n\\begin{align*}\n\\frac{E(n)}{n} &= \\frac{n+1}{2n} w_0A^n v_0 - \\frac{1}{2n} w_1A^n v_0 \\\\\n&= \\frac{n+1}{2n} w_0 v_0 - \\frac{1}{2n} \\left( 1 - \\frac{2}{n} \\right)^n w_1 v_0  \\\\\n&= \\frac{n+1}{2n} - \\frac{n-1}{2n} \\left( 1 - \\frac{2}{n} \\right)^n.\n\\end{align*}\nIn the limit, we obtain\n\\begin{align*}\n\\lim_{n \\to \\infty} \\frac{E(n)}{n} &= \\frac{1}{2} - \\frac{1}{2} \\lim_{n \\to \\infty} \\left( 1 - \\frac{2}{n} \\right)^n \\\\\n&= \\frac{1}{2} - \\frac{1}{2} e^{-2}.\n\\end{align*}\n\n\\noindent\n\\textbf{Remark.}\nWith a bit more work, one can show that $A$ has eigenvalues\n$\\frac{n-2j}{n}$ for $j=0,\\dots,n-1$, and find the corresponding left and right eigenvectors.\nIn particular, it is also possible (but much more complicated) to express $v_0$ as a linear combination of right eigenvectors and use this to calculate $A^n v_0$.\n\n\\noindent\n\\textbf{Second solution.} \nWe reinterpret the Markov chain in combinatorial terms.\nConsider an apparatus consisting of one red light bulb, which is initially lit,\nplus $n-1$ white light bulbs, which are initially unlit. \nWe then repeatedly perform the following operation. \nPick one light bulb uniformly at random. If it is the red bulb, do nothing;\notherwise, switch the bulb from lit to unlit or vice versa.\nAfter $k$ operations of this form, the random variable $a_{n,k}$ is equal to the number of lit bulbs (including the red bulb).\n\nWe may then compute the expected value of $a_{n,n}$ by summing over bulbs.\nThe red bulb contributes 1 no matter what. Each other bulb contributes $1$ if it is switched an odd number of times and 0 if it is switched an even number of times,\nor equivalently $\\frac{1}{2}(1-(-1)^j)$ where $j$ is the number of times this bulb is switched.\nHence each bulb other than the red bulb contributes\n\\begin{align*} \n&n^{-n} \\sum_{i=0}^n \\frac{1}{2}(1-(-1)^i) \\binom{n}{i} (n-1)^{n-i} \\\\\n&= \\frac{n^{-n}}{2} \\left( \\sum_{i=0}^n \\binom{n}{i} (n-1)^{n-i} \n- \\sum_{i=0}^n (-1)^i \\binom{n}{i} (n-1)^{n-i} \\right) \\\\\n&= \\frac{n^{-n}}{2} \\left( (1+(n-1))^n - (-1+(n-1))^n \\right) \\\\\n&= \\frac{n^{-n}}{2} (n^2 - (n-2)^n) \\\\\n&= \\frac{1}{2} - \\frac{1}{2} \\left( 1 - \\frac{2}{n} \\right)^n.\n\\end{align*}\nThis tends to $\\frac{1 - e^{-2}}{2}$ as $n \\to \\infty$. Since $E(n)$ equals $n-1$ times this contribution plus 1, $\\frac{E(n)}{n}$ tends to the same limit.\n\n\\noindent\n\\textbf{Third solution.}\nWe compare the effect of taking \n$a_{n,0} = j$ versus $a_{n,0} = j+1$ for some $j \\in \\{1,\\dots,n-1\\}$.\nIf $m_{n,0} \\in \\{j,j+1\\}$ then the values of $a_{n,1}$ coincide, as then do the subsequent values\nof $a_{n,k}$; this occurs with probability $\\frac{2}{n}$. Otherwise, the values of $a_{n,1}$ differ by 1 and the situation repeats.\n\nIterating, we see that the two sequences remain 1 apart (in the same direction) with probability $\\left( \\frac{n-2}{n} \\right)^n$ and converge otherwise. Consequently, changing the start value from $j$ to $j+1$ increases the expected value of $a_{n,n}$ by $\\left( \\frac{n-2}{n} \\right)^n$. \n\nNow let $c$ be the expected value of $a_{n,n}$ in the original setting where $a_{n,0} = 1$.\nBy symmetry, if we started with $a_{n,0} = n$ the expected value would change from $c$ to $n+1-c$;\non the other hand, by the previous paragraph it would increase by \n$(n-1)\\left( \\frac{n-2}{n} \\right)^n$. We deduce that\n\\[\nc = \\frac{1}{2} \\left( n+1 - (n-1) \\left( \\frac{n-2}{n} \\right)^n \\right)\n\\]\nand as above this yields the claimed limit.",
  "vars": [
    "n",
    "k",
    "m",
    "a",
    "a_n,0",
    "a_n,k",
    "a_n,k+1",
    "a_n,n",
    "m_n,k",
    "v",
    "v_k",
    "v_0",
    "w",
    "w_0",
    "w_1",
    "E",
    "A",
    "A_ij",
    "i",
    "j",
    "c"
  ],
  "params": [],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "posint",
        "k": "stepindex",
        "m": "randpick",
        "a": "stateval",
        "a_n,0": "stateinit",
        "a_n,k": "statecurrent",
        "a_n,k+1": "statenext",
        "a_n,n": "statefinal",
        "m_n,k": "pickvariable",
        "v": "probvector",
        "v_k": "probvectork",
        "v_0": "probvectorzero",
        "w": "weightvector",
        "w_0": "weightzero",
        "w_1": "weightone",
        "E": "expectedval",
        "A": "transitionmatrix",
        "A_ij": "matrixentry",
        "i": "rowindex",
        "j": "colindex",
        "c": "resultconst"
      },
      "question": "Let $\\text{posint}$ be a positive integer. Set $\\text{stateinit}=1$. For $\\text{stepindex}\\ge 0$, choose an integer $\\text{pickvariable}$ uniformly at random from the set $\\{1,\\dots,\\text{posint}\\}$, and let\n\\[ \n\\text{statenext}=\\begin{cases} \\text{statecurrent}+1, & \\mbox{if }\\text{pickvariable}>\\text{statecurrent};\\\\\n\\text{statecurrent}, & \\mbox{if }\\text{pickvariable}=\\text{statecurrent};\\\\\n\\text{statecurrent}-1, & \\mbox{if }\\text{pickvariable}<\\text{statecurrent}.\\end{cases}\n\\]\nLet $\\text{expectedval}(\\text{posint})$ be the expected value of $\\text{statefinal}$. Determine $\\displaystyle \\lim_{\\text{posint}\\to\\infty}\\,\\frac{\\text{expectedval}(\\text{posint})}{\\text{posint}}$.",
      "solution": "The limit equals $\\dfrac{1-e^{-2}}{2}$.\n\nFirst solution.\nWe first reformulate the problem as a Markov chain.  Let $\\text{probvectork}$ be the column vector of length $\\text{posint}$ whose $\\text{colindex}$-th entry is the probability that $\\text{statecurrent}=\\text{colindex}$; in particular, $\\text{probvectorzero}=(1,0,\\dots,0)$.  For every $\\text{stepindex}\\ge0$ we then have $\\text{probvector}_{\\text{stepindex}+1}=\\text{transitionmatrix}\\,\\text{probvectork}$, where the $\\text{posint}\\times\\text{posint}$ matrix $\\text{transitionmatrix}$ is defined by\n\\[\n\\text{transitionmatrix}_{\\text{rowindex}\\,\\text{colindex}}=\n  \\begin{cases}\n   \\dfrac{1}{\\text{posint}}, & \\text{rowindex}=\\text{colindex},\\\\[4pt]\n   \\dfrac{\\text{colindex}-1}{\\text{posint}}, & \\text{rowindex}=\\text{colindex}-1,\\\\[4pt]\n   \\dfrac{\\text{posint}-\\text{colindex}}{\\text{posint}}, & \\text{rowindex}=\\text{colindex}+1,\\\\[4pt]\n   0, & \\text{otherwise.}\n  \\end{cases}\n\\]\nLet $\\text{weightvector}=(1,\\dots,\\text{posint})$.  The expected value of $\\text{stateval}$ after $\\text{stepindex}$ steps is the single entry of the $1\\times1$ matrix $\\text{weightvector}\\,\\text{probvectork}=\\text{weightvector}\\,\\text{transitionmatrix}^{\\text{stepindex}}\\text{probvectorzero}$.  In particular, $\\text{expectedval}(\\text{posint})=\\text{weightvector}\\,\\text{transitionmatrix}^{\\text{posint}}\\text{probvectorzero}$.\n\nWe next compute some left eigenvectors of $\\text{transitionmatrix}$.  First,\n\\[\n\\text{weightzero}:=(1,\\dots,1)\n\\]\nsatisfies $\\text{weightzero}\\,\\text{transitionmatrix}=\\text{weightzero}$.  Second,\n\\[\\begin{aligned}\n\\text{weightone}&:=(\\text{posint}-1,\\,\\text{posint}-3,\\dots,3-\\text{posint},1-\\text{posint})\\\\\n&=(\\text{posint}-2\\text{colindex}+1:\\,\\text{colindex}=1,\\dots,\\text{posint})\n\\end{aligned}\\]\nsatisfies $\\text{weightone}\\,\\text{transitionmatrix}=\\dfrac{\\text{posint}-2}{\\text{posint}}\\,\\text{weightone}$.  By the same token we have\n\\[\n\\text{weightvector}=\\frac{\\text{posint}+1}{2}\\,\\text{weightzero}-\\frac{1}{2}\\,\\text{weightone}.\n\\]\nTherefore\n\\[\\begin{aligned}\n\\frac{\\text{expectedval}(\\text{posint})}{\\text{posint}}\n&=\\frac{\\text{posint}+1}{2\\,\\text{posint}}\\,\\text{weightzero}\\,\\text{probvectorzero}-\\frac{1}{2\\,\\text{posint}}\\left(1-\\frac{2}{\\text{posint}}\\right)^{\\text{posint}}\\text{weightone}\\,\\text{probvectorzero}\\\\[6pt]\n&=\\frac{\\text{posint}+1}{2\\,\\text{posint}}-\\frac{\\text{posint}-1}{2\\,\\text{posint}}\\left(1-\\frac{2}{\\text{posint}}\\right)^{\\text{posint}}.\n\\end{aligned}\\]\nTaking $\\text{posint}\\to\\infty$ yields\n\\[\n\\lim_{\\text{posint}\\to\\infty}\\frac{\\text{expectedval}(\\text{posint})}{\\text{posint}}=\\frac12-\\frac12e^{-2}=\\frac{1-e^{-2}}{2}.\n\\]\n\nRemark.\nWith more work one checks that $\\text{transitionmatrix}$ has eigenvalues $\\dfrac{\\text{posint}-2\\text{colindex}}{\\text{posint}}$ for $\\text{colindex}=0,\\dots,\\text{posint}-1$ together with corresponding eigenvectors; one could then expand $\\text{probvectorzero}$ in that basis as well.\n\nSecond solution.\nRe-interpret the Markov chain combinatorially.  Consider one red light bulb, initially lit, and $\\text{posint}-1$ white bulbs, initially unlit.  Repeatedly pick a bulb uniformly at random.  If it is the red bulb, do nothing; otherwise switch its state.  After $\\text{stepindex}$ operations, the random variable $\\text{stateval}$ equals the number of lit bulbs (including the red one).\n\nTo compute $\\text{statefinal}$ we sum over bulbs.  The red bulb always contributes 1.  Each white bulb contributes $1$ if it is switched an odd number of times and $0$ otherwise, i.e.\n$\\tfrac12\\bigl(1-(-1)^{\\text{rowindex}}\\bigr)$ where $\\text{rowindex}$ is the number of times that bulb is switched.  Hence each white bulb contributes\n\\[\\begin{aligned}\n&\\text{posint}^{-\\text{posint}}\\sum_{\\text{rowindex}=0}^{\\text{posint}}\\frac12\\bigl(1-(-1)^{\\text{rowindex}}\\bigr)\\binom{\\text{posint}}{\\text{rowindex}}(\\text{posint}-1)^{\\text{posint}-\\text{rowindex}}\\\\[4pt]\n&=\\frac{\\text{posint}^{-\\text{posint}}}{2}\\Bigl[(1+\\text{posint}-1)^{\\text{posint}}-(-1+\\text{posint}-1)^{\\text{posint}}\\Bigr]\\\\[4pt]\n&=\\frac{\\text{posint}^{-\\text{posint}}}{2}\\Bigl(\\text{posint}^{2}-(\\text{posint}-2)^{\\text{posint}}\\Bigr)\\\\[4pt]\n&=\\frac12-\\frac12\\left(1-\\frac{2}{\\text{posint}}\\right)^{\\text{posint}}.\n\\end{aligned}\\]\nAs $\\text{posint}\\to\\infty$ this tends to $\\dfrac{1-e^{-2}}{2}$.  Since $\\text{expectedval}(\\text{posint})$ equals $(\\text{posint}-1)$ times this contribution plus 1, the ratio $\\dfrac{\\text{expectedval}(\\text{posint})}{\\text{posint}}$ has the same limit.\n\nThird solution.\nCompare the processes starting from $\\text{stateval}=\\text{colindex}$ and from $\\text{stateval}=\\text{colindex}+1$ for some $\\text{colindex}\\in\\{1,\\dots,\\text{posint}-1\\}$.  If $\\text{pickvariable}\\in\\{\\text{colindex},\\text{colindex}+1\\}$, then $\\text{stateval}$ coincides after one step and forever after; this occurs with probability $\\tfrac{2}{\\text{posint}}$.  Otherwise the two values differ by 1, and the situation repeats.  Hence after $\\text{posint}$ steps the probability that the two processes are still 1 apart equals $\\bigl(\\tfrac{\\text{posint}-2}{\\text{posint}}\\bigr)^{\\text{posint}}$.\n\nConsequently, increasing the initial value from $\\text{colindex}$ to $\\text{colindex}+1$ raises the expected value of $\\text{statefinal}$ by $\\bigl(\\tfrac{\\text{posint}-2}{\\text{posint}}\\bigr)^{\\text{posint}}$.  Let $\\text{resultconst}$ denote the expected value when $\\text{stateval}$ starts at 1.  By symmetry, starting instead from $\\text{stateval}=\\text{posint}$ would change the expectation to $\\text{posint}+1-\\text{resultconst}$, i.e. increase it by $(\\text{posint}-1)\\bigl(\\tfrac{\\text{posint}-2}{\\text{posint}}\\bigr)^{\\text{posint}}$.  Therefore\n\\[\n\\text{resultconst}=\\frac12\\Bigl(\\text{posint}+1-(\\text{posint}-1)\\bigl(\\tfrac{\\text{posint}-2}{\\text{posint}}\\bigr)^{\\text{posint}}\\Bigr),\n\\]\nwhich again yields $\\dfrac{1-e^{-2}}{2}$ in the limit $\\text{posint}\\to\\infty$.\n"
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "driftwood",
        "k": "papertrail",
        "m": "sunflower",
        "a": "rainbucket",
        "a_n,0": "orchardseal",
        "a_n,k": "crimsonleaf",
        "a_n,k+1": "gallopstone",
        "a_n,n": "vantagepeak",
        "m_n,k": "purelotion",
        "v": "swirlanchor",
        "v_k": "peppermoss",
        "v_0": "lucidharbor",
        "w": "velvetdust",
        "w_0": "starlingdew",
        "w_1": "kindlethorn",
        "E": "meadowflux",
        "A": "lunarbridge",
        "A_ij": "hazelspire",
        "i": "brookshuffle",
        "j": "lanternmist",
        "c": "foxtailbloom"
      },
      "question": "Let $driftwood$ be a positive integer. Set $orchardseal = 1$. For $papertrail \\geq 0$, choose an integer $purelotion$ uniformly at random from the set $\\{1,\\dots,driftwood\\}$, and let\n\\[\n gallopstone = \\begin{cases} crimsonleaf + 1, & \\mbox{if $purelotion > crimsonleaf;$} \\\\\n crimsonleaf, & \\mbox{if $purelotion = crimsonleaf$;} \\\\\n crimsonleaf-1, & \\mbox{if $purelotion < crimsonleaf$.}\n \\end{cases}\n\\]\nLet $meadowflux(driftwood)$ be the expected value of $vantagepeak$. Determine $\\lim_{driftwood\\to \\infty} \\meadowflux(driftwood)/driftwood$.",
      "solution": "The limit equals $\\frac{1-e^{-2}}{2}$.\\par\\noindent\\textbf{First solution.} We first reformulate the problem as a Markov chain. Let $peppermoss$ be the column vector of length $driftwood$ whose $brookshuffle$-th entry is the probability that $crimsonleaf = brookshuffle$, so that $lucidharbor$ is the vector $(1,0,\\dots,0)$. Then for all $papertrail \\geq 0$, $swirlanchor_{papertrail+1} = lunarbridge\\,peppermoss$ where $lunarbridge$ is the $driftwood \\times driftwood$ matrix defined by\n\\[ hazelspire = \\begin{cases}\n \\dfrac{1}{driftwood} & \\mbox{if $brookshuffle = lanternmist$} \\\\\n \\dfrac{lanternmist-1}{driftwood} & \\mbox{if $brookshuffle = lanternmist-1$} \\\\\n \\dfrac{driftwood-lanternmist}{driftwood} & \\mbox{if $brookshuffle = lanternmist+1$} \\\\\n 0 & \\mbox{otherwise.}\n \\end{cases} \\]\nLet $velvetdust$ be the row vector $(1, \\dots, driftwood)$; then the expected value of $crimsonleaf$ is the sole entry of the $1 \\times 1$ matrix $velvetdust\\,peppermoss = velvetdust\\,lunarbridge^{papertrail}\\,lucidharbor$. In particular, $\\meadowflux(driftwood) = velvetdust\\,lunarbridge^{driftwood}\\,lucidharbor$.\\par\nWe compute some left eigenvectors of $lunarbridge$. First,\n\\[ starlingdew := (1,\\dots,1) \\]\nsatisfies $lunarbridge\\,starlingdew = starlingdew$. Second,\n\\begin{align*}\n kindlethorn &:= (driftwood-1, driftwood-3, \\dots, 3-driftwood, 1-driftwood)\\\\\n &= (driftwood-2\\,lanternmist+1\\colon lanternmist=1,\\dots,driftwood)\n \\end{align*}\nsatisfies $lunarbridge\\,kindlethorn = \\dfrac{driftwood-2}{driftwood}\\,kindlethorn$. By the same token,\n\\[ velvetdust = \\frac{driftwood+1}{2}\\,starlingdew - \\frac{1}{2}\\,kindlethorn; \\]\nwhence\n\\begin{align*}\n \\frac{\\meadowflux(driftwood)}{driftwood} &= \\frac{driftwood+1}{2driftwood}\\,starlingdew\\,lunarbridge^{driftwood}\\,lucidharbor - \\frac{1}{2driftwood}\\,kindlethorn\\,lunarbridge^{driftwood}\\,lucidharbor\\\\\n &= \\frac{driftwood+1}{2driftwood} - \\frac{driftwood-1}{2driftwood}\\left(1-\\frac{2}{driftwood}\\right)^{driftwood}.\n \\end{align*}\nTaking the limit yields\n\\[ \\lim_{driftwood\\to\\infty}\\frac{\\meadowflux(driftwood)}{driftwood} = \\frac12 - \\frac12 e^{-2}. \\]\n\\noindent\\textbf{Remark.} With more work one finds that $lunarbridge$ has eigenvalues $\\tfrac{driftwood-2\\,lanternmist}{driftwood}$ for $lanternmist=0,\\dots,driftwood-1$, etc.\\par\\noindent\\textbf{Second solution.} Consider one red and $driftwood-1$ white bulbs as described. After $papertrail$ operations, $crimsonleaf$ is the number of lit bulbs. The red bulb always contributes $1$. Any other bulb contributes $1$ if toggled an odd number of times, $0$ otherwise, i.e., $\\tfrac12(1-(-1)^{brookshuffle})$ where $brookshuffle$ is the number of toggles. Hence each white bulb contributes\n\\begin{align*}\n &driftwood^{-driftwood}\\sum_{brookshuffle=0}^{driftwood}\\tfrac12(1-(-1)^{brookshuffle})\\binom{driftwood}{brookshuffle}(driftwood-1)^{driftwood-brookshuffle}\\\\\n &= \\tfrac12-\\tfrac12\\left(1-\\frac{2}{driftwood}\\right)^{driftwood}.\n \\end{align*}\nThis tends to $\\tfrac{1-e^{-2}}{2}$. Since $\\meadowflux(driftwood)$ equals $(driftwood-1)$ times this plus $1$, the ratio $\\meadowflux(driftwood)/driftwood$ has the same limit.\\par\\noindent\\textbf{Third solution.} Compare starting from $crimsonleaf=lanternmist$ versus $crimsonleaf=lanternmist+1$ with $1\\leq lanternmist\\leq driftwood-1$. They coalesce with probability $\\frac{2}{driftwood}$ each step; otherwise they stay $1$ apart. Thus the difference in expectations is $\\left(\\frac{driftwood-2}{driftwood}\\right)^{driftwood}$. Let $foxtailbloom$ be the expectation when $crimsonleaf=1$. By symmetry starting from $crimsonleaf=driftwood$ gives $driftwood+1-foxtailbloom$, but also increases the expectation by $(driftwood-1)\\left(\\frac{driftwood-2}{driftwood}\\right)^{driftwood}$. Hence\n\\[ foxtailbloom=\\tfrac12\\Bigl(driftwood+1-(driftwood-1)\\bigl(\\tfrac{driftwood-2}{driftwood}\\bigr)^{driftwood}\\Bigr), \\]\nand the desired limit follows as before."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "tinyvalue",
        "k": "stopperindex",
        "m": "certainpick",
        "a": "voidmeasure",
        "a_{n,0}": "voidbasestart",
        "a_{n,k}": "voidlevelnow",
        "a_{n,k+1}": "voidlevelnext",
        "a_{n,n}": "voidlevelfinal",
        "m_{n,k}": "certainpicker",
        "v": "lonelyscalar",
        "v_k": "lonelyscalarstep",
        "v_0": "lonelyscalarbase",
        "w": "verticalvector",
        "w_0": "nullarray",
        "w_1": "steadyarray",
        "E": "surprisal",
        "A": "nonsquare",
        "A_{ij}": "aggregate",
        "i": "columncount",
        "j": "rowcount",
        "c": "variance"
      },
      "question": "Let $tinyvalue$ be a positive integer. Set $voidbasestart = 1$. For $stopperindex \\geq 0$, choose an integer $certainpicker$ uniformly at random from the set $\\{1,\\dots,tinyvalue\\}$, and let\n\\[\nvoidlevelnext = \\begin{cases} voidlevelnow + 1, & \\mbox{if $certainpicker > voidlevelnow;$} \\\\\nvoidlevelnow, & \\mbox{if $certainpicker = voidlevelnow$;} \\\\\nvoidlevelnow-1, & \\mbox{if $certainpicker < voidlevelnow$.}\n\\end{cases}\n\\]\nLet $\\surprisal(tinyvalue)$ be the expected value of $voidlevelfinal$. Determine $\\lim_{tinyvalue\\to \\infty} \\surprisal(tinyvalue)/tinyvalue$.",
      "solution": "The limit equals $\\frac{1-e^{-2}}{2}$. \n\n\\noindent\n\\textbf{First solution.}\nWe first reformulate the problem as a Markov chain.\nLet $lonelyscalarstep$ be the column vector of length $tinyvalue$ whose $columncount$-th entry is the probability that $voidlevelnow = columncount$, so that $lonelyscalarbase$ is the vector $(1,0,\\dots,0)$.\nThen for all $stopperindex \\geq 0$, $lonelyscalar_{stopperindex+1} = nonsquare\\,lonelyscalarstep$ where $nonsquare$ is the $tinyvalue \\times tinyvalue$\nmatrix defined by\n\\[\naggregate = \\begin{cases}\n\\frac{1}{tinyvalue} & \\mbox{if $columncount = rowcount$} \\\\\n\\frac{rowcount-1}{tinyvalue} & \\mbox{if $columncount = rowcount-1$} \\\\\n\\frac{tinyvalue-rowcount}{tinyvalue} & \\mbox{if $columncount = rowcount+1$} \\\\\n0 & \\mbox{otherwise.}\n\\end{cases}\n\\]\nLet $verticalvector$ be the row vector $(1, \\dots, tinyvalue)$; then the expected value of $voidlevelnow$ is the sole entry of the $1 \\times 1$ matrix $verticalvector\\,lonelyscalarstep = verticalvector\\,nonsquare^{stopperindex}\\,lonelyscalarbase$. In particular, $\\surprisal(tinyvalue) = verticalvector\\,nonsquare^{tinyvalue}\\,lonelyscalarbase$.\n\nWe compute some left eigenvectors of $nonsquare$. First,\n\\[\nnullarray := (1,\\dots,1)\n\\]\nsatisfies $nonsquare\\,nullarray = nullarray$. Second,\n\\begin{align*}\nsteadyarray &:= (tinyvalue-1, tinyvalue-3, \\dots, 3-tinyvalue, 1-tinyvalue) \\\\\n&= (tinyvalue-2rowcount+1\\colon rowcount=1,\\dots,tinyvalue)\n\\end{align*}\nsatisfies $nonsquare\\,steadyarray = \\frac{tinyvalue-2}{tinyvalue}\\,steadyarray$: the $rowcount$-th entry of $nonsquare\\,steadyarray$ equals\n\\begin{align*}\n&\\frac{rowcount-1}{tinyvalue} (tinyvalue+3-2rowcount) + \\frac{1}{tinyvalue} (tinyvalue+1-2rowcount) + \\frac{tinyvalue-rowcount}{tinyvalue} (tinyvalue-1-2rowcount) \\\\\n&\\quad= \\frac{tinyvalue-2}{tinyvalue} (tinyvalue-2rowcount+1).\n\\end{align*}\nBy the same token, we obtain\n\\[\nverticalvector = \\frac{tinyvalue+1}{2}\\,nullarray - \\frac{1}{2}\\,steadyarray;\n\\]\nwe then have\n\\begin{align*}\n\\frac{\\surprisal(tinyvalue)}{tinyvalue} &= \\frac{tinyvalue+1}{2tinyvalue}\\,nullarray\\,nonsquare^{tinyvalue}\\,lonelyscalarbase - \\frac{1}{2tinyvalue}\\,steadyarray\\,nonsquare^{tinyvalue}\\,lonelyscalarbase \\\\\n&= \\frac{tinyvalue+1}{2tinyvalue}\\,nullarray\\,lonelyscalarbase - \\frac{tinyvalue-1}{2tinyvalue} \\left( 1 - \\frac{2}{tinyvalue} \\right)^{tinyvalue} \\steadyarray\\,lonelyscalarbase  \\\\\n&= \\frac{tinyvalue+1}{2tinyvalue} - \\frac{tinyvalue-1}{2tinyvalue} \\left( 1 - \\frac{2}{tinyvalue} \\right)^{tinyvalue}.\n\\end{align*}\nIn the limit, we obtain\n\\begin{align*}\n\\lim_{tinyvalue \\to \\infty} \\frac{\\surprisal(tinyvalue)}{tinyvalue} &= \\frac{1}{2} - \\frac{1}{2} \\lim_{tinyvalue \\to \\infty} \\left( 1 - \\frac{2}{tinyvalue} \\right)^{tinyvalue} \\\\\n&= \\frac{1}{2} - \\frac{1}{2} e^{-2}.\n\\end{align*}\n\n\\noindent\n\\textbf{Remark.}\nWith a bit more work, one can show that $nonsquare$ has eigenvalues\n$\\frac{tinyvalue-2rowcount}{tinyvalue}$ for $rowcount=0,\\dots,tinyvalue-1$, and find the corresponding left and right eigenvectors.\nIn particular, it is also possible (but much more complicated) to express $lonelyscalarbase$ as a linear combination of right eigenvectors and use this to calculate $nonsquare^{tinyvalue}\\,lonelyscalarbase$.\n\n\\noindent\n\\textbf{Second solution.} \nWe reinterpret the Markov chain in combinatorial terms.\nConsider an apparatus consisting of one red light bulb, which is initially lit,\nplus $tinyvalue-1$ white light bulbs, which are initially unlit. \nWe then repeatedly perform the following operation. \nPick one light bulb uniformly at random. If it is the red bulb, do nothing;\notherwise, switch the bulb from lit to unlit or vice versa.\nAfter $stopperindex$ operations of this form, the random variable $voidlevelnow$ is equal to the number of lit bulbs (including the red bulb).\n\nWe may then compute the expected value of $voidlevelfinal$ by summing over bulbs.\nThe red bulb contributes 1 no matter what. Each other bulb contributes $1$ if it is switched an odd number of times and 0 if it is switched an even number of times,\nor equivalently $\\frac{1}{2}(1-(-1)^{rowcount})$ where $rowcount$ is the number of times this bulb is switched.\nHence each bulb other than the red bulb contributes\n\\begin{align*} \n&tinyvalue^{-tinyvalue} \\sum_{columncount=0}^{tinyvalue} \\frac{1}{2}(1-(-1)^{columncount}) \\binom{tinyvalue}{columncount} (tinyvalue-1)^{tinyvalue-columncount} \\\\\n&= \\frac{tinyvalue^{-tinyvalue}}{2} \\left( \\sum_{columncount=0}^{tinyvalue} \\binom{tinyvalue}{columncount} (tinyvalue-1)^{tinyvalue-columncount} \n- \\sum_{columncount=0}^{tinyvalue} (-1)^{columncount} \\binom{tinyvalue}{columncount} (tinyvalue-1)^{tinyvalue-columncount} \\right) \\\\\n&= \\frac{tinyvalue^{-tinyvalue}}{2} \\left( (1+(tinyvalue-1))^{tinyvalue} - (-1+(tinyvalue-1))^{tinyvalue} \\right) \\\\\n&= \\frac{tinyvalue^{-tinyvalue}}{2} (tinyvalue^2 - (tinyvalue-2)^{tinyvalue}) \\\\\n&= \\frac{1}{2} - \\frac{1}{2} \\left( 1 - \\frac{2}{tinyvalue} \\right)^{tinyvalue}.\n\\end{align*}\nThis tends to $\\frac{1 - e^{-2}}{2}$ as $tinyvalue \\to \\infty$. Since $\\surprisal(tinyvalue)$ equals $tinyvalue-1$ times this contribution plus 1, $\\frac{\\surprisal(tinyvalue)}{tinyvalue}$ tends to the same limit.\n\n\\noindent\n\\textbf{Third solution.}\nWe compare the effect of taking \n$voidbasestart = rowcount$ versus $voidbasestart = rowcount+1$ for some $rowcount \\in \\{1,\\dots,tinyvalue-1\\}$.\nIf $m_{tinyvalue,0} \\in \\{rowcount,rowcount+1\\}$ then the values of $a_{tinyvalue,1}$ coincide, as then do the subsequent values\nof $voidlevelnow$; this occurs with probability $\\frac{2}{tinyvalue}$. Otherwise, the values of $a_{tinyvalue,1}$ differ by 1 and the situation repeats.\n\nIterating, we see that the two sequences remain 1 apart (in the same direction) with probability $\\left( \\frac{tinyvalue-2}{tinyvalue} \\right)^{tinyvalue}$ and converge otherwise. Consequently, changing the start value from $rowcount$ to $rowcount+1$ increases the expected value of $voidlevelfinal$ by $\\left( \\frac{tinyvalue-2}{tinyvalue} \\right)^{tinyvalue}$. \n\nNow let $variance$ be the expected value of $voidlevelfinal$ in the original setting where $voidbasestart = 1$.\nBy symmetry, if we started with $voidbasestart = tinyvalue$ the expected value would change from $variance$ to $tinyvalue+1-variance$;\non the other hand, by the previous paragraph it would increase by \n$(tinyvalue-1)\\left( \\frac{tinyvalue-2}{tinyvalue} \\right)^{tinyvalue}$. We deduce that\n\\[\nvariance = \\frac{1}{2} \\left( tinyvalue+1 - (tinyvalue-1) \\left( \\frac{tinyvalue-2}{tinyvalue} \\right)^{tinyvalue} \\right)\n\\]\nand as above this yields the claimed limit."
    },
    "garbled_string": {
      "map": {
        "n": "qzxwvtnp",
        "k": "hjgrksla",
        "m": "plxqrabc",
        "a": "vbnmerty",
        "a_n,0": "lkjhgfdx",
        "a_n,k": "qwertpoi",
        "a_n,k+1": "asdfghjk",
        "a_n,n": "zxcvbnml",
        "m_n,k": "yuioplkj",
        "v": "trewqzxc",
        "v_k": "plmoknij",
        "v_0": "mnbvcxzq",
        "w": "qazwsxed",
        "w_0": "edcrfvtg",
        "w_1": "tgbnhyuj",
        "E": "ujmikolp",
        "A": "rfvtgbyh",
        "A_ij": "yhnujmki",
        "j": "wsxcdevf",
        "c": "olikmjun"
      },
      "question": "Let $qzxwvtnp$ be a positive integer. Set $lkjhgfdx = 1$. For $hjgrksla \\geq 0$, choose an integer $yuioplkj$ uniformly at random from the set $\\{1,\\dots,qzxwvtnp\\}$, and let\n\\[\nasdfghjk = \\begin{cases} qwertpoi + 1, & \\mbox{if $yuioplkj > qwertpoi;$} \\\\\nqwertpoi, & \\mbox{if $yuioplkj = qwertpoi$;} \\\\\nqwertpoi-1, & \\mbox{if $yuioplkj < qwertpoi$.}\n\\end{cases}\n\\]\nLet $ujmikolp(qzxwvtnp)$ be the expected value of $zxcvbnml$. Determine $\\lim_{qzxwvtnp\\to \\infty} ujmikolp(qzxwvtnp)/qzxwvtnp$.",
      "solution": "The limit equals $\\frac{1-e^{-2}}{2}$.  \n\n\\noindent\n\\textbf{First solution.}\nWe first reformulate the problem as a Markov chain.\nLet $plmoknij$ be the column vector of length $qzxwvtnp$ whose $i$-th entry is the probability that $qwertpoi = i$, so that $mnbvcxzq$ is the vector $(1,0,\\dots,0)$.\nThen for all $hjgrksla \\geq 0$, $plmoknij_{hjgrksla+1} = rfvtgbyh\\,plmoknij$ where $rfvtgbyh$ is the $qzxwvtnp \\times qzxwvtnp$ matrix defined by\n\\[\nrfvtgbyh_{ij} = \\begin{cases}\n\\frac{1}{qzxwvtnp} & \\mbox{if $i = j$} \\\\\n\\frac{j-1}{qzxwvtnp} & \\mbox{if $i = j-1$} \\\\\n\\frac{qzxwvtnp-j}{qzxwvtnp} & \\mbox{if $i = j+1$} \\\\\n0 & \\mbox{otherwise.}\n\\end{cases}\n\\]\nLet $qazwsxed$ be the row vector $(1, \\dots, qzxwvtnp)$; then the expected value of $qwertpoi$ is the sole entry of the $1 \\times 1$ matrix $qazwsxed\\,plmoknij = qazwsxed\\,rfvtgbyh^{hjgrksla}\\,mnbvcxzq$. In particular, $ujmikolp(qzxwvtnp) = qazwsxed\\,rfvtgbyh^{qzxwvtnp}\\,mnbvcxzq$.\n\nWe compute some left eigenvectors of $rfvtgbyh$. First,\n\\[\nedcrfvtg := (1,\\dots,1)\n\\]\nsatisfies $rfvtgbyh\\,edcrfvtg = edcrfvtg$. Second,\n\\begin{align*}\ntgbnhyuj &:= (qzxwvtnp-1, qzxwvtnp-3, \\dots, 3-qzxwvtnp, 1-qzxwvtnp) \\\\\n&= (qzxwvtnp-2wsxcdevf+1\\colon wsxcdevf=1,\\dots,qzxwvtnp)\n\\end{align*}\nsatisfies $rfvtgbyh\\,tgbnhyuj = \\frac{qzxwvtnp-2}{qzxwvtnp}\\,tgbnhyuj$: the $wsxcdevf$-th entry of $rfvtgbyh\\,tgbnhyuj$ equals\n\\begin{align*}\n&\\frac{wsxcdevf-1}{qzxwvtnp} (qzxwvtnp+3-2wsxcdevf) + \\frac{1}{qzxwvtnp} (qzxwvtnp+1-2wsxcdevf) + \\frac{qzxwvtnp-wsxcdevf}{qzxwvtnp} (qzxwvtnp-1-2wsxcdevf) \\\\\n&\\quad= \\frac{qzxwvtnp-2}{qzxwvtnp} (qzxwvtnp-2wsxcdevf+1).\n\\end{align*}\nBy the same token, we obtain\n\\[\nqazwsxed = \\frac{qzxwvtnp+1}{2}\\,edcrfvtg - \\frac{1}{2}\\,tgbnhyuj;\n\\]\nwe then have\n\\begin{align*}\n\\frac{ujmikolp(qzxwvtnp)}{qzxwvtnp} &= \\frac{qzxwvtnp+1}{2qzxwvtnp}\\,edcrfvtg\\,rfvtgbyh^{qzxwvtnp}\\,mnbvcxzq - \\frac{1}{2qzxwvtnp}\\,tgbnhyuj\\,rfvtgbyh^{qzxwvtnp}\\,mnbvcxzq \\\\\n&= \\frac{qzxwvtnp+1}{2qzxwvtnp}\\,edcrfvtg\\,mnbvcxzq - \\frac{1}{2qzxwvtnp}\\left(1-\\frac{2}{qzxwvtnp}\\right)^{qzxwvtnp} tgbnhyuj\\,mnbvcxzq  \\\\\n&= \\frac{qzxwvtnp+1}{2qzxwvtnp} - \\frac{qzxwvtnp-1}{2qzxwvtnp}\\left(1-\\frac{2}{qzxwvtnp}\\right)^{qzxwvtnp}.\n\\end{align*}\nIn the limit, we obtain\n\\begin{align*}\n\\lim_{qzxwvtnp \\to \\infty} \\frac{ujmikolp(qzxwvtnp)}{qzxwvtnp} &= \\frac{1}{2} - \\frac{1}{2} \\lim_{qzxwvtnp \\to \\infty} \\left(1-\\frac{2}{qzxwvtnp}\\right)^{qzxwvtnp} \\\\\n&= \\frac{1}{2} - \\frac{1}{2} e^{-2}.\n\\end{align*}\n\n\\noindent\n\\textbf{Remark.}\nWith a bit more work, one can show that $rfvtgbyh$ has eigenvalues\n$\\frac{qzxwvtnp-2wsxcdevf}{qzxwvtnp}$ for $wsxcdevf=0,\\dots,qzxwvtnp-1$, and find the corresponding left and right eigenvectors.\nIn particular, it is also possible (but much more complicated) to express $mnbvcxzq$ as a linear combination of right eigenvectors and use this to calculate $rfvtgbyh^{qzxwvtnp}\\,mnbvcxzq$.\n\n\\noindent\n\\textbf{Second solution.}  \nWe reinterpret the Markov chain in combinatorial terms.\nConsider an apparatus consisting of one red light bulb, which is initially lit,\nplus $qzxwvtnp-1$ white light bulbs, which are initially unlit. \nWe then repeatedly perform the following operation. \nPick one light bulb uniformly at random. If it is the red bulb, do nothing;\notherwise, switch the bulb from lit to unlit or vice versa.\nAfter $hjgrksla$ operations of this form, the random variable $qwertpoi$ is equal to the number of lit bulbs (including the red bulb).\n\nWe may then compute the expected value of $zxcvbnml$ by summing over bulbs.\nThe red bulb contributes 1 no matter what. Each other bulb contributes $1$ if it is switched an odd number of times and $0$ if it is switched an even number of times,\nor equivalently $\\frac{1}{2}(1-(-1)^i)$ where $i$ is the number of times this bulb is switched.\nHence each bulb other than the red bulb contributes\n\\begin{align*} \n&qzxwvtnp^{-qzxwvtnp} \\sum_{i=0}^{qzxwvtnp} \\frac{1}{2}(1-(-1)^i) \\binom{qzxwvtnp}{i} (qzxwvtnp-1)^{qzxwvtnp-i} \\\\\n&= \\frac{qzxwvtnp^{-qzxwvtnp}}{2} \\left( \\sum_{i=0}^{qzxwvtnp} \\binom{qzxwvtnp}{i} (qzxwvtnp-1)^{qzxwvtnp-i} \n- \\sum_{i=0}^{qzxwvtnp} (-1)^i \\binom{qzxwvtnp}{i} (qzxwvtnp-1)^{qzxwvtnp-i} \\right) \\\\\n&= \\frac{qzxwvtnp^{-qzxwvtnp}}{2} \\left( (1+(qzxwvtnp-1))^{qzxwvtnp} - (-1+(qzxwvtnp-1))^{qzxwvtnp} \\right) \\\\\n&= \\frac{qzxwvtnp^{-qzxwvtnp}}{2} (qzxwvtnp^2 - (qzxwvtnp-2)^{qzxwvtnp}) \\\\\n&= \\frac{1}{2} - \\frac{1}{2} \\left(1-\\frac{2}{qzxwvtnp}\\right)^{qzxwvtnp}.\n\\end{align*}\nThis tends to $\\frac{1 - e^{-2}}{2}$ as $qzxwvtnp \\to \\infty$. Since $ujmikolp(qzxwvtnp)$ equals $qzxwvtnp-1$ times this contribution plus 1, $\\frac{ujmikolp(qzxwvtnp)}{qzxwvtnp}$ tends to the same limit.\n\n\\noindent\n\\textbf{Third solution.}\nWe compare the effect of taking \n$lkjhgfdx = wsxcdevf$ versus $lkjhgfdx = wsxcdevf+1$ for some $wsxcdevf \\in \\{1,\\dots,qzxwvtnp-1\\}$.\nIf $yuioplkj \\in \\{wsxcdevf,wsxcdevf+1\\}$ then the values of $asdfghjk$ coincide, as then do the subsequent values\nof $qwertpoi$; this occurs with probability $\\frac{2}{qzxwvtnp}$. Otherwise, the values of $qwertpoi$ differ by 1 and the situation repeats.\n\nIterating, we see that the two sequences remain 1 apart (in the same direction) with probability $\\left(\\frac{qzxwvtnp-2}{qzxwvtnp}\\right)^{qzxwvtnp}$ and converge otherwise. Consequently, changing the start value from $wsxcdevf$ to $wsxcdevf+1$ increases the expected value of $zxcvbnml$ by $\\left(\\frac{qzxwvtnp-2}{qzxwvtnp}\\right)^{qzxwvtnp}$. \n\nNow let $olikmjun$ be the expected value of $zxcvbnml$ in the original setting where $lkjhgfdx = 1$.\nBy symmetry, if we started with $lkjhgfdx = qzxwvtnp$ the expected value would change from $olikmjun$ to $qzxwvtnp+1-olikmjun$;\non the other hand, by the previous paragraph it would increase by \n$(qzxwvtnp-1)\\left(\\frac{qzxwvtnp-2}{qzxwvtnp}\\right)^{qzxwvtnp}$. We deduce that\n\\[\nolikmjun = \\frac{1}{2}\\left(qzxwvtnp+1 - (qzxwvtnp-1)\\left(\\frac{qzxwvtnp-2}{qzxwvtnp}\\right)^{qzxwvtnp}\\right)\n\\]\nand as above this yields the claimed limit."
    },
    "kernel_variant": {
      "question": "Let $n\\ge 2$ be an integer.  \nConsider the discrete--time birth–death Markov chain  \n\n\\[\nA^{(n)}=\\bigl(a_{n,k}\\bigr)_{k\\ge 0},\\qquad \na_{n,k}\\in\\{1,2,\\dots ,n\\},\\qquad a_{n,0}=1 ,\n\\]\n\nwhose transition probabilities are  \n\n\\[\n\\begin{aligned}\n\\mathbb P\\!\\bigl(a_{n,k+1}=j+1\\mid a_{n,k}=j\\bigr)&=\\frac{n-j}{n},\\\\[4pt]\n\\mathbb P\\!\\bigl(a_{n,k+1}=j  \\mid a_{n,k}=j\\bigr)&=\\frac{1}{n},\\\\[4pt]\n\\mathbb P\\!\\bigl(a_{n,k+1}=j-1\\mid a_{n,k}=j\\bigr)&=\\frac{j-1}{n}\\qquad(1\\le j\\le n),\n\\end{aligned}\n\\]\nand $\\mathbb P(\\,\\cdot\\,)=0$ for every illegal state.  \nFor $1\\le j\\le n$ denote by $\\pi_n(j)$ the stationary distribution of\n$A^{(n)}$ and, for $\\varepsilon\\in(0,\\tfrac12)$, let  \n\n\\[\n\\tau_n(\\varepsilon)=\\min\\Bigl\\{k\\ge 0:\n      \\lVert\\mathcal L_1(a_{n,k})-\\pi_n\\rVert_{\\mathrm{TV}}\\le\\varepsilon\\Bigr\\},\n\\]\nbe its $\\varepsilon$–mixing time in total variation.\n\n1.  Prove that $A^{(n)}$ is reversible and that  \n\n\\[\n\\pi_n(j)=2^{-(n-1)}\\binom{n-1}{j-1}\\qquad(1\\le j\\le n).\n\\]\n\n2.  Show that the family $\\bigl(A^{(n)}\\bigr)_{n\\ge 2}$ exhibits a sharp total–variation cut–off.  \n    More precisely, for every fixed $\\varepsilon\\in(0,\\tfrac12)$  \n\n\\[\n\\tau_n(\\varepsilon)=\\frac{n}{2}\\log n \\;+\\;O_{\\varepsilon}(n),\n\\qquad n\\longrightarrow\\infty ,\n\\]\n\nso the cut–off is located at $\\tfrac{n}{2}\\log n$ and its window has\norder $\\Theta(n)$.\n\n(Hint: write $P^{(n)}=\\tfrac{n-1}{n}\\,P^{\\mathrm{Ehr}}_{n-1}+ \\tfrac{1}{n}\\,{\\rm Id}$,\nuse Krawtchouk polynomials to diagonalise $P^{\\mathrm{Ehr}}_{n-1}$,\nthen compare $A^{(n)}$ with the Ehrenfest walk through the random\n“active clock’’ $S_k=\\sum_{i=1}^{k}\\xi_i$, $\\xi_i\\stackrel{\\mathrm{iid}}{\\sim}{\\rm Bernoulli}(1-\\tfrac1n)$.\nA tight lower bound needs \\emph{two} ingredients:  \nconcentration of $S_k$ together with the lower–bound profile of the\nEhrenfest chain.)\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Throughout we abbreviate  \n\n\\[\nN:=n-1,\\qquad p:=1-\\frac1n=\\frac{N}{n},\\qquad \nS_k:=\\sum_{i=1}^{k}\\xi_i\\quad\\bigl(\\xi_i\\stackrel{\\mathrm{iid}}{\\sim}{\\rm Bernoulli}(p)\\bigr).\n\\tag{1}\n\\]\n\n-----------------------------------------------------------------\n1.  Reversibility and stationary distribution  \n\nPut  \n\n\\[\np_j:=\\frac{n-j}{n},\\qquad q_j:=\\frac{j-1}{n},\\qquad r_j:=\\frac1n .\n\\]\n\nFor $2\\le j\\le n-1$ we have  \n\n\\[\nq_{j}\\,\\pi_n(j)=\n\\frac{j-1}{n}\\,2^{-(N)}\\binom{N}{\\,j-1\\,}\n=\\frac{n-j+1}{n}\\,2^{-(N)}\\binom{N}{\\,j-2\\,}\n=p_{j-1}\\,\\pi_n(j-1),\n\\]\n\nso the detailed–balance equations hold in the bulk.  \nThe boundary cases $j=1$ and $j=n$ are checked identically, hence\n$A^{(n)}$ is reversible and the displayed $\\pi_n$ is its unique\nstationary law.\n\n-----------------------------------------------------------------\n2.  Spectrum of $A^{(n)}$  \n\nLet $P^{(n)}$ be the transition kernel of $A^{(n)}$ and\n$P^{\\mathrm{Ehr}}_{N}$ that of the $(N+1)$–state Ehrenfest walk.  \nA direct computation gives the convex decomposition  \n\n\\[\nP^{(n)}=\\frac{N}{n}\\,P^{\\mathrm{Ehr}}_{N}+\\frac{1}{n}\\,{\\rm Id}.\n\\tag{2}\n\\]\n\nTherefore $P^{(n)}$ and $P^{\\mathrm{Ehr}}_{N}$ share the same\neigenfunctions (the Krawtchouk polynomials) and  \n\n\\[\n\\lambda_r=1-\\frac{2r}{n},\\qquad r=0,1,\\dots ,N ,\n\\tag{3}\n\\]\n\nare the eigenvalues of $P^{(n)}$.\nThe spectral gap is $\\operatorname{gap}_n=2/n$ and the relaxation time\n$t_{\\mathrm{rel}}=n/2$.\n\n-----------------------------------------------------------------\n3.  An exact representation via an “active clock’’  \n\nLet $B^{(N)}=(B^{(N)}_t)_{t\\ge 0}$ be the Ehrenfest chain on\n$\\{0,1,\\dots ,N\\}$ started from $0$, independent of the clock\n$S_{\\bullet}$ in (1).  Define  \n\n\\[\na_{n,k}:=1+B^{(N)}_{\\,S_k},\\qquad k\\ge 0.\n\\tag{4}\n\\]\n\nBecause each $\\xi_i$ either advances the Ehrenfest chain\n($\\xi_i=1$) or keeps the current state ($\\xi_i=0$),\n(4) coincides with the definition of $A^{(n)}$.  Consequently  \n\n\\[\n\\mathcal L_1(a_{n,k})=\\sum_{t=0}^{k}\\mathbb P(S_k=t)\\,\n      \\mathcal L\\!\\bigl(B^{(N)}_t\\bigr),\\qquad k\\ge 0.\n\\tag{5}\n\\]\n\n-----------------------------------------------------------------\n4.  Upper bound on the mixing time  \n\nFor the Ehrenfest walk denote  \n\n\\[\nd_t:=\\bigl\\lVert\\mathcal L\\bigl(B^{(N)}_t\\bigr)-\\pi_N\\bigr\\rVert_{\\mathrm{TV}},\n\\qquad t\\ge 0.\n\\]\n\nThe cut–off theorem of Levin–Peres–Wilmer (Theorem 18.5) asserts that,\nfor every $\\varepsilon\\in(0,\\tfrac12)$,\n\n\\[\n\\tau^{\\mathrm{Ehr}}_{N}(\\varepsilon/2)=\n\\Bigl(\\frac{N}{2}+O_{\\varepsilon}(N)\\Bigr)\\log N .\n\\tag{6}\n\\]\n\nFix $\\varepsilon\\in(0,\\tfrac12)$ and choose  \n\n\\[\nk_{+}:=\\Bigl\\lceil\\tfrac{n}{2}\\log n + C_{\\varepsilon}n\\Bigr\\rceil,\n\\qquad \nT_{+}:=\\Bigl\\lceil\\tfrac{N}{2}\\log N + \\tfrac{C_{\\varepsilon}}{2}\\,N\\Bigr\\rceil,\n\\tag{7}\n\\]\n\nwith $C_{\\varepsilon}>0$ large.  \nSince $\\mathbb E S_{k_{+}}=pk_{+}$ and \n${\\rm Var}(S_{k_{+}})\\le k_{+}$, a Chernoff bound gives  \n\n\\[\n\\mathbb P\\!\\bigl(S_{k_{+}}<T_{+}\\bigr)\n          \\,\\le\\, \\mathrm e^{-c\\,C_{\\varepsilon}n}\n\\tag{8}\n\\]\nfor some absolute $c>0$.\nUsing (5) and the fact that $d_t\\le \\varepsilon/2$ for $t\\ge T_{+}$ we obtain  \n\n\\[\n\\lVert\\mathcal L_1(a_{n,k_{+}})-\\pi_n\\rVert_{\\mathrm{TV}}\n\\le \\mathbb P(S_{k_{+}}<T_{+})+\\frac{\\varepsilon}{2}\n\\le \\varepsilon ,\n\\]\nprovided $C_{\\varepsilon}$ is large enough.  Hence  \n\n\\[\n\\tau_n(\\varepsilon)\\le \\frac{n}{2}\\log n + C_{\\varepsilon}n .\n\\tag{9}\n\\]\n\n-----------------------------------------------------------------\n5.  Lower bound on the mixing time  \n\nThe flaw in the previous version was an incorrect\nmonotone–likelihood–ratio claim.  \nThe argument is now replaced by a much simpler one which relies only on\ntwo facts:\n(i) total variation distance contracts under a Markov kernel, and  \n(ii) the Bernoulli clock $S_k$ is sharply concentrated.\n\n\\medskip\nFix $\\varepsilon\\in(0,\\tfrac12)$ and let  \n\n\\[\nk_{-}:=\\Bigl\\lfloor\\tfrac{n}{2}\\log n-C_{\\varepsilon}n\\Bigr\\rfloor ,\n\\qquad\nT_{-}:=\\Bigl\\lfloor\\tfrac{N}{2}\\log N-\\tfrac{C_{\\varepsilon}}{4}\\,N\\Bigr\\rfloor ,\n\\tag{10}\n\\]\nwith $C_{\\varepsilon}\\ge C_0(\\varepsilon)$ to be fixed.\n\n-----------------------------------------------------------------\n\\textbf{(i) A uniform lower bound for the Ehrenfest distance.}\n\nBy the cut–off theorem (6) and the choice of $T_{-}$ we have  \n\n\\[\nd_{T_{-}}\\ge 1-\\frac{\\varepsilon}{4}.\n\\tag{11}\n\\]\n\nBecause total variation distance contracts under convolution with the\nkernel, $d_t$ is \\emph{non–increasing} in $t$.  Consequently  \n\n\\[\nd_t\\ge 1-\\frac{\\varepsilon}{4}\\qquad(0\\le t\\le T_{-}).\n\\tag{12}\n\\]\n\n-----------------------------------------------------------------\n\\textbf{(ii) Concentration of the active clock.}\n\nAs $\\mathbb E S_{k_{-}}=pk_{-}$ and ${\\rm Var}(S_{k_{-}})\\le k_{-}$,\nChernoff’s bound gives  \n\n\\[\n\\delta_n:=\\mathbb P\\!\\bigl(S_{k_{-}}>T_{-}\\bigr)\n          \\,\\le\\, \\mathrm e^{-c\\,C_{\\varepsilon}n}\n\\tag{13}\n\\]\nfor some absolute $c>0$.\n\n-----------------------------------------------------------------\n\\textbf{(iii) Putting the pieces together.}\n\nSet  \n\n\\[\n\\mu_{k_-}:=\\mathcal L_1(a_{n,k_{-}}).\n\\]\n\nDecompose according to the clock:\n\n\\[\n\\mu_{k_-}\n= \\mathbb P\\bigl(S_{k_-}\\le T_{-}\\bigr)\\,\n       \\mu_{-}+\\mathbb P\\bigl(S_{k_-}>T_{-}\\bigr)\\,\\mu_{+},\n\\]\nwhere  \n\n\\[\n\\mu_{-}:=\\mathcal L\\bigl(B^{(N)}_{\\,S_{k_-}}\\mid S_{k_-}\\le T_{-}\\bigr),\n\\quad\n\\mu_{+}:=\\mathcal L\\bigl(B^{(N)}_{\\,S_{k_-}}\\mid S_{k_-}> T_{-}\\bigr).\n\\]\n\nFrom (12) we obtain  \n\n\\[\n\\lVert\\mu_{-}-\\pi_N\\rVert_{\\mathrm{TV}}\\ge 1-\\frac{\\varepsilon}{4}.\n\\]\n\nUsing this and (13),\n\n\\[\n\\begin{aligned}\n\\lVert\\mu_{k_-}-\\pi_n\\rVert_{\\mathrm{TV}}\n&\\ge \\bigl(1-\\delta_n\\bigr)\\Bigl(1-\\frac{\\varepsilon}{4}\\Bigr)-\\delta_n\\\\\n&\\ge 1-\\frac{3\\varepsilon}{4}-2\\delta_n .\n\\end{aligned}\n\\]\n\nFor $C_{\\varepsilon}$ large enough the exponential term satisfies\n$2\\delta_n\\le \\varepsilon/4$, giving  \n\n\\[\n\\lVert\\mu_{k_-}-\\pi_n\\rVert_{\\mathrm{TV}}\\ge 1-\\varepsilon .\n\\]\n\nTherefore  \n\n\\[\n\\tau_n(\\varepsilon)\\ge \\frac{n}{2}\\log n - C_{\\varepsilon}n .\n\\tag{14}\n\\]\n\n-----------------------------------------------------------------\n6.  Cut–off location and window  \n\nCombining the upper bound (9) and the lower bound (14) yields  \n\n\\[\n\\frac{n}{2}\\log n - C_{\\varepsilon}n\n\\;\\le\\;\n\\tau_n(\\varepsilon)\n\\;\\le\\;\n\\frac{n}{2}\\log n + C_{\\varepsilon}n ,\n\\qquad n\\ge n_0(\\varepsilon).\n\\]\n\nThe window has width $\\Theta(n)=o\\!\\bigl(\\tfrac{n}{2}\\log n\\bigr)$, so\n$\\bigl(A^{(n)}\\bigr)_{n\\ge 2}$ presents a sharp total–variation\ncut–off at time $\\tfrac{n}{2}\\log n$.  \\qed\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.885413",
        "was_fixed": false,
        "difficulty_analysis": "• Higher technical level:  the task is no longer a single–expectation\n  computation but a complete mixing–time analysis, requiring\n  spectral theory of Markov chains, total-variation control,\n  and precise eigenvalue estimates.\n\n• Additional structures:  stationary distribution, spectral gap,\n  orthogonal polynomial eigenbasis (the Krawtchouk polynomials (2)),\n  and total-variation norms all interact.\n\n• Deeper theory:  the solution uses reversibility,\n  L² → TV comparison, and the standard\n  “single–eigenfunction” lower–bound technique.\n\n• More steps:  bounding ‖⋅‖_{TV} from above and below,\n  computing π_n, establishing all eigenpairs,\n  deriving both upper and lower estimates,\n  and taking matching limsup / liminf.\n\nThis enhanced variant is thus substantially harder than merely asking for E[a_{n,n}], demanding full mastery of finite Markov-chain mixing theory and several non-elementary estimates."
      }
    },
    "original_kernel_variant": {
      "question": "For every integer $n\\ge 2$ consider the discrete–time birth–death Markov chain  \n\n\\[\nA^{(n)}=(a_{n,k})_{k\\ge 0},\\qquad a_{n,k}\\in\\{1,2,\\dots ,n\\},\n\\]\nstarted at the left–end point $a_{n,0}=1$ and whose transition\nprobabilities are  \n\\[\n\\begin{aligned}\n\\mathbb P\\!\\bigl(a_{n,k+1}=j+1\\mid a_{n,k}=j\\bigr)&=\\frac{\\,n-j\\,}{n},\\\\\n\\mathbb P\\!\\bigl(a_{n,k+1}=j  \\mid a_{n,k}=j\\bigr)&=\\frac1n,\\\\\n\\mathbb P\\!\\bigl(a_{n,k+1}=j-1\\mid a_{n,k}=j\\bigr)&=\\frac{\\,j-1\\,}{n}\\qquad(j=1,\\dots ,n),\n\\end{aligned}\n\\]\nwith the usual convention that any probability referring to an illegal\nstate equals $0$.\n\nDenote by $\\pi_n$ the stationary distribution of $A^{(n)}$ and, for\n$\\varepsilon\\in(0,\\tfrac12)$, define the (total–variation) $\\varepsilon$–mixing\ntime  \n\\[\n\\tau_n(\\varepsilon)=\\min\\Bigl\\{k\\ge 0:\\;\n      \\lVert\\mathcal L_1(a_{n,k})-\\pi_n\\rVert_{\\mathrm{TV}}\\le\\varepsilon\\Bigr\\}.\n\\]\n\n1.  Prove that $A^{(n)}$ is reversible and  \n    \\[\n        \\pi_n(j)=2^{-(n-1)}\\binom{\\,n-1\\,}{j-1}\\qquad(1\\le j\\le n).\n    \\]\n\n2.  Show that the family $(A^{(n)})_{n\\ge 2}$ exhibits a sharp\n    total–variation cut–off and determine its location and window, namely  \n    for every fixed $\\varepsilon\\in(0,\\tfrac12)$\n    \\[\n        \\tau_n(\\varepsilon)=\\Bigl(\\tfrac n2+o(n)\\Bigr)\\log n ,\n        \\qquad\\text{and the cut–off window has order }\\Theta(n).\n    \\]\n\n(Locating the precise mixing time of a non–uniform\n$n$–state birth–death chain and rigorously verifying the cut–off phenomenon\nrequires orthogonal–polynomial diagonalisation, concentration\ninequalities and a careful two–sided analysis.  An explicit comparison\nwith the classical Ehrenfest walk is particularly convenient.)\n\n────────────────────────────────────────────────────────",
      "solution": "Throughout write  \n\\[\nN:=n-1,\\qquad M:=\\Bigl\\lceil\\tfrac{n+1}{2}\\Bigr\\rceil=\\tfrac n2+O(1).\n\\]\n\n\\textbf{1.  Reversibility, stationary distribution and spectrum}\n\nPut  \n\\[\np_j:=\\frac{n-j}{n},\\qquad q_j:=\\frac{j-1}{n},\\qquad r_j:=\\frac1n ,\n\\]\nso $p_j+q_j+r_j=1$.  For $2\\le j\\le n$,\n\\[\nq_{j}\\,\\pi_n(j)=\\frac{j-1}{n}\\,2^{-(N)}\\binom{N}{j-1}\n           =\\frac{n-j+1}{n}\\,2^{-(N)}\\binom{N}{j-2}=p_{j-1}\\,\\pi_n(j-1),\n\\]\nso the detailed–balance equations hold and $A^{(n)}$ is reversible with\nthe claimed $\\pi_n$.\n\nLet $P^{(n)}$ be the transition kernel of $A^{(n)}$ and\n$P^{\\mathrm{Ehr}}_{N}$ the classical Ehrenfest kernel on $\\{0,1,\\dots ,N\\}$.\nA direct computation gives the \\emph{convex decomposition}\n\\[\nP^{(n)}=\\frac{N}{n}\\,P^{\\mathrm{Ehr}}_{N}\\;+\\;\\frac1n\\,\\mathrm{Id},\n\\tag{1}\n\\]\nhence $P^{(n)}$ shares its eigenfunctions with $P^{\\mathrm{Ehr}}_{N}$,\nnamely the Krawtchouk polynomials.  The corresponding eigenvalues are\n\\[\n\\lambda_r=1-\\frac{2r}{n},\\qquad r=0,1,\\dots ,N.\n\\tag{2}\n\\]\nThe spectral gap equals $\\operatorname{gap}_n=1-\\lambda_1=\\dfrac{2}{n}$,\nand the relaxation time is\n\\[\nt_{\\mathrm{rel}}=\\operatorname{gap}_n^{-1}=\\frac n2.\n\\tag{3}\n\\]\n\n\\textbf{2.  “Lazy clock + Ehrenfest” representation}\n\nWe now give a useful representation that \\emph{exhibits} a small amount\nof laziness.\n\n•  Let $\\bigl(\\xi_i\\bigr)_{i\\ge 1}$ be i.i.d. Bernoulli random variables\nwith parameter\n\\[\np:=\\frac{N}{n}=1-\\frac1n,\n\\]\nand set the \\emph{lazy clock}\n\\[\nS_k:=\\sum_{i=1}^{k}\\xi_i,\\qquad k\\ge 0.\n\\tag{4}\n\\]\n\n•  Independently, let $B^{(N)}=(B^{(N)}_t)_{t\\ge 0}$ be the Ehrenfest\nchain on $\\{0,1,\\dots ,N\\}$ started from $0$.\n\nIndependence of $S_{\\bullet}$ and $B^{(N)}_{\\bullet}$ is imposed by\nconstruction.  It is straightforward to check that the process\n\n\\[\na_{n,k}:=1+B^{(N)}_{\\,S_k},\\qquad k\\ge 0 ,\n\\tag{5}\n\\]\n\nhas exactly the transition probabilities of $A^{(n)}$.\nHence \\emph{all laws and expectations concerning $A^{(n)}$ can be\ncomputed via the independent pair $(S_k,B^{(N)})$}.\n\n\\textbf{3.  Upper bound on the mixing time}\n\nLet\n\\[\nd_t:=\\Bigl\\lVert\\mathcal L\\!\\bigl(B^{(N)}_{t}\\bigr)-\\pi_N\\Bigr\\rVert_{\\mathrm{TV}},\n\\qquad t\\ge 0,\n\\]\nand fix $\\varepsilon\\in(0,\\tfrac12)$.  Denote by\n\\[\nT_+:=\\tau^{\\mathrm{Ehr}}_{N}\\!\\Bigl(\\tfrac\\varepsilon2\\Bigr)\n      =\\Bigl(\\tfrac N2+O_{\\varepsilon}(N)\\Bigr)\\log N\n\\tag{6}\n\\]\nthe $\\varepsilon/2$–mixing time of the Ehrenfest chain\n(Levin–Peres–Wilmer, Thm.\\,18.5).\n\nFor any $k\\ge 0$,\n\\[\n\\Bigl\\lVert\\mathcal L_1\\!\\bigl(a_{n,k}\\bigr)-\\pi_n\\Bigr\\rVert_{\\mathrm{TV}}\n          \\;\\le\\;\\sum_{t=0}^{k}\\Pr\\!\\bigl(S_k=t\\bigr)\\,d_t\n          \\;\\le\\;\\frac\\varepsilon2+\\Pr\\!\\bigl(S_k<T_+\\bigr),\n\\tag{7}\n\\]\nwhere (7) uses $d_t\\le\\varepsilon/2$ for $t\\ge T_+$.\nSince $S_k\\sim\\operatorname{Bin}\\!\\bigl(k,p\\bigr)$, it has mean\n$\\mu_k=pk$ and variance $\\sigma_k^{2}\\le k$.  Chernoff’s bound gives\n\\[\n\\Pr\\!\\bigl(S_k<T_+\\bigr)\\le\n\\exp\\!\\Bigl(-\\frac{(\\mu_k-T_+)^{2}}{2k}\\Bigr).\n\\tag{8}\n\\]\nChoose\n\\[\nk=\\Bigl\\lceil\\tfrac n2\\log n + C_{\\varepsilon}n\\Bigr\\rceil\n\\tag{9}\n\\]\nwhere $C_{\\varepsilon}$ is a large enough positive constant.\nThen $(\\mu_k-T_+)\\asymp n$ while $k\\asymp n\\log n$, hence\nthe exponent in (8) equals $-c_{\\varepsilon}n/\\log n$ with\n$c_{\\varepsilon}>0$.  Consequently\n$\\Pr(S_k<T_+)\\le\\varepsilon/2$, the right–hand side of (7) is $\\le\\varepsilon$\nand\n\\[\n\\tau_n(\\varepsilon)\\;\\le\\;\n\\Bigl(\\tfrac n2+O_{\\varepsilon}(n)\\Bigr)\\log n.\n\\tag{10}\n\\]\n\n\\textbf{4.  Lower bound on the mixing time}\n\nThe main difficulty is to avoid the \\emph{incorrect equality} used\nearlier and to obtain a rigorous lower bound.  Write\n\\[\nk=\\Bigl\\lfloor\\tfrac n2\\log n-2Cn\\Bigr\\rfloor\\qquad(C>0\\text{ to be fixed}),\n\\tag{11}\n\\]\nand set\n\\[\nT_-:=\\Bigl\\lfloor\\tfrac N2\\log N-CN\\Bigr\\rfloor .\n\\tag{12}\n\\]\nSplit the law of $a_{n,k}$ according to $S_k$:\n\n\\[\n\\mathcal L_1\\!\\bigl(a_{n,k}\\bigr)\n     =P(S_k\\le T_-)\\,\\nu_- \\;+\\;P(S_k>T_-)\\,\\nu_+,\n\\tag{13}\n\\]\nwhere $\\nu_\\pm$ are the conditional distributions given\n$S_k\\le T_-$ and $S_k>T_-$.  We need three facts.\n\n(i)  From (11)–(12) and the same Chernoff bound as in\nSection 3 we get, for all $C\\ge C_0(\\varepsilon)$,\n\\[\nP(S_k\\le T_-)\\;\\ge\\;1-\\tfrac{\\varepsilon}{4}.\n\\tag{14}\n\\]\n\n(ii)  The cut–off for the Ehrenfest chain implies\n\\[\nd_t= \\bigl\\lVert\\mathcal L(B^{(N)}_t)-\\pi_N\\bigr\\rVert_{\\mathrm{TV}}\n      \\;\\ge\\;1-\\tfrac{\\varepsilon}{4}\n      \\qquad\\text{for every }t\\le T_-.\n\\tag{15}\n\\]\nTherefore $\\nu_-$ itself satisfies\n\\[\n\\lVert\\nu_--\\pi_n\\rVert_{\\mathrm{TV}}\n        \\ge 1-\\frac{\\varepsilon}{4}.\n\\tag{16}\n\\]\n\n(iii)  For \\emph{any} two probability measures the total variation\ndistance never exceeds $2$, so $\\lVert\\nu_+-\\pi_n\\rVert_{\\mathrm{TV}}\\le 2$.\n\nNow apply the triangle inequality to (13):\n\\[\n\\begin{aligned}\n\\lVert\\mathcal L_1(a_{n,k})-\\pi_n\\rVert_{\\mathrm{TV}}\n&=\\Bigl\\lVert\n    P(S_k\\le T_-)(\\nu_--\\pi_n)\n  + P(S_k>T_-)(\\nu_+-\\pi_n)\\Bigr\\rVert_{\\mathrm{TV}}\\\\\n&\\ge P(S_k\\le T_-)\\lVert\\nu_- -\\pi_n\\rVert_{\\mathrm{TV}}\n     -P(S_k>T_-)\\lVert\\nu_+-\\pi_n\\rVert_{\\mathrm{TV}}\\\\[4pt]\n&\\ge \\Bigl(1-\\tfrac{\\varepsilon}{4}\\Bigr)\\Bigl(1-\\tfrac{\\varepsilon}{4}\\Bigr)\n      -\\tfrac{\\varepsilon}{4}\\cdot 2 \\\\\n&\\ge 1-\\varepsilon \\;>\\;\\varepsilon .\n\\end{aligned}\n\\]\nConsequently\n\\[\n\\tau_n(\\varepsilon)\\;\\ge\\;\n\\Bigl(\\tfrac n2-2C n\\Bigr)\\log n.\n\\tag{17}\n\\]\n\n\\textbf{5.  Location and window of the cut–off}\n\nCombine (10) with (17).  For every fixed\n$\\varepsilon\\in(0,\\tfrac12)$ there exist constants\n$c_1(\\varepsilon),c_2(\\varepsilon)>0$ such that\n\\[\n\\Bigl(\\tfrac n2-c_1(\\varepsilon)n\\Bigr)\\log n\n      \\;\\le\\;\\tau_n(\\varepsilon)\\;\\le\\;\n\\Bigl(\\tfrac n2+c_2(\\varepsilon)n\\Bigr)\\log n .\n\\]\nHence\n\\[\n\\tau_n(\\varepsilon)=\\Bigl(\\tfrac n2+o(n)\\Bigr)\\log n ,\n\\qquad n\\to\\infty.\n\\]\nBecause the two bounds differ by $\\Theta(n)$ steps, the cut–off window\nhas order $n$.  Therefore the family $(A^{(n)})_{n\\ge 2}$ exhibits a\nsharp total–variation cut–off located at $\\tfrac n2\\log n$ with window\n$\\Theta(n)$.  ∎\n\n────────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.668957",
        "was_fixed": false,
        "difficulty_analysis": "• Higher technical level:  the task is no longer a single–expectation\n  computation but a complete mixing–time analysis, requiring\n  spectral theory of Markov chains, total-variation control,\n  and precise eigenvalue estimates.\n\n• Additional structures:  stationary distribution, spectral gap,\n  orthogonal polynomial eigenbasis (the Krawtchouk polynomials (2)),\n  and total-variation norms all interact.\n\n• Deeper theory:  the solution uses reversibility,\n  L² → TV comparison, and the standard\n  “single–eigenfunction” lower–bound technique.\n\n• More steps:  bounding ‖⋅‖_{TV} from above and below,\n  computing π_n, establishing all eigenpairs,\n  deriving both upper and lower estimates,\n  and taking matching limsup / liminf.\n\nThis enhanced variant is thus substantially harder than merely asking for E[a_{n,n}], demanding full mastery of finite Markov-chain mixing theory and several non-elementary estimates."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}