\documentclass{article}
\usepackage{tikz}
\usepackage{subcaption}
\usepackage{amsmath}

\begin{document}

\begin{figure}[h!]
  \centering
  \begin{subfigure}[b]{0.4\textwidth}
    \centering
    \begin{tikzpicture}
      \draw (0,0) grid (2,2);
      \node at (0.5,1.5) {1};
      \node at (1.5,1.5) {2};
      \node at (0.5,0.5) {3};
      \node at (1.5,0.5) {4};
    \end{tikzpicture}
    \caption{Figure 1}
  \end{subfigure}
  \begin{subfigure}[b]{0.4\textwidth}
    \centering
    \begin{tikzpicture}
      \draw (0,0) grid (2,2);
      \node at (0.5,1.5) {1};
      \node at (1.5,1.5) {3};
      \node at (0.5,0.5) {4};
      \node at (1.5,0.5) {2};
    \end{tikzpicture}
    \caption{Figure 2}
  \end{subfigure}
  % \caption{Two figures side by side}
\end{figure}

\begin{align*}
    E[Z_t] & = p\cdot (3\delta_1+\delta_2 + \delta_3) + p\cdot (3\delta_2+\delta_1 + \delta_4) + p\cdot (3\delta_3+\delta_1 + \delta_4) + p\cdot (3\delta_4+\delta_2 + \delta_3) \\
    & = p(5\delta_1 + 5\delta_2 + 5\delta_3 + 5\delta_4)
\end{align*}

\begin{align*}
    E[Z_t] & = p\cdot (3\delta_1+\delta_3 + \delta_4) + p\cdot (3\delta_3+\delta_1 + \delta_2) + p\cdot (3\delta_4+\delta_2 + \delta_1) + p\cdot (3\delta_2+\delta_4 + \delta_3) \\
    & = p(5\delta_1 + 5\delta_2 + 5\delta_3 + 5\delta_4)
\end{align*}

\begin{align*}
    E[Z_t]=\sum_{i \in \mathcal{B}_t} C_i \delta_i \leq \sum_{i \in \mathcal{B}_t} \left(\left(1+\frac{1}{t}\right)^d-1\right)\delta_i
\end{align*}

$\sum_{i \in \mathcal{B}_t} \delta_i \leq d\log{T}+1$

\begin{align*}
    \sum_{t=1}^T E[Z_t] \leq O(d^2 \log{T} + d)  
\end{align*}
\begin{align*}
    \sum_{t=1}^T \left(1+\frac{1}{t}\right)^d-1 \leq  \sum_{t=1}^T 2^\frac{d}{t}-1 \leq T\left(2^{\frac{d}{T}}-1\right) \leq T2^{\frac{d}{T}}(1-2^{-\frac{d}{T}}) \leq \log{2} \cdot d\cdot 2^{\frac{d}{T}}=O(d)
\end{align*}

$2^{-x}>1-x\log{2}$


\begin{align}
    
\end{align}
\end{document}
