Results
We model each stage of the Bitcoin mining game as an independent and parallel Poisson race.
Stage game
.\label{pixi}
The probability of miner \(i\) winning is increasing in \(x_{i}\), given all other miners choose \(x_{-i}\).
Proof.
The expected time for miner \(i\) to successfully complete the puzzle is \(\mathbb{E}(t_{i})=\frac{K}{x_{i}}\). The first derivative of the probability of winning with respect to \(x_{i}\) is
\begin{equation}
\begin{split}\displaystyle\frac{\partial\pi(W_{i};K,x_{i},x_{-i})}{\partial x_{i}}&\displaystyle=\frac{\partial\pi(W_{i};K,x_{i},x_{-i})}{\partial\mathbb{E}(t_{i})}\frac{\partial\mathbb{E}(t_{i})}{\partial x_{i}}\\
&\displaystyle=\frac{\partial\pi(W_{i};K,x_{i},x_{-i})}{\partial\mathbb{E}(t_{i})}\left(\frac{-K}{x_{i}^{2}}\right)\\
&\displaystyle>0\,,\end{split}\nonumber \\
\end{equation}
since we know the first term is negative from (\ref{dpidt}).
∎
\begin{equation}
\par
\pi(W_{i};K,x_{i},x_{-i})<1\,\quad\text{for all}\ x_{i}\,.\par
\nonumber \\
\end{equation}
However, there is no choice of technology for which winning is guaranteed, since the proof-of-work is a random process
Although greater computing technology increases the rate at which computations can be completed and the probability that a miner will win, it does not guarantee that the miner with the most technology will be first to reach \(K\) computations.
.\label{2pixi}
The probability of miner i winning increases in \(x_{i}\) at a decreasing rate.
Proof.
The second derivative of \(\pi(W_{i};K,x_{i},x_{-i})\) with respect to \(x_{i}\) is
\begin{equation}
\begin{split}\displaystyle\frac{\partial^{2}\pi}{\partial x_{i}^{2}}&=\frac{\partial^{2}\pi}{\partial\mathbb{E}(t_{i})^{2}}\left(\frac{\partial\mathbb{E}(t_{i})}{\partial x_{i}}\right)^{2}+\frac{\partial\pi}{\partial\mathbb{E}(t_{i})}\frac{\partial^{2}\mathbb{E}(t_{i})}{\partial x_{i}^{2}}\\
&\displaystyle=\frac{\partial^{2}\pi}{\partial\mathbb{E}(t_{i})^{2}}\left(\frac{-K}{x_{i}^{2}}\right)^{2}+\frac{\partial\pi}{\partial\mathbb{E}(t_{i})}\left(\frac{2K}{x_{i}^{3}}\right).\end{split}\nonumber \\
\end{equation}
The second term is always negative by (\ref{dpidt}). By (\ref{pisoc}), we know that \(\frac{\partial^{2}\pi}{\partial\mathbb{E}(t_{i})^{2}}\) is negative on \(\hat{x}_{i}<\frac{K}{\mathbb{E}(\hat{t}_{i})}\). Therefore, the sign of the second derivative is negative on \(x_{i}<\hat{x}_{i}\), noting that \(\hat{x}_{i}\geq 0\).
For \(x_{i}>\hat{x}_{i}\), the first term is always positive. Since \(\pi(W_{i};K,x_{i},x_{-i})<1\) for all \(x_{i}\), it must be the case that
\begin{equation}
\frac{\partial^{2}\pi}{\partial\mathbb{E}(t_{i})^{2}}\left(\frac{-K}{x_{i}^{2}}\right)^{2}<\frac{\partial\pi}{\partial\mathbb{E}(t_{i})}\left(\frac{2K}{x_{i}^{3}}\right)\,,\nonumber \\
\end{equation}
for \(x_{i}>\hat{x}_{i}\). If not, then this implies that \(\pi(W_{i};K,x_{i},x_{-i})\) is increasing and convex in the region. This contradicts the fact that probability can never exceed \(1\). Therefore, it must be the case that \(\pi(W_{i};K,x_{i},x_{-i})\) is strictly concave on \(x_{i}\)
\begin{equation}
\frac{\partial^{2}\pi(W_{i};K,x_{i},x_{-i})}{\partial x_{i}^{2}}<0\,.\nonumber \\
\end{equation}
∎
.\label{eqmfixedk}
There exists a unique interior Nash equilibrium solution \(x^{*}>0\) to (\ref{u}) if \(U_{i}(x)\geq 0\) for some \(x>0\).
Proof.
The necessary conditions for \(x^{*}\) to be a unique interior Nash equilibrium solution to (\ref{u}) are
\begin{equation}
\label{foc}\frac{\partial U_{i}(x^{*})}{\partial x_{i}}=P\frac{\partial\pi(W_{i};K,x^{*})}{\partial x_{i}}-\frac{\partial c(x^{*})}{\partial x_{i}}=0\\
\end{equation}
\begin{equation}
\label{soc}\frac{\partial^{2}U_{i}(x^{*})}{\partial{x_{i}}^{2}}=P\frac{\partial^{2}\pi(W_{i};K,x^{*})}{\partial{x_{i}}^{2}}-\frac{\partial^{2}c(x^{*})}{\partial{x_{i}}^{2}}<0\,.\\
\end{equation}
The first term of (\ref{foc}) is positive by Lemma \ref{pixi} and is concave in \(x_{i}\) by Lemma \ref{2pixi}. The second term is positive and convex in \(x_{i}\) by assumption. Therefore, there must exist some point \(x^{*}\) for which the first order condition in (\ref{foc}) holds.
To show that \(x^{*}\) is a unique solution, we note that the first term of (\ref{soc}) is negative by Lemma \ref{2pixi}, and the second term is positive by assumption for all \(x_{i}\). Therefore, the second order condition in (\ref{soc}) is satisfied. Since \(U_{i}(x_{i})\) is strictly concave on \(x_{i}\), the interior Nash equilibrium solution \(x^{*}\) is a unique maximum.
∎
We have shown that there exists a unique interior Nash equilibrium solution \(x^{*}=(x^{*}_{1},x^{*}_{2},\dots,x^{*}_{N})\) for every difficulty level \(K\), where the number of miners in the network is fixed. Each Nash equilibrium \((K,x^{*})\) gives a unique expected time for completion of the puzzle \(\delta_{K}\). Therefore, at each stage of the Bitcoin mining game, there exists a unique Nash equilibrium.
Extensive game
To achieve the target time of \(\delta^{*}\) on average, the network periodically adjusts the difficulty level \(K\) of the computational puzzle in each stage of the extensive game. If the average time at which the computational puzzles have been solved in the previous period is below the target \(\delta^{*}\), then the difficulty level \(K\) increases, and vice versa. We assume the miners cannot re-optimize immediately in the short run.
.\label{xk}
The Nash equilibrium technology \(x^{*}\) is increasing in the difficulty level \(K\).
Proof.
Suppose we are initially at some Nash equilibrium \((K,x^{*})\) which has an expected solution time of \(\delta_{K}\). Each miner \(i\) has an equal probability of winning since they all choose the same technology \(x^{*}\) at the Nash equilibrium
\begin{equation}
\label{probw}\pi(W_{i};K,x^{*})=\left[1-\int_{0}^{t_{i}}\gamma_{K,x^{*}(t_{i})}~{}dt_{i}\right]^{N-1}=\frac{1}{N}\,.\\
\end{equation}
Now suppose we increase the difficulty level from \(K\) to \(K+\varepsilon\), where \(\varepsilon>0\). In the short run, miners cannot re-optimize immediately, so their technology remains as \(x^{*}\). The miners still have an equal probability of winning at the new difficulty level \(K+\varepsilon\). Therefore, the integrand of (\ref{probw}) must be the same for \(\gamma_{K,x^{*}}(t)\) and \(\gamma_{K+\varepsilon,x^{*}}(t)\).
Solving for the Nash equilibrium technology \(x^{*}\),
\begin{equation}
\label{x*}\begin{split}\gamma_{K,x^{*}}(t)&\displaystyle=\gamma_{K+\varepsilon,x^{*}}(t)\\
\displaystyle\frac{t^{K-1}}{\Gamma(K)}(x^{*})^{K}e^{-{x^{*}}t}&\displaystyle=\frac{t^{K+\varepsilon-1}}{\Gamma(K+\varepsilon)}(x^{*})^{K+\varepsilon}e^{-{x^{*}}t}\\
\displaystyle\frac{1}{\Gamma(K)}&\displaystyle=\frac{t^{\varepsilon}}{\Gamma(K+\varepsilon)}(x^{*})^{\varepsilon}\\
\displaystyle\therefore x^{*}&\displaystyle=\left[\frac{1}{t^{\varepsilon}}\frac{\Gamma(K+\varepsilon)}{\Gamma(K)}\right]^{\frac{1}{\varepsilon}}\,.\end{split}\\
\end{equation}
Since \(\Gamma(\cdot)\) is convex on \(\mathbb{R}_{++}\), \(\frac{\Gamma(K+\varepsilon)}{\Gamma(K)}\) is increasing in \(K\). Thus, \(x^{*}\) is increasing in \(K\).
∎
.\label{e(t)k}
The expected time required to solve a puzzle \(\delta_{K}\) is strictly monotonically increasing in the difficulty level \(K\), holding all else equal.
Proof.
The expected time for a given miner \(i\) to solve the puzzle at a given difficulty \(K\) is \(\mathbb{E}_{K}(t_{i})=\frac{K}{x_{i}}\). The expected time for the puzzle to be solved by the whole network for the same \(K\) is
\begin{equation}
\mathbb{E}_{K}(t)=\mathbb{E}_{K}(\min\{t_{1},t_{2},\dots,t_{N}\})=\int_{0}^{\infty}\left[1-\int_{0}^{t}\gamma_{K,x^{*}}(t)~{}dt\right]^{N}~{}dt\,.\nonumber \\
\end{equation}
Since the expected time for an individual miner is increasing in \(K\), and the miners are symmetric, the expected time for the network must be increasing in \(K\).
∎
.\label{eqmdelta}
There exists a unique symmetric equilibrium \((K^{*},x^{*})\) given a fixed number of miners \(N\) and fixed target solution time \(\delta^{*}\).
Proof.
Fix \(\delta^{*}\). By Proposition 1.3, \(\mathbb{E}_{K}(t)\) is strictly monotonically increasing in \(K\). Hence, the \(K\) which is associated with \(\delta^{*}\) must be unique. Fixing the \(K=K^{*}\) where \(K^{*}\) gives \(\delta^{*}\), there exists some unique Nash equilibrium level of technology \(x^{*}\) by Proposition 1.1.
Hence, \((K^{*},x^{*})\) is a unique symmetric equilibrium with an average target solution time of \(\delta^{*}\).
∎
With a fixed prize and a fixed number of miners, the gross expected payoff is constant at every Nash equilibrium due to the symmetry of the game. Since the Nash equilibrium technology \(x^{*}\) is increasing in the difficulty level \(K\), the cost of mining increases in \(K\), driving profits towards zero at the equilibrium.
Allowing \(K\) to vary over time in order to maintain a target solution time \(\delta^{*}\) on average, we showed that for each \(N\) and \(P\), there exists a difficulty level \(K^{*}\) and a Nash equilibrium \(x^{*}\) given \(K^{*}\) that form a unique equilibrium of the extensive game. Therefore, with a fixed number of miners \(N\), there is exists a symmetric subgame perfect equilibrium in which each miner plays their Nash equilibrium strategy at every stage of the game.