_{T},\ldots,dy_2|y_1) =\\ \nonumber \mathbb{E}\left[U\left(\sum_{t=2}^T {\psi}_t (X_t-X_{t-1})\right)|X_1=y_1,X_0=z\right]. 
\end{eqnarray}
Now plugging in $z:=y_0$ we obtain \eqref{pp} from \eqref{ppp}, \eqref{pppp} and \eqref{ppppp}. 
\end{proof}





After these preparations we are able to give an explicit solution for the optimal strategies of the wealth process in case of the price is an autoregressive process.


In this Section we prove \autoref{maintheorem}, first we focus on the case where the investor 
uses past information. We consider the case $T=1$, so the wealth process according to 
\eqref{logwealth} takes the form
\begin{equation}
L_1^{\phi}=\phi_1(X_1-X_0)=\phi_1(\beta X_0+\sigma\varepsilon_1).
\end{equation}

We have
\begin{equation}
	\mathbb{E}\left[e^{-\phi_1(\beta X_0+\sigma \varepsilon _1)} | X_0=z\right]
	= e^{-\phi_1\beta z} \mathbb{E}\left[e^{-\phi_1 \sigma \varepsilon_1}\right]=
	e^{-\phi_1\beta z}e^{\frac{\phi_1^2\sigma^2}{2}},
\end{equation}
hence we get
\[
\mathrm{arg}\min _{\phi_1}e^{-\phi_1\beta z+\frac{(\phi_1\sigma)^2}{2}}
 = \mathrm{arg}\min_{\phi_1}e^{\frac{1}{2}\left(\phi_1\sigma-\frac{\beta z}{\sigma}\right)^2-\frac{\beta^2 z^2}{2\sigma^2}}
 =\frac{\beta z}{\sigma^2}=\frac{\beta z}{\sigma^2}\theta_1^1=\hat{\phi}_1^1(z), \]
 because $\theta_1^1=1$. So we proved the first part of \autoref{maintheorem} for $T=1$. Now let's assume that  
 \eqref{strategies} is true for $T-1$, i.e.
 \begin{equation}\label{opp}
\phi^*_t:=\hat{\phi}_t^{T-1}(X_{t-1})\mbox{ with }\hat{\phi}_t^{T-1}(z)=\frac{\beta z}{\sigma^2}\theta_t^{T-1} \quad 1\leq t\leq T-1
 \end{equation}
satisfies \eqref{90} for all $\phi\in\Phi$. 
 We will prove that \eqref{strategies} also holds for $T$. By Lemma \ref{argumentofphistatement},
 for all $\psi\in\Phi$,
 \begin{eqnarray}\nonumber
\mathbb{E}[e^{-L^{\psi}_T}|\mathcal{F}_0] = \mathbb{E}[e^{-\psi_1(X_1-X_0)}
\mathbb{E}[e^{-\sum_{j=2}^T \psi_j(X_j-X_{j-1})}|\mathcal{F}_1]\vert\mathcal{F}_0] &\geq &\\ \nonumber
\mathbb{E}[e^{-\psi_1(X_1-X_0)}
\mathbb{E}[e^{-\sum_{j=2}^T \hat{\phi}_{j-1}^{T-1}(X_{j-1})(X_j-X_{j-1})}|\mathcal{F}_1]\vert\mathcal{F}_0]&=&\\
\mathbb{E}[e^{-\psi_1(X_1-X_0)}
\mathbb{E}[e^{-\sum_{j=2}^T \hat{\phi}_{j}^{T}(X_{j-1})(X_j-X_{j-1})}|\mathcal{F}_1]\vert\mathcal{F}_0],\label{parr} & &
 \end{eqnarray}

 
since $\phi_{j-1}^{T-1}=\phi_j^T$. Now define the trading strategy 
$\omega=(\phi, \hat{\phi}_2^T(X_1), \dots, \hat{\phi_T^T}(X_{T-1}))$ and the function 
$Q_T\,:\,\mathbb{R}^{T+2} \rightarrow\mathbb{R}$ such that
\begin{equation}
Q_T(\phi,X_0,\mathbf{\varepsilon}):=L_T^{\omega},\quad\textrm{where }\mathbf{\varepsilon}=(\varepsilon_1, \dots \varepsilon_T)^{\mathbf{T}}.
\end{equation}
Hence, according to \eqref{parr}, it remains to find $\phi$ which minimizes
\begin{equation}\label{phi1def}
\mathbb{E}\left[e^{-Q_T(\phi,X_0,\mathbf{\varepsilon})}|X_0=z\right].
\end{equation}
If we prove that $\phi=\hat{\phi}_1^{T}(z)$ does the job then we will be able to conclude that the optimal
strategy for time horizon $T$ is indeed as given in \eqref{opp} for $T-1$.


To compute the minimiser $\phi$ we will write $Q_T(\phi, X_0, \varepsilon)$ 
in a sum of a quadratic, a linear and a constant function of $\mathbf{\varepsilon}$.

\begin{eqnarray*}
 	Q_T(\phi,X_0,\mathbf{\varepsilon})&=&\phi(X_1-X_0)+\sum_{j=2}^T\hat{\phi}_j^T(X_{j-1})(X_j-X_{j-1}) \\
 	   &=&\phi(X_1-X_0)+\sum_{j=2}^T\frac{\beta\theta_j^T}{\sigma^2}\underbrace{X_{j-1}(X_j-X_{j-1})}_{A_j}
\end{eqnarray*}
 	   
 	   
For $A_j$ we have
\begin{eqnarray*}
 	A_j&=&\left(\alpha^{j-1}X_0+\sigma\sum_{i=1}^{j-1}\alpha^{j-i-1}\varepsilon_i\right)\left(\alpha^{j}X_0+
 		\sigma\sum_{i=1}^{j}\alpha^{j-i}\varepsilon_i - \alpha^{j-1}X_0-\sigma\sum_{i=1}^{j-1}\alpha^{j-i-1}\varepsilon_i\right) \\
 	&=&\left(\alpha^{j-1}X_0+\sigma\sum_{i=1}^{j-1}\alpha^{j-i-1}\varepsilon_i\right)\left(\alpha^{j-1}\beta X_0+\sigma\varepsilon_j +
 		\sigma\sum_{i=1}^{j-1}\alpha^{j-i-1}\beta\varepsilon_i\right) \\
 	&=&\alpha^{2j-2}\beta X_0^2+\sigma\beta X_0\sum_{i=1}^{j-1}\alpha^{2j-i-2}\varepsilon_i+\sigma X_0\alpha^{j-1}\varepsilon_j+ 
 		\sigma^2\sum_{i=1}^{j-1}\alpha^{j-i-1}\varepsilon_i\varepsilon_j+\sigma\beta X_0\sum_{i=1}^{j-1}\alpha^{2j-i-2}\varepsilon_i \\
 	&\,& +\sigma^2\beta\left(\sum_{i=1}^{j-1}\alpha^{j-i-1}\varepsilon_i\right)^2 \\
 	&=&\underbrace{\alpha^{2j-2}\beta X_0^2}_{B_1(j)}+ \underbrace{2\sigma\beta X_0\sum_{i=1}^{j-1}\alpha^{2j-i-2}\varepsilon_i}_{B_2(j)} 
 		+\underbrace{\sigma X_0\alpha^{j-1}\varepsilon_j}_{B_3(j)}
 		+ \underbrace{\sigma^2\sum_{i=1}^{j-1}\alpha^{j-i-1}\varepsilon_i\varepsilon_j}_{B_4(j)} 
 		+\underbrace{\sigma^2\beta\sum_{i=1}^{j-1}\sum_{k=1}^{j-1}\varepsilon_i\varepsilon_k\alpha^{2j-i-k-2}}_{B_5(j)}. \\
\end{eqnarray*}

Let's substitute this into  $Q_T