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\begin{document}
\textbf{Question 1: Incorrect approach}
\bigskip

\ \\Aproximating $b(t,X(t))$ by the trapezodial rule:


\begin{equation}
dX(t)=a(t,X(t))dt + b(t,X(t))  dW_t
\end{equation}


\begin{equation}
X_{n+1}(t)=X_{n} + a(t_{n},X(t_n))\Delta_t + b(\frac {1}{2} (t_{n+1}+t_n)), \frac {1}{2} (X_{t_{n+1}},X_{t_n})) \Delta W_n
\end{equation}

What is wrong?

\ \\Let's consider the process: $dX(t)= b dW(t)$ for constant $b$.

Using above approximation we obtain 

$
X_{n+1}(t)=X_{n}  + b ( \frac {1}{2} (X_{t_{n+1}},X_{t_n})) \Delta W_n
$
\ \\

$
X_{n+1}(t) (1-\frac {1}{2} b \Delta W_n) =X_{n} (1 + \frac {1}{2} b \Delta W_n)
$


\ \\

$
X_{n+1}(t)  = frac {X_{n} (1 + \frac {1}{2} b \Delta W_n)} {(1-\frac {1}{2} b \Delta W_n)}
$

Computing $E(X_1^2)$ we have

\begin{equation}
E(X_1^2) = (X_0)^2 \int_{\infty}^{\infty} \frac { ((1 + \frac {1}{2} b z))^2} {((1-\frac {1}{2} b )^2
z)} e^{-z^2/ (2 \Delta t) } dz
\end{equation}

where we have used the fact that $E(f(x))=\int(f(x)dp(x))$ where $p(x)$ is the probability density of $x$.

Above integral is of the form 

$\int_{\infty}^{\infty} \frac {(1+x)^2}{(1-x)^2} e^(-x^2) dx$ which does not converge.






\end{document}
