Let $\{W_t\}_{t \ge 0}$ be a standard Brownian motion on a probability space $(\Omega,\cal{F},\Bbb P)$ and let $\{\cal F_t\}_{t \ge 0}$ be the filtration of the Brownian motion.
Determine whether the stochastic process $Y_t=10+t^2+e^{W_t}$ is a martingale with respect to the filtration $\{\cal F_t\}_{t \ge 0}$.
Attempt:
EDIT
Let $s,t \in \Bbb R$ be arbitrary with $0 \le s < t$. Notice that
\begin{align*}
E[Y_t \mid \cal F_s] &= E[10+t^2+e^{4W_t} \mid \cal F_s] \\
&= E[10 \mid \mathcal{F}_s] + E[t^2 \mid \mathcal F_s] + E[e^{4W_t} \mid \mathcal F_s] \\
&= 10 + t^2 + E[e^{4(W_t-W_s+W_s)} \mid \mathcal{F}_s] \\
&= 10 + t^2 + E[e^{4(W_t-W_s)} \cdot e^{4W_s} \mid \mathcal{F}_s] \\
&= 10 + t^2 + e^{4W_s} \cdot E[e^{4(W_t-W_s)}] \\
&= 10 + t^2 + e^{4W_s} \cdot e^{\frac12 \cdot 4^2 \cdot (t-s)} \\
&= 10 + t^2 + e^{4W_s} \cdot e^{8(t-s)} \\
&\ne 10 + s^2 + e^{4W_s} \\
&= Y_s.
\end{align*}
Hence, $Y_t=10+t^2+e^{4W_t}$ is not martingale.
Am I true?
Best Answer
As already pointed out there are several errors in your answer.
Here is a simple way of showing that it is not a martingale: If it is a martingale then $EY_t$ would be independent of $t$. But $EY_t=10+t^{2}+e^{t/2}$.