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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?

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    $\begingroup$ Your third equality is false, but it's completely unexplained so I don't even know where to start with why you think it's true. (e.g. $t^2$ is not random, so $E[t^2 \mid \mathcal{F}_s] = t^2$) $\endgroup$ Commented Oct 19, 2022 at 7:13
  • $\begingroup$ @BrianMoehring Ah, my bad. Yes, $t^2$ is a constant. So, the given process is not martingale? $\endgroup$
    – user1089451
    Commented Oct 19, 2022 at 7:20
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    $\begingroup$ The identity $E[e^{4W_t}\mid\mathcal{F}_s]=e^{4W_s}$ is incorrect. Try writing $e^{4W_t}=e^{4(W_t-W_s)}e^{4W_s}$. Then use that $W_t-W_s$ is independent of $\mathcal{F}_s$ and that $W_s$ is $\mathcal{F}_s$-measurable. $\endgroup$
    – jakobdt
    Commented Oct 19, 2022 at 7:25
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    $\begingroup$ Your third equality is still false, but because $e^{4W_t}$ isn't a martingale. Do you know how to find the mean of a lognormal random variable? If not, then instead of finding the conditional expectation, show that $t \mapsto E[Y_t]$ isn't a constant function. (the only facts you'll need to use are that $e^{4W_t} \geq 0$ and $t^2 \to \infty$) $\endgroup$ Commented Oct 19, 2022 at 7:25
  • $\begingroup$ What about now? $\endgroup$
    – user1089451
    Commented Oct 19, 2022 at 7:32

1 Answer 1

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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 $EY_t=10+t^{2}+e^{t/2}$.

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  • $\begingroup$ Since $EY_t=10+t^2+e^{t/2}$, then $EY_t$ is dependent of $t$? $\endgroup$
    – user1089451
    Commented Oct 19, 2022 at 10:44
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    $\begingroup$ Correct. $EY_t$ depends on $t$ and therefore, $Y_t$ cannot be a martingale. $\endgroup$
    – Kurt G.
    Commented Oct 19, 2022 at 11:18

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