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Show that $X_t = (1+t)^{-1/2} \exp \biggl( \frac{B_t^2}{2(1+t)} \biggr)$, where $B$ is a Brownian motion, is a martingal.

I understand that we need to show that $\mathbb{E}(X_t \vert \mathcal{F}_s) = X_s$ for all $t \ge s \ge 0$, where $\mathcal{F}$ is the canonical filtration of $X$. However, I do not see how to do this computation as we have no density function here. Could you please explain this to me?

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  • $\begingroup$ You have the density function of $B_t$, right? Although it isn't really needed if you have Ito's formula. $\endgroup$
    – user6247850
    Nov 29, 2022 at 21:46

2 Answers 2

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You can explicitly compute the filtered expectation value, given the distribution of the increments of the Brownian motion $W_{st}=B_t-B_s$ and their independence from the increment $W_{0s}=B_s$:

$$\mathbb{E}[X_t|B_{T}, 0\leq T\leq s]=\mathbb{E}\left[(1+t)^{-1/2}\exp\left(\frac{(B_s+W_{st})^2}{2(1+t)}\right)\Bigg|B_s\right]\\=\frac{1}{\sqrt{2\pi(1+t)(t-s)}}\int_{-\infty}^{\infty}dx \exp[(B_s+x)^2/2(1+t)]\exp[-x^2/2(t-s)]$$

This integral is Gaussian. Can you take it from here?

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  • $\begingroup$ Sorry, but I can not follow you. If I undersand you correctly, you substitute $B_t = W_{st} + B_s$ to drop the condition in the expected value. But I do not understand what happens after that (the second line); where does the factor before the integral come from. And I think the $dx$ is misplaced. $\endgroup$
    – 3nondatur
    Nov 30, 2022 at 21:28
  • $\begingroup$ $W_{st}$ is normally distributed $\sim N(0,t-s)$. The factor upfront comes from the normalization of the distribution function. $\endgroup$
    – DinosaurEgg
    Nov 30, 2022 at 22:13
  • $\begingroup$ So you mean to make $W_{st} \sim N(0,t-s)$ into $W_{st} \sim N(0,1)$? Also, where does $exp[-x^2/2(t-s)]$ in the integrand come from and why do you replace $W_{st}$ by $x^2$? $\endgroup$
    – 3nondatur
    Nov 30, 2022 at 22:50
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    $\begingroup$ I trust you have learned in a basic probability class that the probability density for a normal variable is given by $f_{W_{ts}}(x)=(2\pi(t-s))^{-1/2}e^{-x^2/2(t-s)}$ and $\mathbb{E}[g(W_{ts})]=\int_{-\infty}^\infty g(x) f_{W_{ts}}(x)dx$. This is very basic.... $\endgroup$
    – DinosaurEgg
    Nov 30, 2022 at 22:55
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Let $f(t,x)=(1+t)^{-1/2}\exp(\frac{x^2}{2(1+t)})$. Using Itô's formula, $$ df(t,B_t)=\left[\frac{\partial f}{\partial t}(t,B_t)+\frac12\frac{\partial^2 f}{\partial x^2}(t,B_t)\right]dt+\frac{\partial f}{\partial x}(t,B_t)dB_t $$ To prove $f(t,B_t)$ is a Martingale, you just need to show the $dt$ part is zero, which only requires deterministic calculus.

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  • $\begingroup$ Thanks for your answer, but we have not done Ito's formula yet. $\endgroup$
    – 3nondatur
    Nov 30, 2022 at 0:21

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