Showing martingale for a Brownian motion $(W_t)_{t \geq 0}$

Question

I want to show that $\dfrac{e^{W_{t}^2/(1+2t)}}{\sqrt{1+2t}}$ is a martingale with respect to $F_{t}$. We can use that $$E(e^{\alpha X^2/\sigma^2})=\dfrac{\dfrac{\mu^2\alpha}{e^{\sigma^2(1-2\alpha)}}}{\sqrt{1-2\alpha}}.$$ Can somebody help me with this?

Answer 1

There is the following general statement on conditional expectations (see e.g. Brownian Motion - An Introduction to Stochastic Processes by René Schilling & Lothar Partzsch, Lemma A.3, for a proof).

Let $\mathcal{F}$ be a $\sigma$-algebra and $X,Y$ two random variables such that $X$ is $\mathcal{F}$-measurable and $Y$ is independent of $\mathcal{F}$. Then $$\mathbb{E}(f(X,Y) \mid \mathcal{F}) = \mathbb{E}(f(x,Y)) \bigg|_{x=X}$$ for any Borel-measurable function $f$ such that $f(X,Y) \in L^1(\mathbb{P})$.

---

Since $W_t-W_s$ is independent of $\mathcal{F}_s$ and $W_s$ is $\mathcal{F}_s$-measurable, this implies

$$\begin{align*} & \mathbb{E} \left( \exp \left[ \frac{1}{1+2t} (W_t-W_s)^2 + \frac{2}{1+2t} W_s(W_t-W_s) \right] \mid \mathcal{F}_s \right) \\ &= \mathbb{E} \left( \exp ( \frac{1}{1+2t} (W_t-W_s)^2+\frac{2x}{1+2t} (W_t-W_s)) \right) \bigg|_{x=W_s}. \tag{1} \end{align*}$$

Now rewrite the exressopn at the right-hand side in such a way that we get an expression of the form

$$\exp \left( \alpha (Y-\nu)^2 \right)$$

for some Gaussian random variable $Y$. Then you can calculate the right-hand side of $(1)$ using the formula mentioned at the beginning of your question.

---

Solution: Since

$$\frac{1}{1+2t} ((W_t-W_s)^2+2x (W_t-W_s)) = \frac{1}{1+2t} \bigg[ ((W_t-W_s)+x)^2-x^2 \bigg]$$

we have

$$\begin{align*} & \mathbb{E} \left( \exp \left[ \frac{1}{1+2t} (W_t-W_s)^2+\frac{2x}{1+2t} (W_t-W_s) \right] \right) \\ &= e^{x^2/(1+2t)} \mathbb{E}\exp \left( \frac{1}{1+2t} ((W_t-W_s)+x)^2 \right). \end{align*}$$

The random variable $W_t-W_s \sim W_{t-s} \sim N(0,t-s)$ is Gaussian with mean $0$ and variance $t-s$. Therefore, $X := W_t-W_s+x$ is Gaussian with mean $x$ and variance $t-s$. Applying the formula you mentioned at the beginning of your question (with $\alpha = (t-s)/(1+2t)$, $\sigma^2 = t-s$ and $\mu = x$) we find

$$\begin{align*} \mathbb{E}\exp \left( \frac{1}{1+2t} ((W_t-W_s)+x)^2 \right) &= \sqrt{\frac{1+2t}{1+2s}} \exp \left( \frac{t-s}{1+2t} x^2 \left[ (t-s) \frac{1+2s}{1+2t} \right]^{-1} \right) \\ &= \sqrt{\frac{1+2t}{1+2s}} \exp \left(\frac{1}{1+2s} x^2 \right). \end{align*} $$

where we have used that

$$1-2 \alpha = \frac{1+2t}{1+2t} - \frac{2(t-s)}{1+2t} = \frac{1+2s}{1+2t}$$

Consequently, we get

$$ \mathbb{E} \left( \exp \left[ \frac{1}{1+2t} (W_t-W_s)^2+\frac{2x}{1+2t} (W_t-W_s) \right] \right) = e^{-x^2/(1+2t)} \sqrt{\frac{1+2t}{1+2s}} \exp \left(\frac{1}{1+2s} x^2 \right).$$

Combining this with $(1)$ and your calculation proves that the given process is a martingale.