# Martingale wrt natural filtration: $e^{ \sum_{i=1}^t X_i - t/2}$

Let all $X_i$ be standard normal and iid for $i \in [1,T]$, let $X_0 = 0$. Define for each $t \in [0,T]$

$S_t = e^{( \sum_{i=1}^t X_i) - t/2}$

Is this process a martingale wrt its natural filtration? I guess we can write $S_t$ as a product of exponentials of each of the $X_i$, but then I am not sure how to show that most of these factors are measurable with respect to the filtration we are conditioning with (if i can do that, I can pull the out of the expectation), and I am also not sure how to evaluate the remaining expectation $$E_{t-1} e^{X_t}$$ Again, due to independence (is that obvious?), I may be able to reduce this to a regular expectation, but that doesn't seem easy to solve either (if it is to be a martingale, then this should equal $1$).

• By "for each $t\in[0,T]$" did you mean $t\in\{0,1,\ldots,T\}$? Jun 19, 2016 at 14:38

The computations here are relatively straightforward. For $1\leqslant t\leqslant T$ we have \begin{align} \mathbb E[|S_t|] &= \mathbb E\left[\exp\left(\sum_{i=1}^t X_i - \frac t2\right)\right]\\ &=\exp\left(-\frac t2\right)\mathbb E\left[\prod_{i=1}^t \exp(X_i)\right]\\ &=\exp\left(-\frac t2\right)\exp\left(\frac t2\right)\\ &=1 \end{align} so that $S_t$ is integrable, and \begin{align} \mathbb E[S_t\mid \mathcal F_{t-1}] &= \mathbb E\left[\exp\left(\sum_{i=1}^t X_i - \frac t2 \right)\ \bigg\vert\ \mathcal F_{t-1} \right]\\ &= \mathbb E\left[\exp(X_t)\exp\left(-\frac12\right)\right]\mathbb E[S_{t-1}\mid \mathcal F_{t-1}] \\ &= S_{t-1}. \end{align} It follows that $\{S_t\}$ is a martingale with respect to its natural filtration.
• My impression is that proving $E(e^X)=e^{1/2}$ would also be a problem to the OP...