# Steven Shreve: Stochastic Calculus and Finance

The lecture notes have the following theorem:

Let $$\theta\in \mathbb{R}$$ be given and $$B(t)$$ stands for the Brownian motion which is a martingale, then $$Z(t)=exp\{-\theta B(t)-\dfrac{1}{2}\theta^2t\}$$ is also a martingale.

$$\underline{proof:}$$ Let $$0\leq s\leq t$$ be given. Then $$\mathbb{E}[Z(t)|\mathbb{F}(s)]=\mathbb{E}[exp\{-\theta (B(t)-B(s)) -B(s) -\dfrac{1}{2}\theta^2((t-s)+s)\}|\mathbb{F}(s)]\Rightarrow \\ \mathbb{E}[Z(t)|\mathbb{F}(s)]=Z(s)\mathbb{E}[exp\{-\theta (B(t)-B(s)) -\dfrac{1}{2}\theta^2(t-s)\}|\mathbb{F}(s)]\Rightarrow \\ \mathbb{E}[Z(t)|\mathbb{F}(s)]=Z(s)exp\{\dfrac{1}{2}(-\theta)^2\mathbb{Var}(B(t)-B(s))-\dfrac{1}{2}\theta^2(t-s)\}=Z(s)$$ where $$X=B(t)-B(s)\sim N(0,t-s)$$. My question is how can you proove this part $$\mathbb{E}[Z(t)|\mathbb{F}(s)]=Z(s)exp\{\dfrac{1}{2}(-\theta)^2\mathbb{Var}(B(t)-B(s))-\dfrac{1}{2}\theta^2(t-s)\}$$ analytically by using the normal pdf. I am missing something in my effort to proove this part, because no textbook from those that I have does it analytically.

## 1 Answer

Note merely that $$B_t=B_s+(B_t-B_s)$$ which is the sum of independent normally distributed random variables. In particular, $$B_s$$ is $$\mathbb{F}_s$$-measurable and $$B_{t-s}$$ is independent of $$\mathbb{F}_s$$. Thus, \begin{align*} \mathbb{E}_s[Z_t] &= \mathbb{E}_s\left[\exp\left(-\frac{1}{2}\theta^2t+\theta B_t\right)\right] \\ &= \mathbb{E}_s\left[\exp\left(-\frac{1}{2}\theta^2t+\theta B_s+\theta B_{t-s}\right)\right] \\ &= \exp\left(-\frac{1}{2}\theta^2s\right)\exp\left(-\frac{1}{2}\theta^2(t-s)\right)\cdot\mathbb{E}_s\left[\exp\left(\theta B_s+\theta B_{t-s}\right)\right]\\ &= \exp\left(-\frac{1}{2}\theta^2s\right)\cdot\exp\left(-\frac{1}{2}\theta^2(t-s)\right)\cdot\exp\left(\theta B_s\right)\cdot\mathbb{E}\left[\exp\left(\theta B_{t-s}\right)\right]\\ &= Z_s\exp\left(-\frac{1}{2}\theta^2(t-s)\right)\cdot\exp\left(\frac{1}{2}\theta^2 \mathrm{Var}[B_{t-s}]\right) \\ &= Z_s\exp\left(-\frac{1}{2}\theta^2(t-s)+\frac{1}{2}\theta^2 (t-s)\right) \\ &= Z_s. \end{align*}

And you're done and have shown that $$(Z_t)$$ is a martingale (with respect to the natural filtration of the Brownian motion).

You don't need to integrate the normal density. That's just tedious. You should first decompose the Brownian motion as mentioned in the beginning. The $$-\frac{1}{2}\theta^2t$$ can be split up similarly and pulled out of the expectation as it is non-random. Then, you can use measurability and independence (two key properties of the conditional expectation) to pull out $$B_s$$ and reduce the conditional expectation of $$B_{t-s}$$ to a simple expectation. Then, note that $$B_{t-s}\sim N(0,t-s)$$ and that $$\mathbb{E}\left[e^{m+sZ}\right]=e^{m+0.5s^2}$$ - a useful formula to keep in mind.

• thak you! I was trying to proove this with normal density...so i think this is a better way! – Nav89 Jan 21 '20 at 22:09
• @Nav89 You're welcome. This is a standard example illustrating how important it is to understand the properties of the conditional expectation. – Kevin Jan 21 '20 at 22:11
• I have one last question. In the 4th line, how the expectation becomes unconditional, i.e. $\mathbb{E}\left[\exp\left(\theta B_{t-s}\right)\right]$, there is no $s$ subscript any more, why? – Nav89 Jan 21 '20 at 22:59
• @Nav89 $B_{t-s}$ is a Brownian increment over $[s,t]$ hence it is independent of the history over $[0,s]$, $\sigma(B_r)_{0\leq r\leq s}$. – Daneel Olivaw Jan 21 '20 at 23:02
• Because the increment $B_{t-s}$ is independent of $\mathbb{F}_s$. – Kevin Jan 21 '20 at 23:02