I am reading Shreve's Stochastic Calculus for Finance II. He states on pages 110 and 111 that,

$$E[exp(\sigma m-\frac{1}{2}\sigma^2 \tau_m)] = 1$$ $$E[exp(-\frac{1}{2}\sigma^2 \tau_m)] = e^{-\sigma m}$$

I understand that the top equation is a martingale and should thus have a constant expectation but I don't understand why both equations are true. I tried taking the expectations of both equations by taking the integral from 0 to $m$, but that doesn't work since I get

$$ E[exp(-\frac{1}{2}\sigma^2 \tau_m)] = \int_0^m exp(-\frac{1}{2}\sigma^2 \tau_m) dt = \frac{2-2exp{(-\frac{ms^2}{2}})}{s^2} \color{red} \ne e^{\sigma m}$$

How do I prove in detail both expectations?

Here are the pages for reference.

Shreve Stochastic Calculus for Finance Volume II, page 110

Shreve Stochastic Calculus for Finance Volume II, page 111


1 Answer 1


Given that $exp(\sigma m-\frac{1}{2}\sigma^2 \tau_m)$ is a martingale, you just need to substitute $m = 0$ into it to find the value of its expectation, as for any martingale $Z_t$ we have that $E[Z_t] = Z_0$.

If you are really interested in an algebric proof, you need to find first the distribution of $\tau_m$. You can find it on the link below or you can try to prove it by yourself.



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