Exponential Martingales and the Hitting-Time Laplace Transform

Summary

The document discusses two expectation identities involving an exponential martingale evaluated at a Brownian hitting time. The key idea for the first identity is the martingale property: if the process is a martingale, its expectation remains equal to its initial value, which can be found by evaluating it at the starting point. This avoids trying to integrate the exponential expression directly over the hitting-time interval.

For the second expectation, the response says that a direct algebraic derivation requires the distribution of the hitting time, identified as an inverse Gaussian distribution. The document does not provide the distribution’s density or carry out the integration, instead pointing readers to an external derivation. Its explanation is therefore a concise proof strategy rather than a full detailed calculation. Correct use of the identities depends on the stated process being a martingale under the relevant assumptions and on interpreting the hitting time consistently.

Key ideas

• A martingale’s expectation equals its initial value, which establishes the first identity by evaluating the process at the starting point.
• The hitting-time expectation should not be computed by integrating over time as if the stopping time were a deterministic variable.
• A direct algebraic proof of the second identity requires the distribution of the hitting time.
• The cited distribution is inverse Gaussian, but the document does not work through its density or integration.

Tags

# How to take the expectation of an exponential martingale? And an exponential with a random value?


# How to take the expectation of an exponential martingale? And an exponential with a random value?












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.

## Answer by dr_1993 (score 2, accepted)

https://quant.stackexchange.com/a/61564

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.

https://en.wikipedia.org/wiki/Inverse_Gaussian_distribution

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