Why Exponentiating a Brownian Increment Produces a Nonzero Expectation

Summary

The document resolves an apparent conflict between the zero expected value of a Brownian motion increment and a nonzero expectation obtained after exponentiating that increment. The key distinction is that taking an expectation does not generally commute with applying a nonlinear function: the expected value of an exponential is not the exponential of the expected value.

A Brownian increment over an interval is normally distributed with zero mean and variance equal to the interval length. Multiplication by a constant scales its variance, and applying the exponential produces a lognormal random variable. Using the lognormal expectation formula gives an exponential expectation above one when the variance is positive. This is a compact probability explanation relevant to stochastic models in finance. It assumes the standard Brownian distribution and gives no broader trading strategy, empirical evidence, or discussion of discretization and model risk.

Key ideas

• A Brownian motion increment has zero mean, but a nonlinear transformation can have a nonzero expectation.
• The expectation of a function of a random variable generally cannot be found by applying the function to the variable’s expectation.
• A scaled Brownian increment is normally distributed with variance determined by the scale and time interval.
• Exponentiating a normal random variable yields a lognormal variable whose expectation depends on both its mean and variance.

Tags

# What is the expectation of a change in Brownian motion?


# What is the expectation of a change in Brownian motion?












I know $E[W_T-W_t]=0$ but I have a solution which implies this is wrong.

### Question

### Answer

## Answer by actuarialboi9 (score 1, accepted)

https://quant.stackexchange.com/a/63070

$E(X)=\mu$ doesn't necessarily imply $E[f(X)]=f(\mu)$. In this case, if $X \sim N(\mu,\sigma^2)$ then $e^X \sim lnN(\mu,\sigma^2)$ (lognormal distribution) and $$E(e^X)=e^{\mu+\frac{\sigma^2}{2}}$$. We know that $W_T-W_t \sim N(0,T-t)$ therefore $\sigma\gamma(W_T-W_t) \sim N(0,\sigma^2\gamma^2(T-t))$. It is easy to see that $$E[e^{\sigma\gamma(W_T-W_t)}]=e^{\frac{1}{2}\sigma^2\gamma^2(T-t)}$$

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