Normal and Symmetric-Binomial Moment-Generating Functions

Summary

This exchange resolves a confusion about the moment-generating function of a standard normal variable. For a normal variable with mean zero and unit variance, scaling it by a constant gives a normal variable whose variance is the square of that constant. Its exponential expectation is therefore the exponential of half that variance. The response also flags that the expression in the question instead matches a symmetric binomial variable taking values of plus or minus one with equal probability.

The key lesson is to identify the distribution before applying an expectation formula: the average of two exponentials comes from the two-point distribution, while the Gaussian moment-generating function gives a single exponential involving the squared scale. The exchange offers a conceptual correction, not a full account of the textbook passage, and its interpretation of that passage is based on a brief inspection. It concerns probability theory that can support quantitative finance, rather than a trading method or empirical market result.

Key ideas

• A standard normal variable scaled by a constant has variance equal to the square of that constant.
• The normal moment-generating function yields an exponential involving half the scaled variance.
• An equal-probability variable taking values of plus or minus one has a two-term exponential expectation.
• Check the assumed distribution before applying a moment-generating function.

Tags

# expression on page 90 of shreve's stochastic calculus for finance II


# expression on page 90 of shreve's stochastic calculus for finance II












Hi: In the middle of page 90, Shreve has an expression which implies that (I'm using $t$ where he uses $u$ only because I find it confusing to use $u$ and $\mu$ in the same expressions):

$ E[\exp(\dfrac{t}{\sqrt{n}} X_{j})] = \left(\frac{1}{2} \exp(\dfrac{t}{\sqrt{n}}) + \dfrac{1}{2} \exp(-\frac{t}{\sqrt{n}})\right)$

where $X_{j}$ is normal with mean zero and variance equal to one.

I assume that the author is using the expression for the moment generating function of a standardized normal random variable. The confusion I have is that the mgf of a normal with mean $\mu$ and variance $\sigma^2$ is $\exp{(\mu t + \frac{1}{2}\sigma^2 t)}$. Thanks for help.

## Answer by Kurt G. (score 2, accepted)

https://quant.stackexchange.com/a/70304

Are you sure about that formula? What is the expression on p. 90 in Shreve that implies it ?

When $X_j\sim N(0,1)$ then $\frac{t}{\sqrt{n}}X_j$ has mean zero and variance $\frac{t^2}{n}\,.$ Then $$ \textstyle\mathbb E\Big[\exp\Big(\frac{t}{\sqrt{n}}X_j-\frac{t^2}{2n}\Big)\Big]=1. $$ So $$ \textstyle\mathbb E\Big[\exp\Big(\frac{t}{\sqrt{n}}X_j\Big)\Big]=\exp\Big(\frac{t^2}{2n}\Big)\,. $$

Edit

As far as I can tell from briefly looking at Shreve's book p.90 he assumes that $X_j$ is a binomial that takes values in $\pm 1$ with equal probabilities. This means that the formula you are implying is trivial.

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