# 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.

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 $$\textstyle\mathbb E\Big[\exp\Big(\frac{t}{\sqrt{n}}X_j\Big)\Big]=\exp\Big(\frac{t^2}{2n}\Big)\,.$$
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.
• Thanks. Yes, what I stated in my question is not correct. I read backwards more carefully and, as Kurt mentioned, the author is assuming that the $X_{j}$ are coin tosses so binomial and based on the formula I was confused about, it must be that $p = q = \frac{1}{2}$. My apologies for noise and thanks to Kurt for clarification. I was staring at that for hours. Mar 25, 2022 at 17:43