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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.

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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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  • $\begingroup$ 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. $\endgroup$
    – mark leeds
    Mar 25, 2022 at 17:43
  • $\begingroup$ No problem. Glad I could help. $\endgroup$
    – Kurt G.
    Mar 25, 2022 at 17:45

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