Deriving the Exponential Moment of a Normal Variable

# Log-normal risk-neutral price derivation from binomial trees, not clear about step in derivation process


# Log-normal risk-neutral price derivation from binomial trees, not clear about step in derivation process












At page 64 of the book Concepts and practice of mathematical finance, 2nd edition by M. Joshi, paragraph 3.7.2 (Trees and option pricing - A log-normal model - The risk-neutral world behaviour) a quick exercise is presented:

> Show that $$ \mathbb{E}( \exp(\sigma \sqrt{T} N(0,1) ) ) = \exp(0.5 \, \sigma^2 T) $$ where $\mathbb{E}$ indicates the expected value of the expression inside brackets, in which $\sigma$ is the volatility of the underlying asset, $T$ is the option expiration time, and $N(0,1)$ is the normal distribution.

How to verify this relation? Solution is not provided.

For context, this term is useful to simplify the log-normal expected value of the asset at expiry, $$\mathbb{E} (S_T) = \mathbb{E}(S_0 exp{((r - 0.5 \sigma^2) T + \sigma \sqrt{T} N(0, 1))}$$ to $$\mathbb{E} (S_T) = S_0 exp({r T})$$

EDIT: this question reappears as exercise 3.13 at page 72 of Concepts and practice of mathematical finance, 2nd edition. Solution is at the back of the book, and follows the line provided in the accepted answer below.

## Answer by Daneel Olivaw (score 2, accepted)

https://quant.stackexchange.com/a/58391

Let $X\sim N(0,1)$ be a standard normal variable and $\alpha:=\sigma\sqrt{T}$, then by definition of the expectation and the distribution of normal variables: $$\begin{align} \mathbb{E}\left(e^{\alpha X}\right) &=\int_{-\infty}^{\infty}e^{\alpha x}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx \\ &=\int_{-\infty}^{\infty}e^{\alpha x+\frac{1}{2}\alpha^2-\frac{1}{2}\alpha^2}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx \\ &=e^{\frac{1}{2}\alpha^2}\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}\left(x^2-2\alpha x+\alpha^2\right)}dx \\ &=e^{\frac{1}{2}\alpha^2}\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{(x-\alpha)^2}{2}}dx \\[3pt] &=e^{\frac{1}{2}\alpha^2} \end{align}$$ The last step is merely the consequence that the last integral is over the probability density function of a normal variable with mean $\alpha$ and variance $1$ taken with respect to the whole real numbers $\mathbb{R}$, hence it integrates to 1.

Anecdotally, the expression $\mathbb{E}(e^{-\beta X})$ with $\beta:=-\alpha$ is sometimes called the (two-sided) Laplace transform of the random variable $X$.