A random variable Y obeys a normal distribution

$P(Y)=\frac{1}{\sigma \sqrt{2\pi }}\exp [-\frac{(Y-\mu {)}^2}{2{\sigma }^2}]$

The mean value of e^Y is

Explanation:

• The expectation value of a function h(Y) of a random variable Y with a probability density function p(Y) is defined as:

$E[h(Y)]=\int h(Y)p(Y)dY,$ (over all Y).

• Here, we're interested in $h(Y)=e^Y$. So, $E[e^Y]=\int e^{Y}p(Y)dY$, (over all Y).
• Plugging in the given normal distribution for p(Y), we have: $E[e^Y]=\int e^Y(\frac{1}{\sigma \sqrt{2\pi }})\times exp[-\frac{(Y-\mu )²}{(2\sigma ²)}]dY,$ (over all Y).
• Simplify the above expression, we can rewrite it as:

$E[e^Y]=(\frac{1}{\sigma \sqrt{(2\pi )}})\times \int e^{Y-\frac{(Y-\mu )²}{(2\sigma ²)}}dY$, (over all Y).

$E[e^Y]=(\frac{1}{\sigma \sqrt{2\pi }})\times \int e^{\frac{-\mu }{2\sigma ²}-\frac{Y²}{(2\sigma ²)}+Y+\frac{Y\mu }{{\sigma }^2}}dY,$ This is an integral over all Y of an exponential quadratic.

• However, an exponential quadratic integrates into a Gaussian integral, which we know has a value of √π.
• Let's rewrite the exponent as:

$-\frac{\mu }{2\sigma ²}+Y-\frac{Y²}{(2\sigma ²)}+\frac{Y\mu }{{\sigma }^2}=\frac{-(Y-\mu -\sigma ²)²}{(2\sigma ²)}+\mu +\frac{\sigma ²}{2}.$

• So the integral is (up to factors): $\int e^{\frac{-(Y-\mu -\sigma ²)²}{(2\sigma ²)}+\mu +\frac{\sigma ²}{2}}dY$, Which is clearly a normal distribution in the variable $(Y-\mu -\sigma ²)$, with mean $(\mu +\sigma ²)$ and the same $\sigma$.
• Therefore, just from the change of variables $Y^{\prime }=Y-\mu -\sigma ²$, the definite integral from negative infinity to positive infinity of such a normal probability density gives 1.
• So, we end up with: $E[e^Y]=e^{\mu +\frac{\sigma ²}{2}}$.