I am struggling to see the relationship between the left and right side of the equation below:
$$\mathbb E\left[\left( e^K \right)^2\right]=\large{e^{2\sigma^2_K}}$$
with $K\sim \mathcal N\left(0,\sigma_K^2\right)$
I know that you are supposed to show the relationship by the probability density function given a mean and variance, but I do not know how to proceed. Could anyone help me?
The expected value of $(e^K)^2=e^{2K}$ is
$$\large{\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty }^{\infty} e^{2k}\cdot e^{-k^2/(2\sigma^2)}} \,dk$$
$$\large{\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty }^{\infty} e^{4k\sigma^2/(2\sigma^2)}\cdot e^{-k^2/(2\sigma^2)}} \,dk$$
We can put together the exponents in a quadratic expression by completing the squares.
$-(k^2-4k\sigma^2+4\sigma^4)+4\sigma^4=-(k-2\sigma^2)^2+4\sigma^4$
$$\large{\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty }^{\infty} e^{4\sigma^4/(2\sigma^2)}\cdot e^{-(k-2\sigma^2)^2/(2\sigma^2)}} \,dk$$
$$\large{\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty }^{\infty} e^{2\sigma^2}\cdot e^{-(k-2\sigma^2)^2/(2\sigma^2)}} \,dk$$
$$e^{2\sigma^2}\cdot \underbrace{\large{\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty }^{\infty} e^{-(k-2\sigma^2)^2/(2\sigma^2)}} \,dk}_{=1}$$
The expression behind $e^{2\sigma^2}$ is the integral of the pdf of a normal distributed variable. The value is is well known.