My problem is to calculate $E(e^{\lambda X})$, where X has normal distribution $N(\mu, \sigma^2)$. So I have to calculate integral as follows:
$\int_{-\infty}^\infty e^{\lambda x} \frac{1}{\sqrt{2\sigma^2 \pi}} e^{-\frac{(x-\mu)^2}{2 \sigma^2}}$, where $\lambda$ is real constant and I don't know how to do it.
Thanks for any suggestion and help.
First: to calc $E[e^{\lambda X}]$ you have to solve $$\int_{-\infty}^\infty e^{\lambda x} \frac{1}{\sqrt{2\sigma^2 \pi}} e^{-\frac{(x-\mu)^2}{2 \sigma^2}} dx$$
Consider the lower bound!
Then it holds $$e^{\lambda x} e^{-\frac{(x-\mu)^2}{2\sigma^2}} = e^{\frac{2\lambda\sigma^2 x - (x-\mu)^2}{2\sigma^2}} $$
But $$\begin{align*}2\lambda\sigma^2 x - (x-\mu)^2 &= 2\lambda\sigma^2 x - x^2 + 2\mu x - \mu^2 \\ &= -(x^2 - 2(\mu + \lambda\sigma^2)x) - (\mu - \lambda\sigma^2)^2 + (\mu - \lambda\sigma^2)^2 - \mu^2 \\ &=-(x-(\mu - \lambda\sigma^2))^2 + \lambda\sigma^2(2\mu + \lambda\sigma^2)\end{align*}$$
So:
$$e^{\lambda x} e^{-\frac{(x-\mu)^2}{2\sigma^2}} = e^{\frac{2\lambda\sigma^2 x - (x-\mu)^2}{2\sigma^2}} = e^\frac{(x-\mu')^2}{2\sigma^2}e^{\mu\lambda+\frac{\lambda^2\sigma^2}{2}}$$ with $\mu' = \mu - \lambda\sigma^2$
So it holds:
$$\int_{-\infty}^\infty e^{\lambda x} \frac{1}{\sqrt{2\sigma^2 \pi}} e^{-\frac{(x-\mu)^2}{2 \sigma^2}}dx = e^{\mu\lambda+\frac{\lambda^2\sigma^2}{2}} \int_{-\infty}^\infty \frac{1}{\sqrt{2\sigma^2 \pi}} e^{-\frac{(x-\mu')^2}{2 \sigma^2}} dx = e^{\mu\lambda+\frac{\lambda^2\sigma^2}{2}} \cdot 1$$
First do it for $U$ having standard normal distribution: $$\mathbb Ee^{\lambda U}=\frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{\lambda u-\frac12u^2}du=e^{\frac12\lambda^2}\frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac12(u-\lambda)^2}du=\cdots$$
Now do it for $X:=\sigma U+\mu$ wich has normal distribution with mean $\mu$ and variance $\sigma^2$:$$\mathbb Ee^{\lambda X}=\mathbb Ee^{\lambda\sigma U+\lambda\mu}=e^{\lambda\mu}\mathbb Ee^{\lambda\sigma U}=\cdots$$
