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My problem is to calculate $E(e^{\lambda X})$, where X has normal distribution $N(\mu, \sigma^2)$. So I have to calculate integral

$\int_{-\infty}^\infty e^{\lambda x} \frac{1}{\sqrt{2\sigma^2 \pi}} e^\frac{(x-\mu)^2}{2 \sigma^2}$, where $\lambda$ is real constant and I don't know how to do it.

Thanks for any suggestion and help.

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    Complete the square in the exponent. – 2017-02-12
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    Complete the square in the exponential, shift the integration variable, and be careful – 2017-02-12
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    You must integrate over $\mathbb R$ (not over $(0,\infty)$). – 2017-02-12
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    Thanks for suggestions. Ofcourse, I have edited integral. – 2017-02-12

2 Answers 2

2

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:

$$\int_{-\infty}^\infty e^{\lambda x} \frac{1}{\sqrt{2\sigma^2 \pi}} e^\frac{(x-\mu)^2}{2 \sigma^2}dx = e^{\mu\lambda+\frac{\lambda^2\sigma^2}{2}} \int_{-\infty}^\infty \frac{1}{\sqrt{2\sigma^2 \pi}} e^\frac{(x-\mu')^2}{2 \sigma^2} dx = e^{\mu\lambda+\frac{\lambda^2\sigma^2}{2}} \cdot 1$$

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    Thank you very much. Very good explanation. – 2017-02-12
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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 $\sigma^2$:$$\mathbb Ee^{\lambda X}=\mathbb Ee^{\lambda\sigma U+\lambda\mu}=e^{\lambda\mu}\mathbb Ee^{\lambda\sigma U}=\cdots$$

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    Thank you so much for that. Interesting way of doing it first for $N(0,1)$. – 2017-02-12