Expected value of X*exp(X) for X normally distributed

[TaPaKaH]

Assume we have $X\sim\exp(\mu,\sigma^2)$.
How does one compute $\mathbb{E}\left(Xe^X\right)$ and/or what is the outcome value?

 

[CompuChip]

The definition would say that

$$\mathbb{E}[ X e^X ] = \int_{-\infty}^{\infty} x e^{x} e^{-(x - \mu)^2 / \sigma^2} \, dx$$

That's a tricky one, I don't think any of the common integration strategies really work.
WolframAlpha does give me an exact result for the standard distribution $(\mu, \sigma) = (0, 1)$.

Maybe this will also help.

Sorry for the incomplete answer, hoping that this will get you started.

 

[D H]

Rewrite $\exp(x)\exp(-(x-\mu)^2/(2\sigma^2))$ as $\exp(x-(x-\mu)^2/(2\sigma^2)) = \exp((2\sigma^2x-(x-\mu)^2)/(2\sigma^2))$. Now complete the square so that you get $a\exp(-(x-\mu')^2)/(2\sigma^2))$, where $a$ is the exponential of the nasty junk you had to add to complete the square and $\mu'$ has the effect of being a new mean.

 

[CompuChip]

Very nice one, and congratulations on post number 13,000!

 

[blue_raver22]

To compute the expected value of X*e^X, we can use the formula for the expected value of a continuous random variable, which is given by:

\mathbb{E}\left(Xe^X\right) = \int_{-\infty}^{\infty}x*e^x*f(x)dx

Where f(x) is the probability density function of X. In this case, since X is normally distributed, its probability density function is given by:

f(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}

Substituting this into the formula, we get:

\mathbb{E}\left(Xe^X\right) = \int_{-\infty}^{\infty}x*e^x*\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx

To solve this integral, we can use integration by parts, where u = x*e^x and dv = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx. This gives us:

\mathbb{E}\left(Xe^X\right) = \left[x*e^x*\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\right]_{-\infty}^{\infty} - \int_{-\infty}^{\infty}e^x*\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx

The first term in this expression evaluates to 0, since the exponential function goes to 0 as x approaches either infinity. The second term can be simplified using the fact that the integral of the normal density function over its entire support is equal to 1. This gives us:

\mathbb{E}\left(Xe^X\right) = \frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^{\infty}e^x*e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx