Expected value of exp(X^3)] , X~(mu,sigma^2)

Question

Any shortcut in calculating

$$\operatorname E[e^3X ],\qquad X\sim N(μ,\sigma^2)\text{ ?}$$

Insted of having to compute the integral? I'm having a hard time with this integral.

Answer 1

I'm guessing you meant $\operatorname E(e^{3X})$ rather than what you have in the subject line.

Presumably you already know that $\displaystyle\int_{-\infty}^\infty \frac 1 {\sigma\sqrt{2\pi}} e^{-(1/2)((x-\mu)/\sigma)^2} \, dx = 1.$

We seek $\displaystyle \int_{-\infty}^\infty e^{3x} e^{-(1/2)((x-\mu)/\sigma)^2} \, dx.$

We have $e^{3x} e^{-(1/2)((x-\mu)/\sigma)^2} = e ^{3x - (1/2)((x-\mu)/\sigma)^2}.$ We'll work on the exponent. \begin{align} 3x - \frac 1 2 \left( \frac{x-\mu}\sigma \right)^2 & = -\frac 1 2 \cdot \frac{-6x\sigma^2 + (x^2 - 2x\mu + \mu^2)}{\sigma^2} \\[10pt] & = - \frac 1 {2\sigma^2} \left( x^2 - 2(\mu+3\sigma^2) x + \mu^2 \right) \\[10pt] & = - \frac 1 {2\sigma^2}\Big( (x - 2(\mu+3\sigma^2) + (\mu + 3\sigma^2)^2 + \Big[ \mu^2 -(\mu + 3\sigma^2)^2\Big] \Big) \\ & \qquad\qquad (\text{completing the square}) \\[12pt] & = - \frac 1 {2\sigma^2} (x-(\cdots))^2 + \Big( \text{something not depending on $x$} \Big) \end{align} Thus our integral is $$ \frac 1 {\sigma\sqrt{2\pi}} \int_{-\infty}^\infty e^{-(1/2)(x - (\cdots))^2} \, dx \times e^{(\text{something not depending on $x$})}. $$ The value of this integral is $1$ for the same reason the value of the first integral mentioned above is $1.$ The factor after the integral is what you seek.