Expectation of the product of polynomial & exponential transformations of normal r.v

Let $$X \sim \mathcal{N}(\mu, \sigma^2)$$. Are there any (1) general formula and (2) references to the general formula for

$$\mathbb{E} (X^n e^{tX}),\; n \in \mathbb{N}, t \in \mathbb{R}$$

in particular for $$n = 1$$ and $$n = 2$$?

I know $$\mathbb{E}(X^n)$$ are the normal moments ($$\mu$$, $$(\sigma^2 + \mu^2)$$, ... for increasing $$n$$). I also know $$\mathbb{E}(e^{tX})$$ is the moment generating function for a normal, which evaluates to

$$\mathbb{E}(e^{tX}) = \exp\left(\mu t + \frac{1}{2}\sigma^2 t^2\right).$$

Clearly, directly multiplying both will not work as the two parts are dependent. I also considered Delta method and making $$X^n e^{tX}$$ a derivative of something and using the exchangeability of expectation and derivatives, but few pages of calculations in and it does not look promising.

I also looked at the table of normal integrals by Owen (1980), but am unable to find anything of the form (ignoring the constants)

$$\int x^n \exp(tx) \exp\left(-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}\right) \textrm{d}x .$$

• Just complete the square to reduce this to a question about the moments of a Normal distribution.
– whuber
Oct 21 '20 at 13:14
• I have to admit I have not considered that - in an (admittedly more complex) parameterisation I was too determined to take things out the last exponential function without considering whether I should put things in. Thanks for the tip, will report back. Oct 21 '20 at 14:25

As hinted by @whuber (with thanks), the key is to transform the expectation to the (pure) moment of another normal distribution.

We first recognise the expectation in question is, by definition

$$\mathbb{E}_{X \sim \mathcal{N}(\mu, \sigma^2)}(X^n e^{tX}) = \int x^n \exp(tx)\frac{1}{\sqrt{2\pi \sigma^2}}\exp\left(-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}\right) \,\textrm{d}x .$$

We combine the exponential terms on the RHS to obtain

$$\int x^n \frac{1}{\sqrt{2\pi \sigma^2}} \exp\left(-\frac{1}{2}\left[\frac{(x-\mu)^2}{\sigma^2} - 2tx\right]\right) \,\textrm{d}x$$

Completing the square within the square brackets we have

$$\int x^n \frac{1}{\sqrt{2\pi \sigma^2}} \exp\left(-\frac{1}{2}\left[\frac{(x-(\mu+t\sigma^2))^2}{\sigma^2}\right]\right) \exp\left(\mu t + \frac{1}{2}\sigma^2 t^2\right) \,\textrm{d}x$$

The rightmost exponential term can be moved out of the integral. What is left inside is the moment of a different normal $$\mathcal{N}(\mu + t\sigma^2, \sigma^2)$$ by definition. Thus

$$\mathbb{E}_{X \sim \mathcal{N}(\mu, \sigma^2)}(X^n e^{tX}) = \mathbb{E}_{X \sim \mathcal{N}(\mu + t\sigma^2, \sigma^2)}(X^n) \cdot \exp\left(\mu t + \frac{1}{2}\sigma^2 t^2\right)$$

(For completeness) for $$n = 1, 2$$,

$$\mathbb{E}_{X \sim \mathcal{N}(\mu + t\sigma^2, \sigma^2)}(X) = \mu + t\sigma^2$$ $$\mathbb{E}_{X \sim \mathcal{N}(\mu + t\sigma^2, \sigma^2)}(X^2) = Var(X) + \mathbb{E}^2_{X \sim \mathcal{N}(\mu + t\sigma^2, \sigma^2)}(X) = \sigma^2 + (\mu+t\sigma^2)^2.$$