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)

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

  • 1
    $\begingroup$ Just complete the square to reduce this to a question about the moments of a Normal distribution. $\endgroup$
    – whuber
    Commented Oct 21, 2020 at 13:14
  • 1
    $\begingroup$ 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. $\endgroup$
    – B.Liu
    Commented Oct 21, 2020 at 14:25

2 Answers 2


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 $n = 1, 2$,

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


Let $X$ be a random variable with moment generating function (mgf) $m(t),$ where $t$ is a real number. Then, $$E[X^n exp(tX)] = E[\frac{d^n}{dt^n} exp(tX)] = \frac{d^n}{dt^n} m(t),$$ which is the n-th derivative with respect to $t$ of $m(t)$ and $E$ denotes expectation.

In this specific example, $m(t)$ is the mgf of a normal random variable.

  • $\begingroup$ Please use $\LaTeX$ on this site, as it makes reading mathematics much easier. $\endgroup$ Commented Jun 7, 2022 at 14:20

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