I am just trying to understand the following three equations. $\phi(x)$ denotes the standard Gaussian cumulative distribution function and $X$~$N(\mu,\sigma^2)$

(1) $\mathbb{E}[e^{tX}f(X)]=e^{\mu t+\frac{\sigma^2 t^2}{2}}\mathbb{E}f(X+t\sigma^2)$ for all real $t$ and suitable $f$

(2) For any nice function $f$, $\mathbb{E}[f(X)(X-\mu)]=\sigma^2\mathbb{E}[f'(X)]$

(3) $\mathbb{E}\phi(aX+b)=\phi(\frac{a\mu+b}{\sqrt{1+a^2\sigma^2}})$

My approach to see the equality:

In (1) I just used the definition of $\mathbb{E}$, therefore $\mathbb{E}[e^{tX}f(X)]=\int_{-\infty}^{\infty}e^{tx}f(x)p(x)dx$ where $p(x)$ is the probability density function, $p(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}$. The first term on the right hand side $e^{\mu t+\frac{\sigma^2 t^2}{2}}$ is the moment generating function and the result of $\int_{-\infty}^{\infty}e^{tx}p(x)dx$. How can I derive the second term on the right hand side? I do not know how to handle $f(x)$ in thee integral equation.

In (2) I want to prove the quation $\int_{-\infty}^{\infty}(x-\mu)f(x)p(x)dx=\sigma^2\int_{-\infty}^{\infty}f'(x)p(x)dx$. Even if I simplify the LHS I do not see the relation.

In (3) the LH states $\int_{-\infty}^{\infty}\phi(ax+b)p(x)dx=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(\int_{-\infty}^{ax+b}e^{-\frac{z^2}{2}}dz)p(x)dx=?$ Maybe a coordinate tranform helps out, $ax+b=c$

  • $\begingroup$ For (1), try the technique of "completing the square" in the exponent of the exponential term. $\exp(tx)\exp(-(x-\mu)^2/2\sigma^2)$ can be expressed as $\exp(-(x-\lambda)^2/2\nu^2 + g(t))$ for suitable choices of $\lambda, \nu, g(t)$ and you will see that what you need pops out immediately. $\endgroup$ Oct 12, 2013 at 15:19
  • $\begingroup$ You mean I should complete the square in $\mu t+\frac{\sigma^2 t^2}{2}$ ? When I do this I get $t(\sqrt{\mu}+\frac{\sigma^2}{4\sqrt{\mu}})^2-\frac{\sigma^4}{16\mu})$ $\endgroup$
    – Montaigne
    Oct 12, 2013 at 16:58

2 Answers 2


For the second equation, you need to integrate by parts.

Starting with the LHS, $$ \mathbb{E}\{f(X) (X-\mu)\} = \int_{-\infty}^{\infty} (x-\mu) f(x) p(x) dx \\ $$ Using the property for the Gaussian distribution, $p'(x)=-\frac{x-\mu}{\sigma^2} p(x)$ $$ \mathbb{E}\{f(X) (X-\mu)\} = -\int_{-\infty}^{\infty} \sigma^2 f(x) p'(x) dx \\ $$ Integrating by parts,

$$ \mathbb{E}\{f(X) (X-\mu)\} = -\sigma^2 f(x) p(x) |_{-\infty}^\infty + \int_{-\infty}^{\infty} \sigma^2 f'(x) p(x) dx \\ $$

Assuming that the $f(x)$ is a function that is weakly differeniable and won't blow up at $\infty$ you get $$ \mathbb{E}\{f(X) (X-\mu)\}=\sigma^2 \mathbb{E}\{f'(X)\} $$

  • $\begingroup$ very nice! (+1) I was almost ready to make my answer complete, but this is better than what I had in mind. There is one point that I dislike: the $\mu$ are $\sigma^{2}$. It is enough to prove it for $N\left(0,1\right)$ as is made clear in my answer. I 'hate' these parameters and always look for opportunities to get rid of them. $\endgroup$
    – drhab
    Oct 17, 2013 at 12:32

This is almost an answer to your question (I have proofs for (1) and (3) and hope to find one for (2) soon) , and starts by making things easier.

Assume that (1), (2) and (3) are true in the special case $N\left(0,1\right)$.

Let $U\sim N\left(0,1\right)$ and $X=\sigma U+\mu$. Define function $g$ by $u\mapsto f\left(\sigma u+\mu\right)$ where $f$ is 'nice'.

Note that $g'\left(u\right)=\sigma f'\left(\sigma u+\mu\right)$.

We will now show that (1), (2) and (3) are true for $X\sim N\left(\mu,\sigma^{2}\right)$ as well.

(1) $E\left[e^{tX}f\left(X\right)\right]=e^{\mu t}Ee^{\sigma tU}f\left(\sigma U+\mu\right)=e^{\mu t}Ee^{\sigma tU}g\left(U\right)=e^{\mu t}e^{\frac{\sigma^{2}t^{2}}{2}}Eg\left(U+\sigma t\right)=e^{\mu t+\frac{\sigma^{2}t^{2}}{2}}Ef\left(\sigma\left(U+t\sigma\right)+\mu\right)=e^{t\mu+\frac{1}{2}t^{2}\sigma^{2}}Ef\left(X+t\sigma^{2}\right)$.

(2) $Ef\left(X\right)\left(X-\mu\right)=\sigma Ef\left(\sigma U+\mu\right)U=\sigma Eg\left(U\right)U=\sigma Eg'\left(U\right)=\sigma^{2}Ef'\left(\sigma U+\mu\right)=\sigma^{2}Ef'\left(X\right)$.

(3) $E\phi\left(aX+b\right)=E\phi\left(a\sigma U+a\mu+b\right)=\phi\left(\frac{a\mu+b}{\sqrt{1+a^{2}\sigma^{2}}}\right)$.

Proved is now that (1) (2) and (3) hold if for $U\sim N\left(0,1\right)$ the following conditions hold:

(1)' $E\left[e^{tU}f\left(U\right)\right]=e^{\frac{1}{2}t^{2}}Ef\left(U+t\right)$.

(2)' $Ef\left(U\right)U=Ef'\left(U\right)$.

(3)' $E\phi\left(aU+b\right)=\phi\left(\frac{b}{\sqrt{1+a^{2}}}\right)$.

The annoying $\mu$ and $\sigma^{2}$ do not play a part in this.

EDIT: I have a proof for (1)' and (3)'


Applying $v=u-t$ and we find:


$E\varphi\left(aU+b\right)$ can be recognized as $P\left(V\leq aU+b\right)$ where $U,V\sim N\left(0,1\right)$ are independent. So $E\varphi\left(aU+b\right)=P\left(W\leq b\right)$ where $W=V-aU$. Here $W\sim N\left(0,1+a^{2}\right)$ so $W'=\frac{W}{\sqrt{1+a^{2}}}\sim N\left(0,1\right)$. This leads to $E\varphi\left(aU+b\right)=P\left(W\leq b\right)=P\left(W'\leq\frac{b}{\sqrt{1+a^{2}}}\right)=\varphi\left(\frac{b}{\sqrt{1+a^{2}}}\right)$.

  • $\begingroup$ Thank you. I wasn't ready yet (so actually did not deserve this bounty yet), but Karthik filled it up in a very nice way. $\endgroup$
    – drhab
    Oct 17, 2013 at 12:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.