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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, $ax+b=c$

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  • $\begingroup$ 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. $\endgroup$ – Dilip Sarwate Oct 12 '13 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 '13 at 16:58
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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)'

$E\left[e^{tU}f\left(U\right)\right]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{tu-\frac{1}{2}u^{2}}f\left(u\right)du$.

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

$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{tu-\frac{1}{2}u^{2}}f\left(u\right)du=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{tv-\frac{1}{2}\left(v+t\right)^{2}}f\left(v+t\right)dv=\frac{e^{-\frac{1}{2}t^{2}}}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}v^{2}}f\left(v+t\right)dv=e^{-\frac{1}{2}t^{2}}E\left[f\left(U+t\right)\right]$

$E\varphi\left(aU+b\right)$ 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 $E\varphi\left(aU+b\right)=P\left(W\leq b\right)=P\left(W'\leq\frac{b}{\sqrt{1+a^{2}}}\right)=\varphi\left(\frac{b}{\sqrt{1+a^{2}}}\right)$.

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  • $\begingroup$ 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 '13 at 12:35
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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 $$ \mathbb{E}\{f(X) (X-\mu)\}=\sigma^2 \mathbb{E}\{f'(X)\} $$

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  • $\begingroup$ 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 '13 at 12:32

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