TITLE: expectation equations
QUESTION [6 upvotes]: 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, $ax+b=c$

REPLY [5 votes]: 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 
$$
\mathbb{E}\{f(X) (X-\mu)\}=\sigma^2 \mathbb{E}\{f'(X)\}
$$