Suppose $X_1$ and $X_2$ are $N(0,1)$ iid R.V.s.
Compute the characteristic function of $Y=X_1X_2$.
Solution:
For a R.V. X the characteristic function is defined as: $\varphi_X(s)=E[e^{isX}]$.
$\varphi_Y(s)=\varphi_{X_1X_2}(s)=E[e^{isY}]=E[e^{isX_1X_2}]$
Law of total expectation: $E[X]=E[E[X|Y]]$, where X,Y are two R.V.s defined on the same probability space. //en.wikipedia.org/wiki/Law_of_total_expectation
$E[e^{isX_1X_2}]=E[E[e^{isX_1X_2}|X_2]]=E[\varphi_{X_1}(sX_2)]$
$X_1\in N(0,1)$
The characteristic function $\varphi(s)$ for a $N(0,1)$ distributed R.V. is $\varphi(s)=e^{-\frac{s^2}{2}}$.
$\varphi_{X_1}(s)=e^{-\frac{s^2}{2}}$. So, $\varphi_{X_1}(sX_2)=e^{-\frac{(sX_2)^2}{2}}=e^{-\frac{s^2X_2^2}{2}}$.
$E[\varphi_{X_1}(sX_2)]=E[e^{-\frac{s^2X_2^2}{2}}]$
(This next step I don't understand.)
$E[e^{-\frac{s^2X_2^2}{2}}]=(X_2\in N(0,1))=\frac{1}{\sqrt{2\pi}}\int_{\Re}e^{-\frac{s^2x^2}{2}}e^{-\frac{x^2}{2}}dx$
(the pdf of $X_2$ is given by $f_{X_2}(x_2)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$). It seems to me as if the formula: $E[g(X)]=\int_\Omega g(x)f_X(x)\ dx$ has been used to calculate the expectation $E[e^{-\frac{s^2X_2^2}{2}}]$, i.e. $E[g(X_2)]=\int_\Re g(x_2)f_{X_2}(x_2)\ dx_2 =\frac{1}{\sqrt{2\pi}}\int_{\Re}e^{-\frac{s^2x^2}{2}}e^{-\frac{x^2}{2}}dx$. First, why do they use $x$ in the solution, instead of $x_2$ (like I just did in this paragraph)?. Second, mustn't one account for the fact that we have two variables in the expectation, namely: $X_2\ and\ s$. The formula seem to have been used ignoring the variable $s$. Can someone explain why this is and why it is allowed?
As the original question started out, if $X_i$ are iid $N(0,1)$ rvs, and $Y=X_1X_2$ then then (using the law of total expectation) the characteristic function $$\varphi_Y(s)= E[\exp(is X_1 X_2)] = E[\exp(-s^2X_2^2/2)]=\frac 1 {\sqrt{2\pi}}\int_{\mathbb R} e^{-x_2^2/2} e^{-s^2x_2^2/2}\,dx_2. $$ (Here the dummy variable of integration is $x_2$ because it corresponds to the rv $X_2$, but of course any symbol would do.) To finish the problem we only need to work out the integral.
If you make the change of variables $x_2=\frac 1{\sqrt{1+s^2}} u$, so the quadratic term in the exponent of the last integral becomes $-u^2/2$, the integral reduces to $$\varphi_Y(s)= \frac 1 {\sqrt{1+s^2}} \frac 1 {\sqrt{2\pi}}\int_{\mathbb R} e^{-u^2/2}\,du = \frac 1 {\sqrt{1+s^2}}.$$