Characteristic function of product of normal random variables

Solution 1:

Assume that $X$ and $Y$ are two independent standard normal random variables and let us compute the characteristic function of $XY$. One knows that $\mathrm E(\exp(\mathrm itX))=\exp(-\frac12t^2)$ hence $\mathrm E(\exp(\mathrm itXY)\mid Y)=\exp(-\frac12t^2Y^2)$ and $\mathrm E(\exp(\mathrm itXY))=\mathrm E(\exp(-\frac12t^2Y^2))$. Now, $$ \mathrm E(\exp(-{\textstyle\frac12}t^2Y^2))=\frac1{\sqrt{2\pi}}\int_{\mathbb R}\mathrm e^{-\frac12t^2y^2}\mathrm e^{-\frac12y^2}\mathrm dy=\sqrt{\sigma^2}, $$ where $t^2+1=1/\sigma^2$. This proves $$ \mathrm E(\exp(\mathrm itXY))=\frac1{\sqrt{1+t^2}}. $$ Edit The OP mentions the characteristic function of the product of two independent brownian motions, say the processes $(X_t)_t$ and $(Y_t)_t$. The above yields the distribution of $Z_1=X_1Y_1$ and, by homogeneity, of $Z_t=X_tY_t$ which is distributed like $tZ_1$ for every $t\geqslant0$. However this does not determine the distribution of the process $(Z_t)_t$. For example, to compute the distribution of $(Z_t,Z_s)$ for $0\leqslant t\leqslant s$, one could write the increment $Z_s-Z_t$ as $Z_s-Z_t=X_t(Y_s-Y_t)+Y_t(X_s-X_t)+(X_s-X_t)(Y_s-Y_t)$ but the terms $X_t(Y_s-Y_t)$ and $Y_t(X_s-X_t)$ show that $Z_s-Z_t$ is not independent on $Z_t$ and that $(Z_t)_t$ is probably not Markov.