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I would like to find the characteristic function of the product of two independent brownian motions. This boils down to the characteristic function of the product of two normal random variables. This is known to follow the product-normal distribution:

http://mathworld.wolfram.com/NormalProductDistribution.html

but I haven't found an easy proof of this fact. This is proven making use of the Mellin transform, but this questions was in an assignment of a considerably lower level and no mention of the Mellin transform in the lectures either. I would be really happy if someone could help me.

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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.

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