Here's how I approached
Let X∼Norm(0,1) and Y∼Norm(0,1)
Then their joint pdf is
$$ f(x) = \dfrac{\mathrm{e}^{-\frac{x^2}{2}}}{\sqrt{{2\pi}}} * f(y) = \dfrac{\mathrm{e}^{-\frac{y^2}{2}}}{\sqrt{{2\pi}}}$$
$$ f(xy) = \dfrac{\mathrm{e}^{-\frac{x^2+y^2}{2}}}{{{2\pi}}}$$
I tried to rearrange them in the necessary format and compare
$$ f(xy) = \dfrac{\mathrm{e}^{-\frac{x^2+y^2}{2}}}{{{2\pi}}}$$
$$p{_\eta}x =\mathrm{e}^{{\eta}T(x)-A({\eta})}$$
I cannot seem to find $${\eta}$$ and $$A({\eta})$$
$$ f(xy) = \dfrac{\mathrm{e}^{-\frac{\left(x+y\right)^2-2yx}{2}}}{{{2\pi}}}$$
The first four cumulants of T = XY are 0, 1, 0, and 4.
I think $${\eta = -1} \quad and \quad A({\eta}) = 0$$ but that doesn't get me the answer. If someone can point me in the right direction it would help. Thank you.
It is not the joint density of $(X,Y)$ that we need, but the density of the product $XY$. This is given by $f_T(t) = \frac{K_0(|t|)}\pi$, where $K_0$ is the modified Bessel function of the second kind. We compute the moment generating function of $T$: $$ M_T(\theta) = \mathbb E[e^{\theta T}] = \int_{-\infty }^{\infty } \frac{e^{\theta t} K_0(\left| t\right| )}{\pi } \, dt = \frac{1}{\sqrt{1-\theta ^2}},\ \theta>1, $$ and hence the cumulant generating function $$ K_T(t) = \log M_T(t) = \log\left(\frac1{\sqrt{1-\theta^2}}\right) = -\frac{1}{2} \log \left(1-\theta ^2\right). $$ The $n^{\mathrm{th}}$ coefficient in the series for $K_T$ is $\frac{(-1)^n+1}{2 n}$, so taking $n=1,2,3,4$ you have your first four cumulants for $T$.