Let $X \sim \mathcal{N}(\mu_1, \sigma_1)$ and $Y \sim \mathcal{N}(\mu_2, \sigma_2)$ denote two independent normal random variables, then what is the PDF of $Z=(X+Y)^2-X^2=2XY+Y^2$?
The problem is hard due to that $XY$ and $Y^2$ are not independent, since they both depend on $Y$. So I decided to use the bivariate transformation technique to solve the problem.
Here is what I have tried so far. Since $X \sim \mathcal{N}(\mu_1, \sigma_1)$ and $Y \sim \mathcal{N}(\mu_2, \sigma_2)$ are independent, the joint PDF of (X, Y) is \begin{align} f_{X,Y}(x, y) = f_X(x)f_Y(y). \end{align} Now define $U=2XY+Y^2$ and $V=Y$, thus $X=\frac{U-V^2}{2V}$ and \begin{align} f_{U,V}(u, v) = f_{X,Y}(\frac{u-v^2}{2v}, v) = f_X(\frac{u-v^2}{2v})f_Y(v). \end{align} Therefore, we can compute the marginal as \begin{align} f_Z(z) =& f_U(u) \\ =& \int_{-\infty}^{\infty}f_{U,V}(u, v)dv \\ =& \int_{-\infty}^{\infty} f_X(\frac{u-v^2}{2v})f_Y(v) \\ =& \frac{1}{2\pi\sigma_1\sigma_2} \int_{-\infty}^{\infty} \exp\left(-\frac{\left(\frac{u-v^2}{2v} - \mu_1\right)^2}{2\sigma_1^2} -\frac{\left(v - \mu_2\right)^2}{2\sigma_2^2} \right) dv\\ =& \frac{1}{2\pi\sigma_1\sigma_2} \int_{-\infty}^{\infty} \exp\left(-\frac{\left(\frac{u}{2v}-\frac{v}{2} - \mu_1\right)^2}{2\sigma_1^2} -\frac{\left(v - \mu_2\right)^2}{2\sigma_2^2} \right) dv\\ =& \text{I can't sovle it}. \end{align}
As you can see, the integral is too hard and I have totally no idea how to solve it. Did I do something wrong? or it is ok but I just can't solve the integral. em...well
If you write $Z$ as $ \begin{bmatrix} X, Y \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 1 \\ \end{bmatrix} \begin{bmatrix} X \\ Y \\ \end{bmatrix} $, where $X$ and $Y$ are uncorrelated Normal variables, it fits the definition of Generalized Chi-Square.