I am quite new so correct me if I am doing anything wrong, and sorry for my inability to understand how LATEX works.
So my question is regarding the special case when $n=2$, the sample variance formula collapses down to the sum of $(X_1 - \bar{x})^2 + (X_2 - \bar{x})^2$ where $X_i$ follows the standard normal distribution, while $\bar{x}$ follows the normal distribution with mean $0$ and variance $1/2$.
When attempting this question, I thought about trying to first use the convolution formula to find the distribution of $Z=X-Y$, then do a transformation of $Q=Z^2$, and then lastly use convolution again for the sum of the resulting distribution.
However, when attempting the first step I arrived at an integral that is very apparently incorrect. What I arrived at was the integral from negative infinity to positive infinity of $f_x(x)*f_y(Z+X)dx$.
I arrived at this through the consideration of the plane I need to integrate, under $Z=X-Y$, one consider the plane $X-Y$, where $Y=X-Z$ and when considering the $P(X-Y<Z)$, one should integrate from negative infinity to infinity for $y$ and negative infinity to $Z+Y$, and through finding the double integral, and differentiatinf the inner integral, it resulted in the above formula. Which when tried to plug in the distribution, I could not result in a meaningful answer, I would like someone to point out to me where did I go wrong, as I agree that my method of obtaining $Z=X-Y$ seems dodgy.
Thanks a lot for reading, and sorry about the lack of mathematical notations, as I am really not familiar with how one uses it.
The first density (left) in your figure below (as linked in your comment) should correspond to the $\color{magenta}{\text{magenta}}$ term $$\int_{-\infty}^{\infty}f_X(x)\, \color{magenta}{f_Y(z+x)} \, \mathrm{d}x \qquad \text{where} \quad f_Y(t) = \frac1{\sqrt{2\pi}}e^{-\frac{ t^2 }2} \tag{1} \label{Eq01}$$
When you plugin $t = z+x$ in Eq.\eqref{Eq01}, the leading factor $\frac1{\sqrt{2\pi}}$ is not affected, neither is the $2$ in the denominator in the exponent.
Could you explain the "some reason" you said in the comment "For some reason my integral for the question is as so..."? Why is your left term involving $x+z$ scaled by $\sqrt{2}$ overall and by 2 in the exponent?
Note that Eq.\eqref{Eq01} is exactly what you described as (let me quote verbatim)
where your star $*$ is absolutely just a multiplication (and you didn't object to my use of central dot $\cdot$ in my comment). Your entire sentence (starting from "the integral") describes correctly the convolution $f_X \otimes f_Y$. There should be no way for you to have mistakenly viewed the $*$ as meaning yet another convolution $f_X \otimes f_{X+Z}$ ... I guess.