# Joint density of $\chi^2$ distribution and t distribution from a normal distribution N(0,$\sigma^2$)

Let $$X_1$$ and $$X_2$$ be independent N(0,$$\sigma^2$$) random variables. Find the joint $$Y_1 = X_1^2 + X_2^2$$ and $$Y_2 = X_1/\sqrt{X_1^2 + X_2^2}$$. Show that they are independent.

Attempt: Step 1: Find $$x_1 = g(y_1,y_2) = y_2\sqrt{y_1}$$ and $$x_2 = h(y_1,y_2) = \sqrt{y_1(1-y_2^2)}$$

Step 2: Apply the Jacobian Transformation to get $$|J| = \frac{1}{2\sqrt{1-y_2^2}}$$

Step 3: Plug in into $$f_{X_1,X_2}(x_1,x_2)$$ to get $$f(y_1, y_2) = \frac{e^{\frac{y_1}{2\sigma^2}}}{4\pi\sigma^2\sqrt{1-y_2^2}}$$

But I am not sure whether my calculation for Jacobian is correct or not, since getting the $$g(y_1,y_2)$$ and $$h(y_1,y_2)$$ involving the sqrt calculation. I don't know how to deal with it there.

You are quite close. Fixing the missing minus sign in the exponent, we have $$f(y_1, y_2) = \frac{1}{2 \sigma^2} e^{-\frac{y_1}{2\sigma^2}} \cdot \frac{1}{2\pi \sqrt{1-y_2^2}}$$ where $$y_2 \in (-1, 1)$$ and $$y_1 > 0$$. Because this joint density can be written as $$g(y_1) h(y_2)$$, you have shown $$Y_1$$ and $$Y_2$$ are independent. You can check that the first factor is the distribution of $$X_1^2+X_2^2$$ by referring to the density of a chi-squared distribution of degree $$2$$ (scaled by $$\sigma$$). I'm not sure what the name of the distribution of $$Y_2$$ is, but it can't be a $$t$$-distribution since $$-1 \le Y_2 \le 1$$.