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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.

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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 $-1 \le Y_2 \le 1$.

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