# Question on calculating density function of a function of random variables [closed]

I began studying probability theory some time ago and now I've reached multiple random variables and function of random variables and I'm stuck on this particular problem. The issue here is, even though I've solved other problems, I've no idea how to approach the following:

Let $(X_1, X_2)$ be a random variable with PDF (probability density function):

$$f(x_1, x_2)=\frac{1}{2\pi}e^{\frac{x_1^2 + x_2^2}{2}} \in \mathbb{R}$$

and let $X1 = Y_1\cos{Y_2}$ and $X_2 = Y_1\sin{Y_2}$, with $Y_1\gt0$. Prove that $Y_1, Y_2$ are independent and calculate their PDFs.

## closed as off-topic by BruceET, Sahiba Arora, Did, Leucippus, JonMark PerryJan 13 '18 at 4:18

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• I guess you've made a typo, should be $X2=Y_1\sin{Y_2}$ – kludg Jan 12 '18 at 10:29
• Hint: because of rotational symmetry the joint PDF $f_{Y_1,Y_2}(y_1,y_2)$ does not depend on $y_2$ – kludg Jan 12 '18 at 10:49
• And perhaps another type in defining $f(x_1, x_2),$ which is not a density function as it stands. – BruceET Jan 12 '18 at 16:14
• @BruceET Pardon? – theSongbird Jan 12 '18 at 16:20

If you consider the function $f(x_1, x_2)=\frac{1}{2\pi}e^{-\frac{x_1^2 + x_2^2}{2}}$ on the plane $(x_1,x_2)$, it depends only on the distance from the center $y_1=\sqrt{x_1^2+x_2^2}$ . It is intuitively clear from here that the marginal PDF of $Y_2$ (polar angle) is uniform on $[0,2\pi]$ and $Y_1$ does not depend on $Y_2$.
The probability that $Y_1$ takes value in a small ring $(y_1,y_1+dy_1)$ where $y_1^2=x_1^2+x_2^2$ is equal to $$f_{Y_1}(y_1)dy_1=f(x_1,x_2)\cdot 2\pi y_1\cdot dy_1$$ so $$f_{Y_1}(y_1)=f(x_1,x_2)\cdot 2\pi y_1=y_1e^{-\frac{y_1^2}{2}}$$