Let the random vector $(x,y)$ be uniformly distributed within the unit circle $B = \{ (x,y) : x^2 + y^2 \leq 1 \}$. Then the probability density function is
$$ f(x,y) = \begin{cases} \frac{1}{\pi} \quad &(x,y) \in B \\ 0 \quad &\text{otherwise} \end{cases}$$
Now, the marginal distributions are
$$\tilde{f}(x) = \int_{\mathbb{R}} f(x,y) \, dy = \frac{2}{\pi} \sqrt{1-x^2}$$
and similar for $\tilde{f}(y)$.
Question is:
Let $y = y_0 > 0$ be fixed. What is the probability distribution function for the random variables $x^4$, $|x|$ and $-x$?
Any ideas?
Best Answer
When they say:"Let $y=y_0>0$ be fixed" it is evident to me that they are interested in the conditional distribution as fixing y the support of the marginal $X$ changes.
The problem is trivial because the conditional distribution $X|Y=y$ is uniform.
$$f_{X|Y=y_0}(t)=\frac{1}{2\sqrt{1-y_0^2}}\mathbb{1}_{\Big[-\sqrt{1-y_0^2};\sqrt{1-y_0^2}\Big]}(t)$$
Clarified this, I think you will have no problems to continue by your own
Example 1:
In the case $Z=|X|y_0|$ the distribution function is
$$F_Z(z)=F_{X|Y}(z)-F_{X|Y}(-z)=\frac{z}{\sqrt{1-y_0^2}}$$
...better
$$F_Z(z)=\frac{z}{\sqrt{1-y_0^2}}\mathbb{1}_{\Big[0;\sqrt{1-y_0^2}\Big]}(z)+\mathbb{1}_{\Big[\sqrt{1-y_0^2};+\infty\Big)}(z)$$
(again an Uniform Distribution)
Example 2
$Z=X^4$
$$F_Z(z)=\mathbb{P}[X^4\leq z]=\mathbb{P}\Big[- z^{\frac{1}{4}}\leq X\leq z^{\frac{1}{4}}\Big]=F_X\Big(z^{\frac{1}{4}}\Big)-F_X\Big(-z^{\frac{1}{4}}\Big)$$