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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?

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  • $\begingroup$ It's not clear what you mean by "let $y$ be fixed". Do you mean the conditional probability? $\endgroup$ Commented Oct 29, 2020 at 14:23
  • $\begingroup$ Good question. The question is posed as above, but I guess the conditional probability is not what is meant here. I think we are looking at the problem from the moment $y$ was fixed and now try to find out what the probbility distribution for the variables $x$, $x^4$, etc. are. $\endgroup$ Commented Oct 29, 2020 at 14:29

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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.

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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

$Z=X^4$

$$F_Z(z)=\mathbb{P}[X^4\leq z]=\mathbb{P}\Big[- z^{\frac{1}{4}}\leq X\leq z^{\frac{1}{4}}\Big]=F_X\Big(z^{\frac{1}{4}}\Big)-F_X\Big(-z^{\frac{1}{4}}\Big)$$

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  • $\begingroup$ In this case I have to calculate: $$ f(x|y=y_0) = \frac{f(x,y)}{\tilde{f}(y)}$$, which is $$f(x|y=y_0) = \frac{1}{2\sqrt{1-y_0^2}}$$. But this term is $x$-independent. Wouldn't that mean, that the random variables $x^4$, $|x|$ and $-x$ have a density function of zero everywhere? $\endgroup$ Commented Oct 29, 2020 at 15:42
  • $\begingroup$ @Octavius : in the meantime you were writing the comment I edited my answer with more details. The fact that in your conditional density there is no $X$ is better for you....it's a Uniform density in the support that I showed. The X will appear as soon as you calculate the CDF $\endgroup$ Commented Oct 29, 2020 at 15:50
  • $\begingroup$ I see, so my CDF is $$F(x) = \frac{x}{2\sqrt{1-y_0^2}} + \frac{1}{2} \quad .$$ But how do I get the CDF for the random variables $x^4$, $|x|$, $-x$? The "transformation law" for the densities would be $$g(\tilde{x}) = \Big| \frac{\mathrm{d}h^{-1}}{\mathrm{d}\tilde{x}} \Big| \cdot f\big(h^{-1}(\tilde{x})\big)$$ where $f(x)$ is the density of the variable $X$, and $g(\tilde{x})$ is the density of the variable $\tilde{X} = h(X)$. With constant density as written above I would have density zero for the other variables. $\endgroup$ Commented Oct 29, 2020 at 16:05
  • $\begingroup$ @Octavius : ahah ... I edited the answer again with more and more details...your formula is correct but with non monotonic transformation it must be updated. in general you can pass via CDF...it is the general method. I wrote down the case of $|X|$. Reading the text you posted, they ask you the distribution function, thus my method is faster. $\endgroup$ Commented Oct 29, 2020 at 16:15
  • $\begingroup$ Understood. Thank you very much. $\endgroup$ Commented Oct 29, 2020 at 17:55

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