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I have two joint PDF-s and three questions below.

I have a random vector' $(U,V)'$ and it's PDF. I need to find the marginal PDF-s for both $U$ and $V$. I tried deploying the idea from this anwer, but I still have some questions.

This is the joint PDF:

$$ f_{U,V}(u,v)=\begin{cases} \frac{12}{4\pi-3\sqrt{3}}, \hspace{2mm} |2u|\leq \sqrt{3}, \hspace{2mm} 1\leq 2v\leq 2\sqrt{1-u^2}\\ 0 \hspace{12mm} \text{elsewhere} \end{cases} $$

The biggest problem for me is finding the support i.e the bounds for integration.

I think I have correctly found the marginal $f_U(u):$ $$f_{U}(u)= \int_{v=\frac{1}{2}}^{v=\sqrt{1-u^2}}f(u,v)dv= \frac{12}{4\pi-3\sqrt{3}} \int_{v=\frac{1}{2}}^{v=\sqrt{1-u^2}} 1dv= \frac{12\sqrt{1-u^2}-6}{4\pi-3\sqrt{3}} $$

The problem appears with $f_V(v)$. From the PDF plot below we can see that the area is symmetric around $0$. Therefore we can find integral with respect to $u$ from $0$ until the curved line above it, and the whole area would be twice the value?

  1. Is it correct that I can only look at the upper bound: $2v\leq 2\sqrt{1-u^2} \implies v^2\leq 1-u^2 \implies u^2 \leq 1-v^2 \implies u\leq \sqrt{1-v^2} $

and now

$$ f_V(v) = 2\int_{u=0}^{u=\sqrt{1-v^2}}f(u,v)du= 2\frac{12}{4\pi-3\sqrt{3}} \int_{u=0}^{u=\sqrt{1-v^2}} 1du= \frac{24\sqrt{1-u^2}}{4\pi-3\sqrt{3}} $$

  1. Is it correct that for the expected values the integration areas are now $EX=\int_{-\frac{\sqrt3}{2}}^{\frac{\sqrt3}{2}} xf_X(x)$ as well as $EY=\int_{\frac{1}{2}}^1 yf_Y(Y)$?

  2. I have another example where the PDF is as $f_{X,Y}(x,y)=\begin{cases} \frac{3}{10}xy, \hspace{2mm} 0<x<2, \hspace{2mm} 0<y <\sqrt{x(4-x)}\\ 0 \hspace{12mm} \text{elsewhere} \end{cases}$

and for the $f_Y(y)$ I can find integration area as looking at the upper bound again, because lower is $0$. So $y = \sqrt{x(4-x)} \implies y^2 = 4x-x^2 \implies y^2 + 4 = 4x-x^2 +4 \implies (x^2-4x+4) + y^2 = 4 \implies (x-2)^2=4-y^2 \\ \implies x-2 =\sqrt{4-y^2} \implies x = \sqrt{4-y^2}+2$. Now, I would integrate $f_Y(y)=\int_0^{\sqrt{4-y^2}+2}f_{X,Y}(x,y)dx$, but I already can see that $x$ should be maximum $2$ so the answer is incorrect. My textbook says is should be integrated as $f_Y(y)=\int_{2-\sqrt{4-y^2}}^2f_{X,Y}(x,y)dx$. Where am I going wrong with this one?

Thanks a bunch!

graph

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  • $\begingroup$ Oh my, I forgot a $\pi$.... This doesn't change anything really, but will update $\endgroup$
    – statistic
    Jan 25, 2022 at 14:27

1 Answer 1

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Your work for $1$ and $2$ is mostly correct except some typos.

$ \displaystyle f_U(u) = \frac{12\sqrt{1-u^2}-6}{4\pi-3\sqrt{3}}, ~2 |u| \leq \sqrt3$

Also, $ \displaystyle f_V(v) = \frac{24\sqrt{1-v^2}}{4\pi-3\sqrt{3}}, 1/2 \leq v \leq 1$

So, $ \displaystyle E[U] = \int_{-\sqrt3/2}^{\sqrt3/2} u \cdot f_U(u) ~ du~,$ and similarly for $E[V]$

For the third question, the below diagram may help understand the limits of integration to obtain marginal of $Y$.

enter image description here

Note that $~(x-2)^2 \leq 4-y^2$ $$\implies 2 - \sqrt{4-y^2} \leq x \leq 2 + \sqrt{4-y^2}$$

But the support of the given pdf has $x \leq 2$. So for a given $y$, $2 - \sqrt{4-y^2} \leq x \leq 2$

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