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Let X and Y have a joint uniform distribution on the region described by 0≤y≤1-$x^2$; -1≤x≤1.

Find E[X] and E[Y]

What I've tried

The graph will look like this. enter image description here

I know i need to find the marginal densities, and here's what i've tried

$f_X(x)$=$\int_{0}^{1-x^2}f(x,y)dy$

E[X]=$\int_{-1}^{1}$$xf_X(x)dx$

$f_Y(y)$=$\int_{-1}^{1}f(x,y)dx$

E[Y]=$\int_{0}^{1-x^2}$$yf_Y(y)dy$

however when I plug in $1-x^2$ as the density the answers for the expected values come out wrong. I'm using this because I assume the density to be used should be $y=1-x^2; -1≤x≤1$

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  • $\begingroup$ You can compute the expectations directly using the joint distribution, which is uniform. In order for the domain to have probability one, the uniform PDF should be constant: $f_{X,Y}(x,y) = 1/Area($domain$)$. $\endgroup$ Commented Sep 24, 2015 at 17:04
  • $\begingroup$ Is the joint function correct? are my limits of integration correct? Thank you for your help. $\endgroup$
    – Osuynonma
    Commented Sep 24, 2015 at 17:27

1 Answer 1

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HINT

The joint density is given: It is said that $X,Y$ are uniformly distributed over the parabolic area.

Since

$$\int_{-1}^1 1-x^2 \ dx=\frac43.$$

The joint density

$$f(x,y)= \begin{cases} \frac34,& \text{ if } -1 \le x \le 1 \text{ and } 0\le y \le 1-x^2\\ 0, &\text{ otherwise.} \end{cases} $$

So,

$$f_X(x)=\frac34\int_0^{1-x^2}\ dy=\frac34(1-x^2)$$ if $-1\le x\le 1.$ and

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