how to find domain of marginal pdf when its two variables domain are dependent

Question

I have a pdf $f(x,y)=1/π, 0< x^2+ y^2 <1$； 0, e.w.

Here, we can see $-\sqrt{1-x^2} < y < \sqrt{1-x^2}$

So, the marginal pdf of $X$ is $$\int_{-\sqrt{1-x^2}}^\sqrt{1-x^2} 1/πy \, dy\,.$$

and finally, I have $f(x)= 2/π \sqrt{1-x^2}$.

I do not know how to define the range of $f(x)$, can I present it in $y$ form like ：

$-\sqrt{1-y^2} < x < \sqrt{1-y^2}$?

Looks not solving anything. Can any one tell me how to find the domain of $f(x)$?

@Glen_b I add " 0, e.w" in the question. Is this a pdf now? or you mean my result of the marginal pdf of X is wrong? p.s. thanks for your edit. I'll learn how to show the mathematical formula next time. — S.F. Yeh, Commented Oct 23, 2019 at 1:47
There was a typo earlier where it is writtent ahat $f(x,y)=1/2$ rather than $1/\pi$. — Siong Thye Goh, Commented Oct 23, 2019 at 2:21
@S.F.Yeh for getting your mathematics done See math.meta.stackexchange.com/questions/5020/… — Glen_b, Commented Oct 23, 2019 at 3:01

Siong Thye Goh · Accepted Answer · 2019-10-23 02:14:20Z

1

The marginal distribution is $$\int_{-\sqrt{1-x^2}}^\sqrt{1-x^2} \frac1\pi\, dy=\frac{2\sqrt{1-x^2}}{\pi}.$$

We do not multiply $y$ in the integral.

The support would be from $-1$ to $1$. These are the $x$ values that the unit disk can take.

answered Oct 23, 2019 at 2:14

Siong Thye Goh

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$\begingroup$ This looks correct (so upvoted) but for what looks rather like routine bookwork/assignment type questions we usually give hints and guidance rather than explicit solutions. See the help center under Homework $\endgroup$
– Glen_b
Commented Oct 23, 2019 at 3:03
$\begingroup$ I did not relate it to the unit disk. thanks for your hint $\endgroup$
– S.F. Yeh
Commented Oct 23, 2019 at 5:23

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