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

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 $$f(x)$$?

• Your pdf isn't a pdf -- it doesn't integrate to 1. – Glen_b -Reinstate Monica Oct 23 '19 at 1:28
• @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 Oct 23 '19 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 Oct 23 '19 at 2:21
• @S.F.Yeh for getting your mathematics done See math.meta.stackexchange.com/questions/5020/… – Glen_b -Reinstate Monica Oct 23 '19 at 3:01

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.