Marginal density functions

Asked Mar 10 '15 at 19:45

Active Mar 10 '15 at 20:04

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Given a disc with center at origin and radius one, where

$$f(x,y) = 1/\pi$$

$$\sqrt{x^2+y^2} =1$$

the marginal density function of $X$ is $$2/π * \sqrt{1-x^2}$$

Can the marginal density function of $Y$ be extrapolated without calculation and taken to be the same over $x=1$ and $x=-1$, or is it equal to $2/π$ ?

edited Mar 10 '15 at 20:04

asked Mar 10 '15 at 19:45

Edward

Shouldn't $\sqrt{x^2+y^2}=1$? – Ben Lansdell Mar 10 '15 at 19:48
By symmetry of the disk, X and Y have the same PDF. – Did Mar 10 '15 at 20:07
but is it still correct to say that the marginal density function is equal to 2/pi? – Edward Mar 10 '15 at 20:08
I'm not sure I understand why it would be equal to $2/\pi$? – Ben Lansdell Mar 11 '15 at 21:32

Marginal density functions

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