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I have the following density function: $f(x,y) = \frac{2}{\pi}$ for $x^2 + y^2 \leq 1$ and $y > x$.

I figured out that this represents half of the unit circle (the upper half when cut along the line $y=x$). I would like to find the marginal density of Y. To do this, I understand that I will have to integrate over x, as the marginal density is usually $f_y(y) = \int{f(x)dx}$, which in this case would be $f_y(y) = \int{\frac{2}{\pi}dx}$. I am having trouble understanding how to find the values over which to integrate. Could someone please clarify this for me? I don't think I have a strong grasp on the intuition behind this.

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To find the (marginal) density of $Y$, we "integrate out" $x$. Draw the circle, and the line $y=x$. The geometry is everything. Your post shows that you are aware that the density function is $\frac{2}{\pi}$ in the part of the disk that is "above" the line $y=x$, and $0$ elsewhere.

First deal with negative $y$. The smallest negative value of $y$ is $-\frac{1}{\sqrt{2}}$. In that part of the world, $x$ travels from $-\sqrt{1-y^2}$ to $y$. So for $-\frac{1}{\sqrt{2}}\le y\lt 0$ we have $$f_Y(y)=\int_{x=-\sqrt{1-y^2}}^y \frac{2}{\pi} \,dx=\frac{2}{\pi}\left(y+\sqrt{1-y^2} \right).$$

Now look at $y\ge 0$. For $0\le y\lt \frac{1}{\sqrt{2}}$, the variable $x$ travels from $-\sqrt{1-y^2}$ to $y$: we get exactly the same expression as for negative $y$. So in fact we could have done it in one step for all $y$ with $\frac{1}{\sqrt{2}}\le y\lt \frac{1}{\sqrt{2}}$.

The geometry changes in the interval $\frac{1}{\sqrt{2}}\le y\le 1$. There $x$ travels from one side of the circle to the other. that is, from $-\sqrt{1-y^2}$ to $\sqrt{1-y^2}$. Thus in that interval we have $$f_Y(y)=\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{2}{pi}\,dx=\frac{4}{\pi}\sqrt{1-y^2}.$$

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  • $\begingroup$ Thank you! This makes so much more sense now. $\endgroup$ Oct 24, 2013 at 5:39
  • $\begingroup$ You are welcome. Many continuous bivariate distribution problems come down to getting the geometry right. Probability has many instances of cdf, pdf given by different formulas for different parts of the domain. $\endgroup$ Oct 24, 2013 at 5:43

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