Computing the $f_{Y|X}=\frac{1}{2\sqrt{1-x^2}}$ over the unit circle

Question

Assume the random variable $(X,Y)$ is uniformly distributed on the disc $B=\{(x,y)\in\mathbb{R}^2:x^2+y^2\leqslant 1\}$ Determine the conditional distribution:

From a previous answer I have been able to compute the following density conditional distribution $f_{Y|X}=\frac{1}{2\sqrt{1-x^2}}$

I want to compute the cumulative conditional distribution. However I do not know how to define the boundaries for the double integral of $\frac{1}{2\sqrt{1-x^2}}$.

Question:

How should I solve the problem?

Thanks in advance!

Answer 1

Here is a simpler (but geometric) view of the problem.

• The joint pdf of $X$ and $Y$ can be thought of as defining a surface above the $x$-$y$ plane with the property that the volume trapped between the surface and the $x$-$y$ plane is $1$. What is the surface for your problem? If you think of the volume of the solid defined by $f_{X,Y}(x,y)$ and the $x$-$y$ plane as being $1$, what does the solid loo like in your problem?
• The value of the marginal pdf of $X$ at $x_0$ is found by integrating the function $f_{X,Y}(x_0,y)$ with respect to $y$ from $-\infty$ to $\infty$. Here, $x_0$ is a constant (say, $0.6$) and not a variable. Think of $f_{X,Y}(0.6,y)$ as the cross-section of the $f_{X,Y}(x,y)$ solid at $x=0.6$ so that $f_X(0.6)$ is given by the area of the cross-section being looked at. What does this cross-section look like for your solid?
• The conditional pdf of $Y$ given that $X$ equals $0.6$ is defined as $\displaystyle \frac{f_{X,Y}(0.6,y)}{f_X(0.6)}$. It is a function of $y$ (which occurs only in the numerator -- the denominator is just a constant. Well, you already know what $f_{X,Y}(0.6,y)$ looks like as a function of $y$ (you don't? go back to the previous paragraph) and so $f_{Y\mid X}(y\mid X = 0.6)$ is just a scaled version of the function $f_{X,Y}(0.6,y)$ with the scaling factor being $\displaystyle \frac 1{f_X(0.6)}$. For your question, what does $f_{X,Y}(0.6,y)$ look like as a function of $y$? You still didn't do the previously-suggested thought problem? Do it now.

Finally, if you do things right, you will come up with the answer that the conditional pdf is a uniform density on an interval that will be immediately obvious without any need for computations of any kind.