1
$\begingroup$

Suppose that $X$ and $Y$ are uniformly distributed on $0\lt |x| + |y| \lt 1$. How do I find $P(Y\gt 1/4 ~\vert ~X=1/2)?$

  • 1
    Kolmogorov says: "the concept of a conditional probability with regard to an isolated hypothesis whose probability equals 0 is inadmissible." You could interpret your conditional probability at the limit of the probabilities obtained by conditioning with respect to the events |X-\frac{1}{2}|<\frac{1}{n}, but proceed with caution!2012-01-05

2 Answers 2

1

The intuition says that the answer is $1/4$. Computation is really unnecessary, but for the sake of more complicated situations let us do it.

We are told, I think, that the pair $(X,Y)$ is uniformly distributed on the square with corners $(1,0)$, $(0,1)$, $(-1,0)$, and $(0,-1)$. This square has area $2$, so our joint density function $f_{X,Y}(x,y)$ is $1/2$ inside the square and $0$ elsewhere.

First we find the marginal density $f_X(x)$. This is $\int_y f_{X,Y}(x,y)\,dy.$ In our case (for positive $0\le x< 1$), we are integrating from $y=x-1$ to $y=1-x$, and we get $1-x$. (More precisely, $f_X(x)=1-|x|$ on the interval $|x|<1$).

Next we find the conditional density of $Y$ given $X=x$, that is, $f_Y(y|X=x)$. To do this, divide the joint density by the marginal density. Take in particular $x=1/2$. Since $f_X(1/2)=1/2$, we have $f_Y(y|X=1/2)=1$ for $-1/2.

Finally, for our answer, "integrate" the constant function $1$ from $y=1/4$ to $y=1/2$. We get $1/4$. (The process would be more interesting with a problem in which the answer is not intuitively clear.)

  • 0
    Again one could take $x=1/2$. The conditional density is still easy to compute, using the same procedure (it is linear), but the answer is no longer geometrically clear.2012-01-05
0

Ben Derrett's comment allows conditioning here so agreeing with yoyo,

if $Y$ is uniformly distributed such that $\frac{1}{2}+|Y| \lt 1$,

then $Y$ is uniformly distributed on $(-\frac{1}{2},\frac{1}{2})$,

and so $\Pr(Y\gt \frac{1}{4}|X=\frac{1}{2}) = \frac{1}{4}$.

  • 0
    For the given joint density, is $Y$ really uniformly distributed on $\left(-\frac{1}{2},\frac{1}{2}\right)$? This is the _conditional_ density of $Y$ given $X=\frac{1}{2}$, right?2012-01-05