Determine probabilites from PDF

Question

Two random variables $X,Y$ have the following density:

$f_{X,Y}(x,y)=\begin{cases} x+y,\ \ (x,y)\in [0,1]\times [0,1]\\ 0,\ \ else\\ \end{cases}$.

I want to compute $P(X=1/2,Y=1/2), P(X=Y), P(X\le Y), P(X\le 1/2, Y\le 1/2)$.

It should be simple, but somehow I am stuck.

For $P(X\le 1/2, Y\le 1/2)$ I got $P(X\le 1/2, Y\le 1/2)=\int^{1/2}_0\int^{1/2}_0(x+y)\mathrm{d}x\mathrm{d}y= 1/8$

I conclude that $P(X=1/2,Y=1/2)$ is zero, as the measure of countably many points is zero. — EpsilonDelta, May 15 '17 at 11:50
Right, in general if you try to calculate the probability of a zero-measure event it will give you zero — caverac, May 15 '17 at 12:14
I have also computed $P(X\le Y)=\int^y_0\int^1_0 (x+y)\mathrm{d}y\mathrm{d}x=y/2(y+1)$ but I should obtain a number as result, not a function, so not sure what is wrong here. — EpsilonDelta, May 15 '17 at 12:17

score 0 · Answer 1 · answered May 15 '17 at 19:36

0

$$P(X = 1/2, Y=1/2)=0$$

$$P(X = Y)=0$$

$$P(X \leq Y)=\int_0^1\int_0^{x}(x + y)dydx=\frac{1}{2}$$

$$P(X \leq 1/2,Y \leq 1/2)=\int_0^{\frac{1}{2}}\int_0^{\frac{1}{2}}(x+y)dxdy =\frac{1}{8} $$

answered May 15 '17 at 19:36

PiE

I got the same, except $P(X=Y)$. Can u elaborate on $P(X=Y)$? Mathematica gives me a different answer for it, too. – EpsilonDelta May 16 '17 at 10:16
Because there are infinitely many possible values a continuous variable could take for $X$ and $Y$. There is zero chance of picking any exact values s.t. $X = Y$. – PiE May 16 '17 at 10:29

1 Answers1