Let say $A$ and $B$ are two uniform random variables independent over $[0,10]$ and:

$X = max(A-1, 0)$

$Y = max(B-2, 0)$

So that $X$ and $Y$ have their density function respectively:

$F_{X}(x) = \frac{x+1}{10}$, $0 \le x \le 9$

$F_{Y}(y) = \frac{y+2}{10}$, $0 \le y \le 8$

I would like to calculate the probability that $X+Y \le 5$.

So here is my solution:

$f_{X}(x) = f_{Y}(y)=\frac{1}{10}$.


$P(X+Y\le 5) =E_{Y}[P(X \le 5 -y)|Y=y]=E_{Y}[P(X \le 5 -y)]=\int_{0}^{5}\frac{6-y}{10}\frac{1}{10}dy=\frac{7}{40}$

But the correct answer should be $\frac{59}{200}$

I think that there is an error in my solution but I could not find it out. Can you take a look at it and explain to me?

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    $\begingroup$ You erred at the very beginning: neither $X$ nor $Y$ have density functions. When a random variable $U$ has a density, then $\Pr(U=u)=0$ for all numbers $u.$ But $\Pr(X=0)=1/10$ and $\Pr(Y=0)=2/10$ contradict that. $\endgroup$ – whuber Jan 9 at 20:59
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    $\begingroup$ Plotting the event in question can help the intuition. Here is a plot created by Wolfram Alpha. $\endgroup$ – whuber Jan 9 at 23:08

It's wrong because you ignored the cases where $X=0$ and/or $Y=0$. That's why $f_X(x)\neq 1/10$ and $X$ is not continuous. A way to calculate is using total probability theorem: $$P(A)=P(A|X=0,Y=0)P(X=0,Y=0)+...+P(A|X\neq0,Y\neq 0)P(X\neq 0,Y\neq 0)$$

Where $A$ is defined as the event $X+Y\leq 5$. The last case can be calculated easily via a 2d sketch of conditional distributions on $[0,9]\times [0,8]$ with $f_{X,Y|X\neq 0,Y\neq0}(x,y)=1/72$.


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