# Probability of sum of 2 variables - Convolution

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:

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

So:

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

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?

• 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. – whuber Jan 9 at 20:59
• Plotting the event in question can help the intuition. Here is a plot created by Wolfram Alpha. – 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 $$f_{X,Y|X\neq 0,Y\neq0}(x,y)=1/72$$.