# How to find the probability that the average of X and Y is less than 15?

Suppose that the length of time that Tom Cruise runs in a movie is a continuous value uniformly distributed between 10 and 20 minutes. Let X and Y be the lengths of the running of two randomly chosen Tom Cruise movies. X and Y are independent. Consider the following events:

A = {X < Y}

B = {The average of X and Y is less than 15}

(a) (5 points) Show that events A and B are independent. Include a diagram of the sample space and the events to support your answer.

I don't need help drawing the diagram of the sample space/events. I am mostly confused about finding the probability of B and proving independence when the probability is continuous.

I am trying to prove A and B independent. I understand that for 2 events to be independent, P(A∩B)=P(A)P(B).

I found P(A) to be $\frac{1}{2}$.

So, B is $P( \frac{X+Y}{2} <15)=P(X+Y-30<0)$

I am confused on how to find the probability of B from there... I know X and Y are uniformly distributed and independent but I'm kind of confused on how to turn what I have into that form as seen here: let x and y be uniformly distributed independent random variables on [0 ,1].the probability that the distance between x and y is less than 1/2 is?

I'm also confused on how to find A∩B for continuous values, as in this situation.

• You can do this geometrically. Graph the line $X+Y=30$ in the rectangle that gives all possible values of $(X,Y)$. Compute the portion of the relevant area. – lulu Sep 15 '18 at 19:07

The density of the sum $X+Y$ of two independent random variables $X,Y$ is a convolution of the densities of $X$ and $Y$. The density of the uniform distribution on $[10,20]$ is $0.1\chi_{[10,20]}$, where $\chi_A$ stands for a characteristic (i.e. an indicator) function of a set $A$. Hence we have $$f_{X+Y}(x)=\int_{-\infty}^{\infty}0.01\chi_{[10,20]}(x-t)\chi_{[10,20]}(t)\,\text{d}t.$$ Your job is to compute the integral $$P(B)=\int_{-\infty}^{30}f_{X+Y}(t)\,\text{d}t.$$
Since the characteristic function does vanish outside $[10,20]$, we have $$f_{X+Y}(x)=\int_{10}^{20}0.01\chi_{[10,20]}(x-t)\,\text{d}t.$$
A simple computation proves that $$f_{X+Y}(x)=\begin{cases}0.01(x-20)&\text{for }20\le x\le 30\\ 0.01(40-x)&\text{for }30<x\le 40\\ 0&\text{otherwise}\end{cases}$$ Then out integral is rather simple to compute. $$P(B)=\int_{-\infty}^{30}f_{X+Y}(t)\,\text{d}t=\int_{20}^{30}f_{X+Y}(t)\,\text{d}t=0.5$$ and this is the probability we are looking for.