How to get probability of $X \le Y + Z$ for $X, Y, Z$ all from the same specific distribution?

Question

I'm struggling to get the probability of $\mathbb{P}[X\le Y+Z]$ where all three are from the same distribution with the following pdf:

$$f_X (x) = \frac{2(x-a)}{(b-a)^2}$$ where $0<a<b$. (Notice the pdf above comes from taking the maximum of two random variables uniformly distributed between $a,b$).

However, I know that I need to use convolution to compute $\mathbb{P}[X\le Y+Z]$ where $X,Y,Z$ are i.i.d. But I admit that I'm stuck on this convolution step. How do I convolve these and get $\mathbb{P}[X\le Y+Z]$?

Answer 1

Edit: Updated the answer to the new question.

To expand on my answer in the comments, you can check my Mathematica notebook here: https://www.wolframcloud.com/obj/d4a194bf-fb09-4ce3-abe6-4238d62e53e0

If $$p(x) = \frac{2(x-a)}{(a-b)^2} * \mathbb1[a<x<b]$$

You get the convolution (and pdf for the sum) $$ q(y)=\int_{-\infty}^\infty p(x)p(y-x)dx = \begin{cases} -\frac{2 (2 a-y)^3}{3 (a-b)^4)} & a<b, 2 a<y, a+b\ge y\\ \frac{2 (2 b-y) (6 a^2-2 b^2-6 a y+2 b y+y^2)}{3 (a-b)^4} & a<b, a+b<y, 2 b>y. \end{cases} $$ Just using the formula from the Wikipedia page says. The only difficulty is getting all the cases of the integration right, but Mathematica helps with that.

Answer 2

This is a problem that is conceptually easy, but computationally messy, and best solved with the help of a computer algebra system. As you have noted, the solution must obviously be a function of parameters $a$ and $b$.

Given that $X$, $Y$ and $Z$ are independent random variables, their joint pdf $f(x,y,z)$ is simply the product of the individual pdf's:

We seek $P(X < Y+Z)$:

where I am using the Prob function from the mathStatica package for Mathematica, where as full disclosure, I note that I am one of the authors. All done.

Note that this probability is always close to 1. Here is a plot of the probability, with arbitrarily $a = 1$, as $b$ increases:

It appears the smallest the probability can be is $\frac{17}{18}$:

Answer 3

$$P(X \leq Y+Z) = \int_{a}^b\int_{a}^b F(y+z) f(y) f(z) \ dy \ dz$$

$$F(x) = \int_a^x f(y) dy = \int_a^x \frac{2(y-a)}{(b-a)^2} dy = \frac{2(\frac{x^2-a^2}{2} - a(x-a))}{(b-a)^2} = \frac{(x-a)^2}{(b-a)^2}$$

Now $$P(X \leq Y+Z) = \int_{a}^b\int_{a}^b F(y+z) f(y) f(z)$$

is an integration of polynomials in $y,z$. Try to take it from here on.