If there are three identical random variable X,Y and Z, find the probability of P(X+Y>Z).

Question

I was trying to solve from the simplest case of uniform distribution and try to convolute the three distributions. I am new to this area and would like to ask: firstly, how to answer the question itself. secondly, if I define the uniform distribution as a piecewise function: f(x)= 1(0=x<=1), how shall I convolute according to the question (X+Y-Z).I was thinking firstly convolute the function upon itself and then convolute with the inverse of itself f(x)=-1(0<=x<=1). But I got stuck in the second step.

Answer 1

I'm writing this assuming that the answer to my question above (are the variables independent?) is yes. I propose that the clearest way of solving this isn't to start with convolutions, but rather to let them drop out from some other integration.

Since $X, Y, Z$ are all uniform on $[0,1]$ and are independent, we can easily compute their joint density: it's the product of the individual densities, so it's $1$ on the cube $[0, 1] \times [0,1] \times [0,1]$ and $0$ elsewhere.

With that in mind, to find any probability, you integrate the density function over the desired region. In this case, that's the region of $[0,1]^3$ for which $x + y > z$: $$\mathbb P(X + Y > Z) = \int_0^1 \int_0^1 \int_0^{\min\{x + y, 1\}} 1 \, \textrm d z \, \textrm d y \, \textrm d x$$ This sheds some light on why the convolution approach isn't going to be as easy here, which is because the geometry of the cube forces that awkward $\min$ condition on the upper limit of the inner integral.

Note that in all cases, writing the integrals in this way forces the convolutions to drop out naturally. This seems easier to me, but your mileage may vary.