# probability – Question About Sum of \$3\$ Uniform Independent Random Variables and Convolution

I’m currently working on a problem where I add $$3$$ uniform independent random variables $$x_1, x_2, x_3$$, each from $$(0,1)$$, and find the probability that $$x_1+x_2+x_3geq 1$$ and $$x_1+x_2 < 1$$.

I approached it by finding the pdf $$f_{x_1+x_2}(z)$$ by convolution and then treating $$x_1+x_2$$ like a random variable from $$(0,2)$$ and using convolution to find the pdf $$f_{(x_1+x_2)+x_3}(z)$$.

There are many posts that establish
$$f_{x_1+x_2}(z) = begin{cases}0 & z< 0 \ z & 0 leq z leq 1 \ 2-z & 1 < z leq 2 \ 0 & z> 2 end{cases}$$
And I have worked that math out and am fine with the reasoning.

Then I do the following.
$$f_{(x_1+x_2)+x_3}(z) = int_{-infty}^{infty} f_{x_1+x_2}(t)f_{x_3}(z-t)dt$$
Since I want $$P(x_1+x_2+x_3 geq 1) = 1 – P(x_1+x_2+x_3 < 1)$$, all I need is the case where $$0 leq z leq 1$$. Well, $$f_{x_1+x_2}(t)$$ is nonzero when $$t in (0,2)$$, and $$f_{x_3}(z-t)$$ is nonzero when $$0 leq z-t leq 1$$ or $$t leq z leq t+1$$. We know $$zleq t+1$$ is redundant since $$z leq 1$$ and $$t geq 0$$, so by combining conditions we have $$f_{x_1+x_2}(t)f_{x_3}(z-t)$$ is nonzero iff $$0. In fact, since $$0 < t leq z leq 1$$, $$f_{x_1+x_2}(t) = t$$. And, since $$0 leq z-t leq 1$$, $$f_{x_3}(z-t) = 1$$. Thus,
$$f_{(x_1+x_2)+x_3}(z) = int_{0}^{z} t dt = frac{z^2}{2} text{ for } 0 leq z leq 1$$
Also,
$$P(x_1+x_2+x_3 < 1) = int_0^1 f_{(x_1+x_2)+x_3}(z) dz = int_0^1 frac{z^2}{2} dz = frac{1}{6}$$
This implies $$P(x_1+x_2+x_3 geq 1) = frac{5}{6}$$. To find the probability that their sum is greater than or equal to $$1$$ and the sum of $$x_1+x_2$$ is not, we multiply the probabilities. Using $$P(x_1+x_2 < 1) = int_0^1 f_{x_1+x_2}(z)dz = int_0^1 z dz = frac{1}{2}$$, we should have that this probability is $$frac{5}{6}*frac{1}{2} = frac{5}{12}$$ or about $$0.4166ldots$$

I ran a Python simulation of this to see if it was about right, but ended up getting around $$frac{1}{3} = 0.33ldots$$. I’m confident the simulation is right, but I can link or add the code if that would be helpful. I believe that I possibly went wrong in treating $$x_1+x_2$$ like a random variable; is there some condition to convolution for finding pdfs that I violated? Or is it simply an error in the reasoning?

Thank you for all help, I really appreciate it.