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 $x_1+x_2+x_3\geq 1$ 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 $z\leq t+1$ 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<t\leq z $. 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 $0.4166\ldots$
I ran a Python simulation of this to see if it was about right, but ended up getting around $\frac{1}{3} = 0.33\ldots$. 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.
Best Answer
You approach of finding the pdf for the sums is valid. It is unnecessarily complicated. We can simply do it this way:
$\begin{align}\mathsf P(X_1+X_2<1)~&=~\int_0^1\mathsf P(X_1<1-t) f_{X_2}(t)\,\mathrm d t\\[1ex]&=~\int_0^1\int_0^{1-t} 1\,\mathrm d s\,\mathrm d t\\[1ex]&=1/2\\[2ex]\mathsf P(X_1+X_2+X_3 > 1) ~&=1-\int_0^1\mathsf P(X_1+X_2<1-u)\,f_{X_3}(u)\,\mathrm d u\\&=1-\int_0^1\int_0^{1-u}\int_0^{1-u-t} 1\,\mathrm d s\,\mathrm d t\,\mathrm d u\\[1ex]&=5/6\end{align}$
Now the joint probability is not the product of these probabilities. The random variables $(X_1+X_2)$ and $(X_1+X_2+X_3)$ are not independent. Notice that the latter must be at least as great as the former; this means they cannot be independent.
Rather, use:
$$\small\begin{align}\mathsf P(X_1+X_2<1, X_1+X_2+X_3>1) &=\mathsf P(1-X_3<X_1+X_2<1) \\[1ex]&=\int_0^1 \mathsf P(1-u<X_1+X_2<1)\,f_{X_3}(u)\,\mathrm d u\\[1ex]&=\int_0^1 \mathsf P(X_1+X_2<1)-\mathsf P(X_1+X_2<1-u)\,\mathrm d u\\[1ex]&~~\vdots\end{align}$$
Give it a shot.