1
$\begingroup$

How to find $P(X_1 + X_2 + X_3 + X_4 \geq 3)$ for uniformly distributed independent random variables $X_1$, $X_2$, $X_3$, $X_4\sim U(0,1)$?

It follows from independence that their cumulative density function is 1, but I'm struggling with integration space.

$\endgroup$
  • 1
    $\begingroup$ (Big) Hint: By symmetry, this is equal to $$P(X_1+X_2+X_3+X_4\le1)$$ Now, compute the latter... $\endgroup$ – Did Dec 4 '18 at 15:23
  • $\begingroup$ @Did I can't figure that out. Looks like it gets us to equation $$F(1) = 1 - F(3)$$ but I can't recall any formal symmetry connected with that, $\endgroup$ – DeuzharNickens Dec 4 '18 at 17:07
  • 1
    $\begingroup$ Sub-hint: $(1-X_1,1-X_2,1-X_3,1-X_4)$ is distributed like $(X_1,X_2,X_3,X_4)$. $\endgroup$ – Did Dec 4 '18 at 17:36
  • $\begingroup$ @Did Oh I see, thanks! $\endgroup$ – DeuzharNickens Dec 4 '18 at 17:54
1
$\begingroup$

Once we get to the step proposed by @Did, we can obtain the solution easily using geometric probability. Our probability here would be the hyper-volume covered by $$x_1+x_2+x_3+x_4 \leq 1 \text{ and } 0\leq x_i \leq 1$$ is exactly the hyper-volume of an $4$-dimensional simplex, which is $\dfrac{1}{4!} = \dfrac{1}{24}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.