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Assume that $X,Y$ and $Z \sim U(0,1)$. Assume further $X,Y$ and $Z$ are mutually independent.
What is $P(X+Y+Z>1)$?

I've written down the density function of $X,Y$ and $Z$. I've drawn a picture. I understand the range of a new variable $U=X+Y+Z$ should be between $0$ and $3$. But I'm stuck on how to find the distribution of $U$.

Kian
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  • I've written down the density function of $X$, $Y$ and $Z$. I've drawn a picture. I understand the range of a new variable $U = X + Y + Z$ should be between 0 and 3. But I'm stuck on how to find the distribution of $U$. – Kian Mar 12 '15 at 14:43
  • Oh, do you know $f_{X,Y,Z} = f_X f_Y f_Z$? – AlexR Mar 12 '15 at 14:45
  • https://math.stackexchange.com/questions/769545/volume-of-t-n-x-i-ge0x-1-cdotsx-n-le1/ – StubbornAtom Nov 26 '19 at 18:05

3 Answers3

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The PDF of $U(0,1)$ is $f(x) = \chi_{[0,1]}(x)$. In other words, $P(X\le a) = a$ if $a\in[0,1]$ and $0$ else. Now $$P(X+Y+Z > 1) = 1 - P(X+Y+Z \le 1)$$ What do you know about the joint PDF of iid random variables?

We know $f_{X,Y,Z}(x,y,z) = f_X(x) f_Y(y) f_Z(z)$ so in this case we have $$P(U \le a) = \int_0^a \int_0^{a-x} \int_0^{a-x-y} 1\cdot 1\cdot 1 \ \mathrm dz \ \mathrm dy \ \mathrm dx$$ Can you take it from here?

AlexR
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  • I am struggling to see how you use $f_{X,Y,Z}(x,y,z) = f_X(x) f_Y(y) f_Z(z)$ to get $P(U \le a) = \int_0^a \int_0^{a-x} \int_0^{a-x-y} 1\cdot 1\cdot 1 \ \mathrm dz \ \mathrm dy \ \mathrm dx$ – Kian Mar 12 '15 at 15:00
  • The area we integrate over is the set $$A:={(x,y,z)| x+y+z \le a} \subset [0,1]^3$$ Thus $P(U\le a) = P(A) = \iiint_A f_{X,Y,Z}$ – AlexR Mar 12 '15 at 15:04
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$$P(X+Y+Z\leq 1)=\mu\{(x,y,z)\in[0,1]^3:0\leq x+y+z\leq 1\}=\frac{1}{6}$$ since the middle term is the volume of a simple pyramid.

Jack D'Aurizio
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You don't actually need to explicitly find the distribution of $U$, or to do any calculus. The integral that AlexR gave is just the volume of a certain three-dimensional figure (a subset of the unit cube) which you should be able to find by basic geometry.

Michael Lugo
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