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Let $X_1,X_2,X_3,X_4$ and $X_5$ be uniformly distributed random variables over $(0,1)$.

What is the probability that $X_1+X_2+X_3+X_4+X_5 < 1$? I would like to know the shortest way to do this.

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closed as off-topic by dustin, user99914, Dario, Did, Claude Leibovici Feb 6 '15 at 7:52

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    $\begingroup$ See: math.stackexchange.com/questions/1128396/… for a geometric representation ... or check out the Irwin-Hall distribution at the link provided $\endgroup$ – wolfies Feb 6 '15 at 7:18
  • $\begingroup$ The wiki page for Irwin Hall distribution provides the probability distribution function for sum of up to 5 variables. However it doesn't provide the method for derivation. $\endgroup$ – Achal Premi Feb 10 '15 at 9:30
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Hint: Can you describe the region $\{(x_1,x_2,x_3,x_4,x_5) \in [0,1]^5 : x_1+x_2+x_3+x_4+x_5<1\}$ as a geometric object? What is this object's volume? Try with $5$ replaced by $2$ or $3$ first if you don't see it.

Hint 2: it looks like you don't yet have intuition about the relationship of expectations and integrals (that okay! you have to start somewhere). Consider that if a random variable $X$ taking values in $\mathbb{R}^d$ has density $f$, then $$ P( X \in A ) = \int_A f(x)dx $$ for nice sets $A$. Note that the set $\{X_1+\cdots + X_5 < 1\}$ is the set where the random variable $X:=(X_1,\ldots,X_5)$ is in the set $A:=\{(x_1,x_2,x_3,x_4,x_5) \in [0,1]^5 : x_1+x_2+x_3+x_4+x_5<1\}$. Take for granted that this set is nice enough. Assuming the $X_i$ are independent, we know the density of $X$ is $f(x) = 1_{x_1,\ldots,x_5 \in [0,1]}$. Check that integrating this over $\mathbb{R}^n$ gives $1$. In particular $$ P(X_1+\cdots+X_5 < 1) = P(X \in A) = \int_{A}1_{x_1,\ldots,x_5 \in [0,1]} 1dx = \int_{A} 1 dx. $$ Now, we know that the integral of $1$ over a region is the volume of that region. Thus to calculate the integral, we can try to geometrically recognize what kind of shape $A$ is and just calculate it's volume.

Lets looks at the shape in a few different dimensions, replacing $5$ here with $n$.

$n=1$: $[0,1)$ thats an interval with length $1$.

$n=2$: $\{(x,y) \in [0,1]^2: x+y<1\}$ is a triangle with area $1/2$.

$n=3$: $\{(x,y,z) \in [0,1]^2: x+y+z<1\}$ is a tetrahedron with volume $1/(2\cdot 3)$.

In general the shape is a simplex with volume $1/n!$. To calculate this we can just do the integral $$ \int_0^1\int_0^{1-x_1}\int_0^{1-x_1-x_2}\cdots \int_0^{1-x_1-x_2\ldots-x_{n-1}} 1 dx_n\cdots dx_3 dx_2 dx_1 = \frac{1}{n!}. $$

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  • $\begingroup$ @Eupraxis1981 It's the standard textbook approach for multivaritate uniform distributions in most textbooks e.g. Pitman's Probability. $\endgroup$ – Fizz Feb 6 '15 at 10:00
  • $\begingroup$ @RespawnedFluff Yes, and....? I was simply commenting on the poster's speed and for not resorting to convolutions. $\endgroup$ – user76844 Feb 6 '15 at 12:11
  • $\begingroup$ I was able to derive for x_1 + x_2 + x_3 < 1. By following: let X_3 = u. So P(x_1+x_2 < 1-u)= 0.5*(1-u)^2 P(x_1 + x_2 + x_3 < 1) = Integration u=0 to u=1 [(1-u)^2 du]. Is the same approach required to calculate P(x_1 + x_2 + x_3 + x_4 < 1) ? $\endgroup$ – Achal Premi Feb 10 '15 at 9:37

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