Let X1, X2,...,Xn be n mutually independent random variables, each of which is uniformly distributed on the integers from 1 to k. Let Y denote the minimum of the Xi's. Find the distribution of Y.

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- Aug 2nd 2009, 11:16 AMmorganforDistribution problem
Let X1, X2,...,Xn be n mutually independent random variables, each of which is uniformly distributed on the integers from 1 to k. Let Y denote the minimum of the Xi's. Find the distribution of Y.

- Aug 2nd 2009, 06:18 PMmatheagle
The continuous case is easier, but the same logic works here.

$\displaystyle P(Y=1)=P({\rm at \; least \; one\;} X_i=1)=1-P({\rm all\;} X_i>1)=1-\biggl({k-1\over k}\biggr)^n$

Then it gets messy.

$\displaystyle P(Y=2)=P({\rm at \; least \; one\;} X_i=2\; {\rm and \; none \; are \; equal \; to \; one}) $

Here you have to think about all the possible strings of length n of numbers 1 through k.

All $\displaystyle k^n$ sequences are equally likely.

To have exactly one 2 (and no 1's) that's $\displaystyle n\biggl({1\over k}\biggr)\biggl({k-2\over k}\biggr)^{n-1}$

To have exactly two 2's (and no 1's) that's $\displaystyle {n\choose 2}\biggl({1\over k}\biggr)^2\biggl({k-2\over k}\biggr)^{n-2}$

Then exactly three 2's....

The sum of all of these will give you P(Y=2).

The last is easy.

$\displaystyle P(Y=k)=P(X_1=k, \ldots ,X_n=k) =P(X_1=k) \cdots P(X_n=k)=\biggl({1\over k}\biggr)^n$ - Aug 3rd 2009, 12:08 AMMoo
Hello,

Here is the good ol'trick of the "cumulative" probability ! (better than**bald**eagle's method :D)

For any j, integer between 1 and k :

$\displaystyle \mathbb{P}(Y\geqslant j)=\mathbb{P}(\forall i=1..n ~,~ X_i\geqslant j)=\left(\frac{k-j+1}{k}\right)^n$

Similarly, for any j, integer between 1 and n-1, we have :

$\displaystyle \mathbb{P}(Y\geqslant j+1)=\left(\frac{k-j}{k}\right)^n$

If j=k, this probability equals 0. So this formula works for any j, integer between 1 and k.

Since Y has integer values, we can easily see that $\displaystyle \mathbb{P}(Y=j)=\mathbb{P}(Y\geqslant j)-\mathbb{P}(Y\geqslant j+1)$

So finally,

$\displaystyle \boxed{\mathbb{P}(Y=j)=\left(\frac{k-j+1}{k}\right)^n-\left(\frac{k-j}{k}\right)^n}$