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.
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.
$\displaystyle P(Y=k)=P(X_1=k, \ldots ,X_n=k) =P(X_1=k) \cdots P(X_n=k)=\biggl({1\over k}\biggr)^n$
Hello,
Here is the good ol'trick of the "cumulative" probability ! (better than baldeagle's method )
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,
$\displaystyle \boxed{\mathbb{P}(Y=j)=\left(\frac{k-j+1}{k}\right)^n-\left(\frac{k-j}{k}\right)^n}$