Mean and variance of first order statistic from a uniform distribution - min(x)

Question

So here are the first two moments about the mean and the origin of the first order statistic of a uniform random variable in the interval $[0, 1]$. Maybe someone can derive the moment generating function and post it as a answer.

CDF:

$P\left(Y_{\left(1\right)}>y\right)=1-P\left(Y_{\left(1\right)}\le y\right)=P\left(Y_1>y\right)P\left(Y_2>y\right)\ldots=\left(1-y\right)^n$

$P\left(Y_{\left(1\right)}\le y\right)=1-\left(1-y\right)^n$

PDF:

$f\left(y\right)=n\left(1-y\right)^{n-1} $

First moment about the origin:

$E\left(Y_{\left(1\right)}\right)=\int_{0}^{1}{ny\left(1-y\right)^{n-1}dy}=n\int_{0}^{1}{y\left(1-y\right)^{n-1}dy}=n\left[-y\frac{\left(1-y\right)^n}{n}+\frac{1}{n}\int{\left(1-y\right)^ndy}\right]_0^1=-\left[y\left(1-y\right)^n+\frac{\left(1-y\right)^{n+1}}{n+1}\right]_0^1=\frac{1}{n+1}$

Second moment about the origin:

$E\left({Y_{\left(1\right)}}^2\right)=\int_{0}^{1}{ny^2\left(1-y\right)^{n-1}dy}=n\int_{0}^{1}{y^2\left(1-y\right)^{n-1}dy}=n\left[-y^2\frac{\left(1-y\right)^n}{n}+\frac{2}{n}\int{y\left(1-y\right)^ndy}\right]_0^1=-\left[y^2\left(1-y\right)^n-2\int{y\left(1-y\right)^n}dy\right]_0^1=-\left[y^2\left(1-y\right)^n-2\int{y\left(1-y\right)^n}dy\right]_0^1=-\left[y^2\left(1-y\right)^n-2\left(-\frac{1}{n+1}\left(y\left(1-y\right)^{n+1}+\frac{\left(1-y\right)^{n+2}}{n+2}\right)\right)\right]_0^1=-\left[y^2\left(1-y\right)^n+\frac{2}{n+1}\left(y\left(1-y\right)^{n+1}+\frac{\left(1-y\right)^{n+2}}{n+2}\right)\right]_0^1=\frac{2}{(n+1)(n+2)}$

Second moment about the mean

$V\left(Y_{\left(1\right)}\right)=\frac{2}{\left(n+1\right)\left(n+2\right)}-\frac{1}{\left(n+1\right)^2}=\frac{2\left(n+1\right)-\left(n+2\right)}{\left(n+2\right)\left(n+1\right)^2}=\frac{n}{\left(n+2\right)\left(n+1\right)^2}$

EDIT: I have edited this post to make it more useful other people searching it in the future. The intention now is to illustrate the mean and variance of the first order statistic of a uniform random variable.

Answer 1

The derivative with respect to $y$ of $1-(1-y)^n$ is $+n(1-y)^{n-1}$ not $-n(1-y)^{n-1}$, and densities should always be non-negative

Step by step:

• The derivative of $y^n$ is $+ny^{n-1}$
• The derivative of $(1-y)^n$ is $-n(1-y)^{n-1}$ by the chain rule
• The derivative of $-(1-y)^n$ is $+n(1-y)^{n-1}$
• The derivative of $1-(1-y)^n$ is $+n(1-y)^{n-1}$