Finding the PDF of Y, where Y = min X

Question

Have $ X_{1},X_{2},\cdots,X_{10}$ random sample from a distribution with PDF:

$$ f(x;\theta) = e^{ - (x- \theta) },\, \theta \leq x \lt \infty $$

Know that $ \hat{\theta}_{MLE} = Y = min(X_{i},\;i=1,2,\cdots,10) $

How do I go about finding the PDF of Y?

Answer 1

Consider the general case. Assume $X_{1},X_{2},\ldots ,X_{n}$ are IID random variables with cumulative distribution function $F_{X}(x)$ and density $f_{X}(x)$.

The order statistics are: $$X_{(1)}<X_{(2)}<\cdots<X_{(n)}$$

where

$$X_{(1)}=\text{min}(X_{1},X_{2},\ldots ,X_{n})$$

Let $Y=X_{(1)}$. The distribution function of the minimum can be derived as follows:

$$\begin{align} F_{Y}(y)&=\text{Pr}(Y\leq y)\\ &=1-\text{Pr}(Y>y)\\ &=1-\text{Pr}(\text{min}(X_{1},X_{2},\ldots ,X_{n})>y)\\ &=1-\text{Pr}(X_{1}>y)\cdot\text{Pr}(X_{2}>y)\cdots\text{Pr}(X_{n}>y)\\ &=1-(1-F_{X}(y))^{n} \end{align}$$

The density can be found by applying the chain rule to the distribution function:

$$f_{Y}(y)=n\cdot f_{X}(y)\cdot(1-F_{X}(y))^{n-1}$$

We know:

$$\begin{align} F_{X}(y)&=\int_{\theta}^{y}e^{-(u-\theta)}du\\ &=\Big[-e^{-(u-\theta)}\Big]_{\theta}^{y}\\ &=1-e^{-(y-\theta)} \end{align}$$

So,

$$f_{Y}(y)=10 e^{-10(y-\theta)}$$