I am trying to solve the question which is below. However, I cannot find the expectation value because of the function that I found cannot be integrated. I am not sure if I am doing a mistake or not. If I do, please show my mistake.
QUESTION
Let $(X_1, \ldots , X_n)$ be a random sample of i.i.d. random variables distributed as:
$$f (x, \theta) =\begin{cases}e^{\theta−x},& x ≥ θ\\0,&\text{otherwise}\end{cases}$$
- Find the distribution of the statistics $X_{(1)}= \min_i\{X_i\}$.
- Find $E(X_{(1)})$ and $\text{Var}(X_{(1)})$.
MY SOLUTION
$F_\min(x) = F_{(1)}(x) = 1-P(X_\min > x)\\
= 1-P(X_1>x, X_2>x,…,X_n>x)\\
= 1-(1-F_1(x))(1-F_2(x))…(1-F_n(x))\\
= 1-(1-F(x))^n$
$f_\min(x) = -\frac{d}{dx}[(1-F(x))^n] = n[1-F(x)]^{n-1}f(x)$
where the $F(x) =-\exp(-x+\theta)$ and $f(x) =\exp(-x+ \theta)$
So $f_{x_{(1)}}(x) = n[1+\exp(-x+\theta)]^{n-1}\exp(-x+\theta)$
and $E[X_{(1)}] =\int xn[1-exp(-x+\theta)]^{n-1}(-\exp(-x+\theta)) dx$
Best Answer
You have made an error in calculating $F(x)\ne-e^{\theta-x}$.
This is because $F(x)=P(X_i\le x)=\int_\theta^xf(x,\theta)dx=\color{red}1-e^{\theta-x}$ when $x>\theta$ and $0$ otherwise. You missed the $1$.