TASK:
Suppose $X_i (i=1,2,3,…,n)$ are i.i.d random variables with PMF: $f(x, \theta ) = exp(\theta-x), x>\theta$. Is the estimate unbiased: $\mu = \frac{1}{n} + $min$(X_i)$?
Answer:
First, we find the CDF for $\mu$: $F(x) = P(\mu < x) = P($min$(X_i) < x - \frac{1}{n}) = 1 - P($min$(X_i)\geq x - \frac{1}{n}) = 1 - \prod_{i=1}^{n} P(X_i \geq x - \frac{1}{n}) = 1 - \prod_{i=1}^{n} (1 - P(X_i < x - \frac{1}{n})) = [1]$.
Next: $P(X_i < x - \frac{1}{n}) = \int_{\theta}^{x-\frac{1}{n}} exp(\theta - x)dy = -exp(\theta - x + \frac{1}{n}) + 1$, so
$[1] = 1 - \prod_{i=1}^{n} (1+ exp(\theta - x + \frac{1}{n}) - 1) = 1 -exp(\theta * n - x * n + 1)$.
Now we find PMF for $ \mu: f(x) = F'(x) = n * exp(\theta * n - x * n +1)$.
EDITED:
We can find mean: $E(\mu) = \int_{\theta}^{+\infty}x * n * exp(\theta * n - x * n +1) dx = e * (\theta + \frac{1}{n})$
So mean with $f(x, \theta)$ is: $E(\theta) = \int_{\theta}^{+\infty}x * exp(\theta - x)dx = \theta + 1$.
Finally, $E(\mu) \neq E(\theta)$, so the estimate isn't unbiased.
That's it?
Thank you for any help.
To show an estimator is unbiased, we need to show that $\mathbb{E}(\mu)=\theta$, where $\mu$ is the estimator and $\theta$ is the parameter.
Let's call $Y:= \min(X_1, \dots, X_n)$.
$$ F_Y(y) = 1-P(X_1>y, \dots, X_n>y)=1-[1-F_X(y)]^n=1-e^{n(\theta - y)} \\ \Rightarrow f_Y(y) = ne^{n(\theta-y)}, \quad y>\theta $$
The expectation then becomes:
$$ \mathbb{E}(\mu) = \frac{1}{n}+\int_{\theta}^{+\infty}y f_Y(y)dy = \frac{1}{n}+\theta +\frac{1}{n}=\frac{2}{n}+\theta \ne \theta $$
hence $\mu$ is biased.