let $X$ be a random variable with density
$$f_x(x) = e^{\alpha-x} \: x>a$$ (a) what is the maximum likelihood estimator for alpha
(b) make the estimator in (a) unbiased, and evaluate its variance
so without much explaination, it should be obvious that the likelihood function is maximized when $\alpha = X_{(1)}$, i.e. the first ordered element in the random sample of random variables.
e.g. $$ln(L_X(X)) = ln\left(\prod\limits_{i=1}^n{e^{\alpha-x_i}}\right) = n\alpha - \sum\limits_{i=1}^n{x_i}$$ (which is naturally maximized at the maximum possible value for $\alpha$, which is $X_{(1)}$)
my question is: how can I evaluate the bias of $X_{(1)}$? This is what I have so far:
first, I need to calculate the expected value of $X_{(1)}$, since the definition of bias is: $$B[\Gamma] = E[\Gamma]-\gamma$$
to calculate $E[X_{(1)}]$, I need the pdf of $X_{(1)}$, which I derive using the following reasoning:
$$P(X<t) = \int_{\alpha}^t{e^{\alpha-x}dx} = 1-e^{\alpha-t}$$
$$P(X_{(1)}<t) = P(X_{(2)}>t)P(X_{(3)}>t)...P(X_{(n)}>t) = (e^{\alpha-x})^{n-1}$$
so the pdf of $X_{(1)}$ is:
$$\frac{\partial P(X_{(1)}<t)}{\partial x} = (n-1)(e^{\alpha-x})^{n-2}$$
and so, the expected value of $X_{(1)}$ is:
$$E[X_{(1)}] =
\int_\alpha^\infty{x(n-1)(e^{\alpha-x})^{n-2}dx} =
\frac{(n-1)(\alpha(n-2)+1)}{(n-2)^2}$$
which isn't exactly the most meaningful expression. The answer in the back of the book suggests that $X_{(1)}-\dfrac{1}{n}$ is an unbiased estimator, suggesting the $E[X_{(1)}] = \alpha+\dfrac{1}{n}$, however, I have no clue how they came up with this...
any help or hints would be greatly appreciated!
so it turns out I made a pretty bozo mistake: When deriving the CDF, I should have noticed that:
$$P(X_{(1)}>x) = P(X_{1}>x)P(X_{2}>x)P(X_{3}>x)\cdots P(X_{n}>x) = (e^{\alpha-x})^{n}$$
so the pdf of $X_{(1)}$ is:
$$-\frac{\partial P(X_{(1)}>x)}{\partial x} = n(e^{\alpha-x})^{n}$$
and so, the expected value of $X_{(1)}$ is:
$$E[X_{(1)}] = \int_\alpha^\infty{x n(e^{\alpha-x})^{n}dx} = \frac{1}{n}+\alpha$$