# Thread: how do u show this MLE is biased

1. ## how do u show this MLE is biased

$X_{1},...,X_{n} ,$ independent exponential distribution random variables, with parameter $\theta.$

show the MLE is biased.

2. Originally Posted by silversand
$X_{1},...,X_{n} ,$ independent exponential distribution random variables, with parameter $\theta.$

show the MLE is biased.
The first thing you need to do is calculate the maximum likelihood estimator. Have you done this?

Then you need to show that it's expected value is not equal to $\theta$.

3. ## ...ok...

yes. i have done this.
i found the MLE is just $\frac{n}{\sum_{1}^{n}{x_{i}}}$

and how can i calculate its expectation ?

i know sum of Xi is just Erlang Distribution, and i know its p.d.f. $\frac{\theta^{n}x^{n-1}e^{-\theta x}}{(n-1)!}$

so if i integrate it, $\int_{0}^{\infty}{\frac{\theta^{n}{x^{n}e^{-\theta x}}}{(n-1)!}dx}$

then i got some messy terms cancel out, but then i just found its expectation : $\frac{n}{\theta}$. Therefore, this MLE is unbiased.
which is not like the question asks.

4. Originally Posted by silversand
yes. i have done this.
i found the MLE is just $\frac{n}{\sum_{1}^{n}{x_{i}}}$

and how can i calculate its expectation ?

i know sum of Xi is just Erlang Distribution, and i know its p.d.f. $\frac{\theta^{n}x^{n-1}e^{-\theta x}}{(n-1)!}$

so if i integrate it, $\int_{0}^{\infty}{\frac{\theta^{n}{x^{n}e^{-\theta x}}}{(n-1)!}dx}$

then i got some messy terms cancel out, but then i just found its expectation : $\frac{n}{\theta}$. Therefore, this MLE is unbiased.
which is not like the question asks.
If the MLE is biased then you'll get that $E(\hat{\theta}) \neq \theta$. Your answer agrees with my very rough calculation on a scrap bit of napkin but I'm not convinced you got it using a valid approach.

Here is an outline of an approach I considered (there is probably some clever theorem that gives the required result in two lines that will be brought to my attention):

Let $X = X_1 + X_2 + \, .... \, + X_n$. Then the pdf of $X$ is $f(x) = \, ....$

Let $U = \frac{X}{n}$. Then the pdf of $U$ is $g(u) = n f(nu) = \, ....$

Let $W = \frac{1}{U}$. To get the pdf of $W$, first consider the cdf:

$H(w) = \Pr(W < w) = \Pr \left( \frac{1}{U} < w \right) = \Pr \left( U > \frac{1}{w} \right)$ $= \int_{1/w}^{+ \infty} g(u) \, du$ $= 1 - \int^{1/w}_{0} g(u) \, du = \, ....$

therefore the pdf of $W$ is $h(w) = \frac{dH}{dw} = \frac{1}{w^2} \, g \left(\frac{1}{w} \right)$.

Now calculate $E(\hat{\theta}) = E(W) = \int_0^{+\infty} \frac{1}{w} \, g\left(\frac{1}{w} \right) \, dw$.

By exploiting known results from the Erlang distribution the calculation here should not be too arduous.

I got what you got but even when I'm being careful my carelessness is legendary. So be warned.

5. ## help

$g(\frac{1}{w})=\frac{n^{n}w^{1-n}\theta^{n}e^{-\frac{\theta n}{w}}}{(n-1)!}$

6. ## Problem

A problem:

how can u find a bijection satisfies the invariance property of MLE ?