1. ## MLE estimation

Suppose $\displaystyle X$ is exponentially distributed with density $\displaystyle f_{X}(x) = e^{-x}1_{x>0}.$ Consider $\displaystyle Z = \frac{h}{X},$ where $\displaystyle h > 0.$ Our goal is to estimate $\displaystyle h.$

(a) Find the density function of $\displaystyle Z.$
(b) Let $\displaystyle Z1,...,Zn$ be an i.i.d. sample from $\displaystyle f_{Z}(z),$ the density of $\displaystyle Z.$ Find the MLE of $\displaystyle h,$ denoted by $\displaystyle \hat{h}.$
(c) Find the asymptotic variance of $\displaystyle \hat{h}.$

2. Hello,

Use the Jacobian transformation to get the pdf of Z.

You should get $\displaystyle f_Z(z)=e^{-h/z}\cdot \frac{z^2}{h} \cdot 1_{z>0}$

The likelihood function is then the pdf of the n-tuple, which gives :

$\displaystyle L(\bold{z};h)=\exp\left(-h\sum_{i=1}^n \frac{1}{z_i}\right) \cdot \frac{\left(\prod_{i=1}^n z_i\right)^2}{h^n}$ (where all the zi are >0, and where $\displaystyle \bold{z}=(z_1,\dots,z_n)$)

So the log likelihood function is $\displaystyle \log L=-h\sum_{i=1}^n \frac{1}{z_i}-n\log h+\eta(\bold{z})$

The function $\displaystyle \eta$ is not important, since we will differentiate the log likelihood function with respect to h.

Now find its maximum and you'll get $\displaystyle \hat h$

3. Originally Posted by Moo
$\displaystyle L(\bold{z};h)=\exp\left(-h\sum_{i=1}^n \frac{1}{z_i}\right) \cdot \frac{\left(\prod_{i=1}^n z_i\right)^2}{h^n}$
Thank you for your wonderful answer. Quick question though, why is this not the summation of the $\displaystyle z_i?$

Also, notable shortcut: instead of taking the Jacobian. Consider the cdf, $\displaystyle P(Z\leq z),$

$\displaystyle => Z$ ~ $\displaystyle exp(h/z)$

Or, wait does this not work?

4. Originally Posted by Anonymous1
Thank you for your wonderful answer. Quick question though, why is this not the summation of the $\displaystyle z_i?$
Because in the pdf, it's 1/z, so you're summing 1/z_i. You just have to know that e^x e^y = e^(x+y)

Also, notable shortcut: instead of taking the Jacobian. Consider the cdf, $\displaystyle P(Z\leq z),$
You can, but you have to find the cdf of X first.

$\displaystyle => Z$ ~ $\displaystyle exp(h/z)$
What is z ??? It's nothing near from a parameter...

You should read again some basics...

5. Originally Posted by Moo
You should read again some basics...
I'll admit I go to fast to keep up with myself sometimes, but I do know these things...., buried somewheres or another.