0
$\begingroup$

I'm working with estimators. Let X be the random sample from $f(x,\theta)$ = $\theta$ $e^{-\theta x}$, $x \geq 0$

Find the estimator by the method of moments for $\theta$ and by the maximum likelihood method....

I'm trying to use parametric method of moments but I guess very wrong. What is the right procedure?

$\endgroup$
2
  • $\begingroup$ What do you mean under "parametric method of moments"? $\endgroup$
    – NCh
    Mar 26, 2017 at 4:59
  • $\begingroup$ it's a "revised" Pearson's method of moments. But i'm still unable to find a right procedure for this thing. $\endgroup$
    – Tim Carter
    Mar 26, 2017 at 5:22

1 Answer 1

2
$\begingroup$

Outline for Method of Moments Estimation (MME).

Let $X_i \stackrel{iid}{\sim}\mathsf{Exp}(rate = \theta).$ Then $E(X_i) = 1/\theta,$ so $E(\bar X) = 1/\theta.$ Setting sample mean equal to population mean, gives the MME of $\theta$ to be $\tilde \theta = \frac{1}{\bar X}.$ I have no idea what the 'revised MME' means.

Outline for Maximum Likelihood Estimation (MLE).

The likelihood function is $$L(\theta) = \prod_{i=1}^n f_X(x_i|\theta) = \theta^n e^{-\theta T},$$ where $T = \sum_{i=1}^n X_i.$ Taking logarithms we get $\ell(\theta) = n\log(\theta) - \theta T.$ And setting the derivative $\ell^\prime(\theta) = 0$ gives $\hat \theta = \frac{n}{T}= \frac{1}{\bar X}.$

Note: This is just to get you started. Please compare the above with the notation and explanations in your textbook. There are some gaps to fill and some explanations to provide before this exercise is done. By the way, $E(1/\bar X) \ne \theta,$ so neither the MME nor the MLE is unbiased. (Sometimes MME's are unbiased, but not when the algebra to solve for them involves nonlinear operations like taking the reciprocal.)

You might want to see the Wikipedia pages on MME and on exponential distributions.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .