4
$\begingroup$

$X_1, \ldots, X_n$ iid ~ Pois($\lambda$). Suppose, you don't know the value of each $X_i$, but you know if $X_i = 0$ or not for every i.

Find MLE for $\lambda$. Does MLE always exist?

I know how to find MLE. My problem is that I don't understand how to deal with the limited knowledge of the sample.

I used to think that we don't care about what sample we have when creating an (not only ML, but any) estimator. So, my answer is MLE would be the same. Am I right?

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ Start by writing down the probability mass function of the data. $\endgroup$
    – Scortchi ♦
    Commented Jul 9, 2014 at 11:21
  • $\begingroup$ So, for those $X_i = 0$, pmf would be $e^{-\lambda}$, and now proceed as usual? $\endgroup$
    – Yal dc
    Commented Jul 9, 2014 at 11:26
  • 1
    $\begingroup$ & $\Pr(X_i > 0)=$ ? $\endgroup$
    – Scortchi ♦
    Commented Jul 9, 2014 at 11:30
  • $\begingroup$ as usual: $e^{-\lambda} \frac{\lambda^{X_i}}{{X_i}!}$ $\endgroup$
    – Yal dc
    Commented Jul 9, 2014 at 11:31

1 Answer 1

Reset to default
4
$\begingroup$

You don't know the precise value of $x_i$ when it's different from zero. Define indicator variables $Y_i$, taking value 0 when $X_i=0$, or 1 otherwise.

$$\Pr(Y_i=0)=\Pr(X_i=0)=\mathrm{e}^{-\lambda}$$

What about $\Pr(Y_i=1)$ ?

$$\Pr(Y_i=1)=\Pr(X_i>0)=1- \mathrm{e}^{-\lambda}$$

Now write the likelihood in terms of the observations $y_i$ & the parameter $\lambda$:

$$f(\vec{y};\lambda) = \prod_{i=1}^n {(1-\mathrm{e}^{-\lambda})^{y_i}\cdot (\mathrm{e}^{-\lambda})^{1-y_i}}\\= (1-\mathrm{e}^{-\lambda})^{\sum_{i=1}^n y_i}\cdot (\mathrm{e}^{-\lambda})^{n-\sum_{i=1}^n y_i}$$

It should look familiar when you think of how it might be reparametrized.

Improve this answer
$\endgroup$
5
  • $\begingroup$ It looks like, I didn't write it clearly: could you explain, why the information about the sample affects MLE? I expect sample to affect estimate, not estimator. $\endgroup$
    – Yal dc
    Commented Jul 9, 2014 at 11:44
  • 2
    $\begingroup$ Sorry, I've no idea what you mean by that. How could you even use "the same" estimator - $\frac{\sum(X_i)}{n}$ - in this case when you can't distinguish $X=1$ from $X=2$? $\endgroup$
    – Scortchi ♦
    Commented Jul 9, 2014 at 12:12
  • $\begingroup$ When "Suppose there is a sample x1, x2, …, xn" is written, it is assumed that all $x_i$ are known? $\endgroup$
    – Yal dc
    Commented Jul 9, 2014 at 13:17
  • 1
    $\begingroup$ Typically, but not when it's immediately followed by:-"Suppose you don't know the value of each $X_i$" [my emphasis]. $\endgroup$
    – Scortchi ♦
    Commented Jul 9, 2014 at 14:36
  • $\begingroup$ Ok, then we have found the source of confusion: I thought the only thing that matters for MLE is iid. Now, all calculations you wrote make sense to me. Thank you, Scortchi, for your help. $\endgroup$
    – Yal dc
    Commented Jul 9, 2014 at 14:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.