0
$\begingroup$

I am trying to understand the metrics of a good estimator. For example, the Bernoulli probability of success takes the parameter p. But for $X_1,\ldots, X_n \overset{\textrm{iid}}{\sim}\textsf{Ber}(p^2), $ how would you estimate the $ p^2?$

How would you find $\hat {p^2}= \frac1n\sum_{i=1}^n X_i =\bar X? $ Or similarly $Y_1,\ldots, Y_n \overset{\textrm{iid}}{\sim}\textsf{Ber}(p), $ how would you find, working forwards not backwards, that$\hat p = \frac1n \sum_{i=1}^n Y_i^2 = y^2 $ is a GOOD estimator? I am not following these examples.

$\endgroup$
2
  • 1
    $\begingroup$ Please edit your question to use Mathjax (mathjax.org) for the math; it makes it SO much easier for us to read! $\endgroup$
    – jbowman
    Commented Apr 16, 2019 at 20:26
  • $\begingroup$ Are you assuming a Bernoulli $p$ or Bernoulli $p^2$ distribution, as written? In the latter case $p^2$ is merely a strange way of specifying the usual Bernoulli parameter and therefore is estimated as always--usually as the mean of the data. Maybe you're trying to estimate $p=\sqrt{p^2}$? Or maybe you're in the former case and trying to estimate $p^2$ for a Bernoulli $p$ distribution?? $\endgroup$
    – whuber
    Commented Sep 5 at 13:56

1 Answer 1

0
$\begingroup$

There are a number of ways to evaluate an estimator, in the cases you have mentioned unbiasedness is a good one to start with.

The bias of an estimator $\hat p$, estimating a parameter $p$, is defined as

$$Bias(\hat p) = \mathbb{E}(\hat p) - p .$$

An unbiased estimator is an estimator with a bias of zero. For a lot of cases (and before one learns the other ways of evaluating an estimator), unbiased means good.

In your first example you are estimating the parameter $p^2$ with the estimator $\hat{p^2} = \frac{1}{n}\sum_{i=1}^{n} X_i$, where $X_i \overset{i.i.d}{\sim} Ber(p^2)$. Using the definition above, and noting $\mathbb{E}(X_i) = p^2$, you should find the bias of $\hat{p^2}$ to be zero.

In the second example you are estimating the parameter (which I have intentionally renamed to avoid confusion with the first example) $q$ with the estimator $\hat{q} = \frac{1}{n} \sum_{i=1}^{n} Y_i^2$, where $Y_i \overset{i.i.d}{\sim} Ber(q)$. Again using the definition of bias, and noting $\mathbb{E}(Y_i^2) = q$, you should find the bias of $\hat{q}$ to be zero.

$\endgroup$

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