Compute the Mean Squared Error of $$

Question

The famous biologist Henk de Rijn has come up with a skill test for chimpanzees. We assume that each chimpanzee has a probability $$ to pass the test, independent from previous attempts and from other monkeys. Let $_$ be the number of attempts that chimpanzee $$ needs to pass. Furthermore, $ = \overline{X}_$ is an estimator for $1/$. Compute the mean squared error of $$ .

So let $T$ be an estimator for a parameter $1∕$. The $MSE$ of $T$ is $MSE(T)=\operatorname{Var}(T) + (E[T] − 1∕)^2$, where$(E[T] − 1∕)^2$ is the bias.
In this case the $E[T]=E[\overline{X}_]=\mu$ and $\operatorname{Var}(T)=\operatorname{Var}(\overline{X}_)=\dfrac{\sigma^2}{n}$.
Now my question is: which distribution should I use?
I think that I think that is a Binomial but it also make sense to use a Geometric Distribution, so then the estimator $T$ will be unbiased and the $\displaystyle\operatorname{MSE}(T)=\operatorname{Var}(T)=\frac{1-p}{p^2}*\frac{1}{n}$, is it right?

Strange, I see all the value inside the dollars – Luke Marci Jan 24 '19 at 08:19 — Luke Marci, Jan 24 '19 at 08:19

score 1 · Answer 1 · answered Jan 23 '19 at 21:56

Each $X_i$ follows a Geometric distribution.

ie $X_i ~ Geom(p)$, and $T=\frac{X_1+...+X_n}{n}$, and the $X_i$s are independent and identically distributed, ie $E(T)=\frac{1}{p}$. Therefore, $MSE(T)$ is just $Var(T)$, and since the variance of a sum is the sum of the variances when dealing with independent random variables, then $Var(T)=Var(\frac{X_1+...+X_n}{n})=\frac{1}{n^2}$$(Var(X_1)+...+Var(X_n))=\frac{1}{n}\frac{p}{(1-p)^2}$.

So yes, it is correct.

The Variance for a geometric distribution is not $\frac{1-p}{p^2}$? — Luke Marci, Jan 24 '19 at 08:30

Compute the Mean Squared Error of $$

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