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Let $X\sim Geo(p)$ be a random variable, and $X_1,...,X_n$ a random random sample from the distribution of $X.$
Prove that $\frac{1}{2n}\sum^n_{i=1}X_i^2-\frac{1}{2}\bar X$ is an estimator in the methods of moments for $Var(X)$.

Basically what I've tried to do is try to express $Var(X)$ as a function of $E(X)=\mu_1,E(X^2)=\mu_2$.
I did these two attempts:
$Var(X)=E(X^2)-[E(X)]^2=\mu_2-\mu_1^2=g_1(\mu_1,\mu_2)$.
$Var(X)=\frac{1-p}{p^2}=\frac{1-\frac{1}{\mu_1}}{\frac{1}{\mu_1^2}}=\mu_1(\mu_1-1)=g_2(\mu_1)$.

Both these functions led to two different estimators.
But $g_1$ led to: $\frac{1}{n}\sum^n_{i=1}X_i^2-(\bar X)^2$. Which somehow looks like the estimator I'm asked to prove.

But I got stuck here, I would appreciate any help in how to prove this. Thanks in advance!

Pwaol
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1 Answers1

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You have found two different method of moments estimators

  • $g_1$ leads to $\frac{1}{n}\sum\limits^n_{i=1}X_i^2-(\bar X)^2$
  • $g_2$ leads to $(\bar X)^2 - \bar X$

The average of these is therefore also a method of moments estimator and is

  • $\frac{\left(\frac{1}{n}\sum\limits^n_{i=1}X_i^2-(\bar X)^2\right)+\left((\bar X)^2 - \bar X\right)}2=\frac{1}{2n}\sum\limits^n_{i=1}X_i^2-\frac12\bar X$

using $Var(X)=\frac12\mu_2-\frac12\mu_1 = \frac{2-p}{2p^2}-\frac{1}{2p}=\frac{1-p}{p^2}$ and is what you were asked to prove

Henry
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  • Thank you so much, I couldn't think of that! I'm wondering right now, I could use the average of estimators always? since I will get a function of $Var(X)$ always, so I could do the average of these $3$ and get a new estimator also? – Pwaol Nov 16 '21 at 18:10
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    You could (the average of method of moments estimators is a method of moments estimator), but it would not help here since the average of $a$ and $b$ and $\frac{a+b}{2}$ is $\frac{a+b}{2}$. But for example $\frac35a+\frac25b$ would also be a method of moments estimator – Henry Nov 16 '21 at 19:11