Suppose I have a discrete random variable $X$ which follows a geometric distribution on $x=0,1,2,...$ and I take a random sample from this distribution of size $n$. What is the asymptotic distribution of $\bar X$?

I already know that $E(X)=\frac{1-p}{p}$ and $V(X)=\frac{1-p}{p^2}$.

This seems like an application of central limit theorem, so I'm sure that $\bar X$ converges to a normal distribution. However, the part that's tripping me up is calculating the mean and variance of the normal distribution that it's converging to. How do you do this?


Note that $$ E\bar{X}=E\left(n^{-1}\sum_{i=1}^nX_{i}\right) =n^{-1}\sum_1^nEX_i =n^{-1}(nEX)=EX $$ since the $X_i$ are identically distributed. Similarly, $$ V(\bar{X})=V\left(n^{-1}\sum_{i=1}^nX_{i}\right) =n^{-2}\sum_1^nV(X_i) =n^{-2}(nV(X)) =n^{-1}V(X) $$ since the $X_i$ are independent and identically distributed.

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    $\begingroup$ If we're considering asymptotic convergence, then we're considering the limit as $n$ goes to infinity. So if that is the calculated mean and variance of the normal distribution that it's converging to, then how can we say: $X$~$N(E(X),\frac{V(X)}{n})$ when n is suddenly part of one of the parameters of the distribution? I'm not understanding. Or does this mean that the variance of the normal distribution that it converges to is zero? $\endgroup$ – jippyjoe4 Feb 14 '18 at 23:03
  • $\begingroup$ @jippyjoe4 : In MathJax one writes $X\sim N(E(X), \frac{V(X)} n),$ not $X$~$N(E(X), \frac{V(X)} n),$ i.e the binary relation symbol should be inside of MathJax, not outside. $\endgroup$ – Michael Hardy Feb 14 '18 at 23:35
  • $\begingroup$ Thanks for the tip. But I still don't understand what distribution, exactly, $\bar X$ is converging to from this answer. $\endgroup$ – jippyjoe4 Feb 15 '18 at 4:47

You have $$ \operatorname E( \, \overline X \, ) = \operatorname E\left( \frac {X_1+\cdots+X_n} n \right) = \frac 1 n \left( \operatorname E(X_1) + \cdots + \operatorname E(X_n) \right) = \frac 1 n \cdot n \operatorname E(X) = \operatorname E(X) $$ and \begin{align} \operatorname{var}\left( \sqrt n \cdot \overline X \right) = \operatorname{var} \left( \frac{X_1 + \cdots + X_n}{\sqrt n} \right) = \frac 1 n \left(\operatorname{var}(X_1) + \cdots + \operatorname{var}(X_n) \right) = \frac 1n \cdot n \operatorname{var}(X) \end{align}

So $\displaystyle \dfrac{\overline X - \operatorname E(X)}{\sqrt n}$ has expected value $0$ and variance $1,$ and approaches $N(0,1)$ as $n\to\infty.$

  • $\begingroup$ So then I understand what the mean of $\bar X$ converges to as $n$ goes to infinity; but what happens to the variance? Remember, I'm concerned specifically with the asymptotic distribution of $\bar X$, not any other function of it. $\endgroup$ – jippyjoe4 Feb 15 '18 at 4:49

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