Let $X$ be a random variable with the following distribution. $P[X = 1] = 1-a-a^2$, $P[X = 2] = a^2$, $P[X = 3] = a$.

I want to construct an asymptotically normal estimator for parameter $$\frac{1}{a+1}$$ for this distribution.

My attempts. It is not hard to see that $EX = (a+1)^2$. So, given a vector $(X_1,\ldots, X_n)$ of independent random variables, having the same distribution we can use the following estimator $$ \sqrt{\frac{1}{\frac{1}{n}\sum_{j=1}^nX_j}}. $$ But I don't understand how can we prove that this estimator is asymptotically normal.

My intuition says me that we should use the estimator which is the sum of some i.d random variables and use the central limit theorem, but I don't know how to construct this sum for good estimation $\frac{1}{a+1}$.


1 Answer 1


In order for (one version) of the Central Limit Theorem to apply, we need $E[X^2] < \infty$. We can check that quickly. \begin{align} E[X^2] &= 1^2 P[X=1] + 2^2 P[X=2] + 3^2 P[X=3]\\ &= 1 - a - a^2 + 4(a^2) + 9a\\ &= 3a^2 + 8a + 1\\ &< \infty \end{align}

We will also need the variance \begin{align} Var[X] &= E[X^2] - E[X]^2\\ &= 3a^2 + 8a + 1 - (a+1)^4\\ &= 4a - 3 a^2 - 4 a^3 - a^4 \end{align}

Central Limit Theorem: Let $\{X_i\}$ be a sequence of iid random variables such that $E[X_i^2] < \infty$. Given $E[X_i] = \mu$ and $Var[X_i] = \sigma^2$, let $\bar{X}_n = \sum_{i=1}^n X_i$. Then $$\sqrt{n}(\bar{X}_n - \mu) \overset{d}{\to} \operatorname{Normal}(0, \sigma^2)$$

Then by the central limit theorem we have $$\sqrt{n}(\bar{X}_n - (a+1)^2) \overset{d}{\to} \operatorname{Normal}(0, 4a - 3 a^2 - 4 a^3 - a^4 )$$

Okay, but we are interested in $\dfrac{1}{a+1}$. Luckily we have another theorem.

The Delta Method: Let $\{Y_n\}$ be a sequence of random variables such that $$\sqrt{n}(Y_n - \mu) \overset{d}{\to} \operatorname{Normal}(0, \sigma^2)$$ Then for a function $g(\cdot)$ such that $g^\prime(\mu)$ exists and is non-zero valued we have $$\sqrt{n}(g(Y_n) - g(\mu)) \overset{d}{\to} \operatorname{Normal}(0, \sigma^2 g^\prime(\mu)^2)$$

So the function $g(x) = \dfrac{1}{\sqrt{x}}$ will work. We have \begin{align} g^\prime(x) &= -\dfrac{1}{2}x^{-3/2}\\ g^\prime((a+1)^2) &= -\dfrac{1}{2}[(a+1)^2]^{-3/2}\\ &=-\dfrac{1}{2}(a+1)^{-3}\\ g^\prime((a+1)^2)^2 &= \dfrac{1}{4}(a+1)^{-6} \end{align}

So finally let ${Y}_n = \sqrt{\frac{1}{\frac{1}{n}\sum_{j=1}^nX_j}}$. Then by the delta method we have

$$\sqrt{n}(Y_n - \dfrac{1}{a+1}) \overset{d}{\to} \operatorname{Normal}(0, \dfrac{(4a - 3 a^2 - 4 a^3 - a^4)}{4(a+1)^6} )$$


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