Convergence in distribution for a maximum likelihood estimator

probability theorystatisticsweak-convergence

I need some help with the following exercise. Let $X_1,\dots,X_n$ normal random variables i.i.d. with known mean $\mu$ and unknown variance $\sigma^2$.

  1. Find $\sigma$ maximum likelihood estimator $T_n$;
  2. Prove that $T_n$ is consistent;
  3. Find the asymptotic distribution of $\sqrt{n}(T_n-\sigma)$.

Now, 1. and 2. are very easy:

  1. some calculation and the invariance principle for ML estimators show us that $T_n=\sqrt{\frac1n\sum_{k=1}^n(X_i-\mu)^2}$;
  2. for the weak law of large numbers $T_n^2$ converges in probability to $\sigma^2$, so $T_n\rightarrow \sigma$ in probability.

I'm actually stuck with the third request. I've tried applying the Cramer theorem, but I failed in finding a function $F(x)\in L^1(\mathbb{R})$ such that
$$
\forall\sigma>0\quad\left|\frac{d}{d\sigma}f(\sigma;x)\right|\leq F(x),\qquad \text{where }f\text{ is the density function of }X_1.
$$

I also tried finding the $T_n$ characteristic function $\phi_{T_n}$ and computing
$$
\lim_{n\to+\infty}\phi_{\sqrt{n}(T_n-\sigma)},
$$

but I gained nothing. Can someone help me?


Cramer theorem can be found here: Mathematical Methods of Statistics – Harald Cramer, Princeton University Press, 1946, page 500.

Best Answer

Notice that since $\mu$ is known, $$\sum_{i=1}^n \frac{(X_i-\mu)^2}{\sigma^2}=\frac{nT_n^2}{\sigma^2}\sim \chi^2_n$$

So by classical CLT, you have $$\sqrt n(T_n^2-\sigma^2)\stackrel{L}\longrightarrow N\left(0,2\sigma^4\right)$$

Now simply apply 'Delta method' to get the asymptotic behaviour of $\sqrt n(T_n-\sigma)$.


There is another way to get a direct answer using a general result about (some) maximum likelihood estimators, namely:

$$\sqrt n(T_n-\sigma)\stackrel{L}\longrightarrow N\left(0,\frac{1}{I(\sigma)}\right)\,,$$

where $I(\sigma)=\mathbb E_{\sigma}\left[\frac{\partial}{\partial\sigma}\ln f_{\sigma}(X)\right]^2$ is the Fisher information in $X\sim N(\mu,\sigma^2)$ having pdf $f_{\sigma}$.

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