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Let be $X$ a population with normal distribution. Using likelihood function I get the below expression $\hat{\sigma_X}^2 = \sum_{i=1}^{n}{\dfrac{(X_i-\mu)^2}{n}}$ for variance.

I want prove that expression is consistent, i.e. $E[\hat{\sigma_X}^2]=\sigma_X^2$.

I begin ...

$E[\hat{\sigma_X}^2]=\dfrac{1}{n}E[\sum_{i=1}^{n}{(X_i-\mu)^2}]$

$\dfrac{1}{n}(E[(X_1-\mu)^2]+E[(X_2-\mu)^2]\cdots E[(X_n-\mu)^2])$

I don't know what else to do.

pdta: $\mu=\dfrac{1}{n}\sum_{i=0}^{n}{X_i}$

juaninf
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  • Do you intend the parameter $\mu$ to be a fixed constant, of a random variable, function of $X_i$? – Sasha Aug 31 '12 at 23:31
  • yes, $\mu=\dfrac{1}{n}\sum_{i=0}^{n}{X_i}$ – juaninf Aug 31 '12 at 23:58
  • In that case, it is a well known result that $\mathsf{E}\left(\hat{\sigma}_X^2\right) = \frac{n-1}{n} \sigma_X^2$. – Sasha Sep 01 '12 at 00:02
  • but, in mi lecture would have to be $E[\hat{\sigma_X}^2] = \sigma_X^2$ and not $\mathsf{E}\left(\hat{\sigma}_X^2\right) = \frac{n-1}{n} \sigma_X^2$...(for the consistent)... then this estimative is not consistent? – juaninf Sep 01 '12 at 00:08
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    The proper term is $\hat{\sigma}_X^2$ is a biased estimator. Consistency has to do with large $n$ limit. $\hat{\sigma}_X^2$ is consistent, since the expectation approaches the population value for large $n$. – Sasha Sep 01 '12 at 00:10
  • but then is not biased, because $E[\hat{\sigma_X}^2]$ is not equal to $\sigma_X^2$ ... or not? – juaninf Sep 01 '12 at 00:14
  • sorry, now understand, please will be able to development this expression $\mathsf{E}\left(\hat{\sigma}_X^2\right) = \frac{n-1}{n} \sigma_X^2$... What steps remain? ... – juaninf Sep 01 '12 at 00:24

1 Answers1

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To show the desired (weak) consistency, you have to show that $\frac{1}{n}\sum_{i=1}^n (X_i - \mu)^2$ converges in probability to $\sigma^2$ as $n$ tends to infinity.

To that end, note that $$ \frac{1}{n}\sum_{i=1}^n (X_i - \mu)^2 = \frac{1}{n} \sum_{i=1}^n X_i^2 - 2 \cdot \mu \cdot \frac{1}{n} \sum_{i=1}^n X_i + \mu^2. $$ Assuming $X_i \overset{\mathrm{i.i.d.}}{\sim} \mathcal N(\mu, \sigma^2)$ for $i \in \{1, \ldots, n\}$, the weak law of large numbers gives $$ \frac{1}{n} \sum_{i=1}^n X_i \overset{p}{\underset{n \to \infty}\longrightarrow} \mathbb E[X_1]\;\; \text{and} \,\; \frac{1}{n} \sum_{i=1}^n X_i^2 \overset{p}{\underset{n \to \infty}\longrightarrow} \mathbb E[X_1^2]. $$ With $\mathbb E[X_1] = \mu$, the continuous mapping theorem yields $$ \frac{1}{n} \sum_{i=1}^n X_i^2 - 2 \cdot \mu \cdot \frac{1}{n} \sum_{i=1}^n X_i + \mu^2 \overset{p}{\underset{n \to \infty}\longrightarrow} \mathbb E[X_1^2] - 2 \cdot (\mathbb E[X_1])^2 + (\mathbb E[X_1])^2. $$ Finally, $\mathbb E[X_1^2] - 2 \cdot (\mathbb E[X_1])^2 + (\mathbb E[X_1])^2 = \mathbb E[X_1^2] - (\mathbb E[X_1])^2 = \mathbb V[X_1] = \sigma^2.$

Strong consistency follows analogously by using the strong law of large numbers and replacing convergence in probability by almost sure convergence.

statmerkur
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