Consistency of an estimator

Question

I am trying to prove $\widehat{\sigma} = \frac{1}{n} \sum_{i=1}^n \widehat{u}_i^2$ is a consistent estimator for $\sigma= \operatorname E[u_i^2\mid X_i]$ assuming $\operatorname E[u_i\mid X_i]=0$ where $Y_i=\beta_0 + \beta_1 X_i + u_i$ and $\widehat{u}_i = Y_i-(\widehat{\beta}_0 + \widehat{\beta}_1 Xi)$

Answer 1

It would be important to know what you have already tried or what knowledge you have, in order to give an answer that is appropriate to your level.

Anyway, remember that $\hat{\sigma}$ is a consistent estimator (in weak sense) of $\sigma$ if and only if $$\hat{\sigma} \xrightarrow[n\to \infty]{\:\mathcal{P}\:} \sigma.$$

Two possible paths to prove this are:

• showing that $MSE_\sigma(\hat\sigma)\xrightarrow[n\to\infty]{}0$, or equivalently that $$E_\sigma(\hat\sigma)\xrightarrow[n\to\infty]{}\sigma$$ (that is, $\hat\sigma$ is asymptotically unbiased) and that $$Var_\sigma(\hat\sigma)\xrightarrow[n\to\infty]{};0$$ or
• see if you can apply the Law of Large Numbers (LLN): this might work, for instance, if your estimator is the mean of certain $Z_i$ variables (here $\hat\sigma$ is the mean of the $\hat{u}_i^2$) satisfying the hypotheses of that theorem/property/law.

For this case, it may help you to remember that $\hat{\beta}_k$, $k=0,1$, are unbiased estimators of $\beta_k$, $k=0,1.$ Use that property to prove that $E(\hat{u}_i|X_i)=0$. Then try to calculate the expectation and variance of your estimator, or try to apply the LLN as told before (I don't know if this will work, since $E(\hat u_i^2)\neq \sigma$; in fact —spoiler alert— $E(\hat \sigma)=\tfrac{n-2}n \sigma$.)

It may also be useful to rewrite $$\hat{u}_i=Y_i-\hat\beta_0-\hat\beta_1 X_i=Y_i-\beta_0-\beta_1 X_i+\beta_0+\beta_1 X_i-\hat\beta_0-\hat\beta_1 X_i=$$ $$=u_i+(\beta_0-\hat\beta_0)+(\beta_1-\hat\beta_1) X_i,$$ among other possibly useful ideas or hints.