Show that the MLE converges to the true value (consistency)

Let $Y_i=\alpha_0+\beta_0 X_i + \epsilon_0$, where $\epsilon_i \sim N(0, \sigma_0^2)$ and $X_i \sim N(\mu_x,\tau_0^2)$ are independent.

The data $(X_i, Y_i)$ are generated from $Y_i=\alpha_0+\beta_0 X_i + \epsilon_0$.

I have found the maximum likelihood estimator for each parameter by using $$L_n({X_i, Y_i};\, \alpha, \beta, \mu_x, \sigma^2, \tau^2) = \prod_{i=1}^n f(X_i, Y_i)=\prod_{i=1}^n f_x(X_i)f_{.|X_i}(Y_i),$$ differentiating it with respect to the parameter, setting it equal to zero, and solving for the parameter.

For example, I get that $\hat{\alpha}_{MLE}=\bar{Y_n}-\beta \bar{X_n}$.

But how do I show that $\hat{\alpha}_{MLE}=\bar{Y_n}-\beta \bar{X_n}$ converges to its true value?

I do not know how to find the true value of $\hat{\alpha}_{MLE}$ or how to begin showing convergence.

Thank you.

• I have tried to edit your question for readability, but the notation at the end of the displayed equation (not changed) is unclear to me. Please re-edit as required. – BruceET Dec 3 '17 at 17:19

You have by the law of total expectation that \begin{equation} E[Y_i] = E[E[Y_i | X_i]] = E[\alpha_0 + \beta_0\mu_x + \epsilon_i] = \alpha_0 + \beta_0\mu_x \end{equation}

You also have by the law of large numbers that $\bar{X_n} \overset{P}{\rightarrow} \mu_x$. By Slutsky's theorem, we have that:

\begin{equation} \bar{Y_n} -\beta_0\bar{X_n} \overset{P}{\rightarrow} Y - \beta_0\mu_x \end{equation}

Where $Y$ is the limiting distribution of $\bar{Y_n}$. Again using the law of large numbers we have that $\bar{Y_n} \overset{P}{\rightarrow} Y = E[Y_i] = \alpha_0+\beta_0\mu_x$.

Plugging these into the previous expression we have that \begin{equation} \bar{Y_n} - \beta_0\bar{X_n} \overset{P}{\rightarrow} \alpha_0 +\beta_0\mu_x - \beta_0\mu_x = \alpha_0 \end{equation}

• Thank you very much for your clear answer. Do you know if it is reasonable that $\hat\sigma_{MLE}\overset{P}{\rightarrow} 0$ and $\hat\tau_{MLE}\overset{P}{\rightarrow} 0$? – Silvia Rossi Dec 4 '17 at 5:43
• No, the MLE for $\sigma^2$ is $\hat{\sigma}^2=\frac{1}{n}\sum_i(X_i - \bar{X})^2$, which has sampling distribution: \begin{equation} \frac{n\hat{\sigma}^2}{\sigma^2} \sim \chi^2_{n-1} \end{equation} Which implies that the expectation of $E[\hat{\sigma}^2] = \frac{\sigma (n-1)}{n}$. By the law of large numbers $\hat{\sigma}^2 \overset{P}{\rightarrow} \sigma^2$. – Ryan Warnick Dec 4 '17 at 5:56
• Ah okay, something definitely went wrong for me, thank you again :) – Silvia Rossi Dec 4 '17 at 6:01
• No problem, good luck. – Ryan Warnick Dec 4 '17 at 6:09