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I intended to derive the large sample distribution of the intercept of the least squares estimator $\beta_0$, but after many tries I could not obtain the solution that is given in the book of Stock&Watson as is depicted below.

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I have been able to derive the large sample distribution of the slope $\beta_1$ myself, so we could treat that as a known random variable.

$\hat{\beta}_1 \sim N(\beta_1,\sigma_{\hat{\beta_1}}^{2})$

$\hat{\beta}_0$ have I earlier defined as $\hat{\beta}_0 = \bar{Y} - \hat{\beta}_1 \bar{X}$

The approach I chose was to treat $\hat{\beta}_0$ as a function of $\hat{\beta}_1$

We know that $\bar{Y}$ and $\bar{X}$ are consistent estimators of their true mean $\mu_X$ and $\mu_Y$ by means of LLN, so the equation becomes the following (given n is large). The assumptions of LLN are assumed to be all statisfied.

$\hat{\beta}_0 = \mu_Y - \hat{\beta}_1 \mu_X$

$\hat{\beta}_0 \sim N(\mu_Y - \beta_1 \mu_X,\frac{\sigma_{\hat{\beta_1}}^{2}}{(n\sigma_x^2)^2})$

Could anyone shed some light to this issue I am facing. I hope someone can help me out!

EDIT

$\bar{X}\sim N(\mu_X,\sigma_x^2/n)$ (assuming the case N is large and $X_i$ is i.i.d., $\bar{X}$ is a consistent estimator, and approximately normally distributed due to CLT)

$\bar{Y}\sim N(\mu_Y,\sigma_y^2/n)$

$\sigma_{\hat{\beta_1}}^2=\frac{1}{n}\frac{var((X_i-\mu_x)u_i)}{(var(x_i))^2}$

$u_i=Y_i-\beta_0-\beta_1 X_i$ (the unobserved error term of the simple linear regression model)

EDIT 2:

$\hat{\beta}_0 = \bar{Y} - \hat{\beta}_1 \bar{X}$

$var(\hat{\beta}_0) = var(\bar{Y} - \hat{\beta}_1 \bar{X})$

$ var(\hat{\beta}_0) = var(Y)/n + \bar{X}^2 var(\hat{\beta}_1)$

Note that X is regarded as nonrandom: $var(Y) = var( \beta_0 + \beta_1X + u) = var(u)$

$var(\hat{\beta}_0) = \frac{var(u)}{n} + \bar{X}^2 var(\hat{\beta}_1)$

$var(\hat{\beta}_0) = \frac{var(u)}{n} + \frac{\bar{X}^2}{n} \frac{var((X-\mu_x)u)}{var(x)^2}$

$var(\hat{\beta}_0) = \frac{1}{n} \Big( var(u)+ \bar{X}^2 \frac{var((X-\mu_x)u)}{var(x)^2} \Big)$

$var(\hat{\beta}_0) = \frac{1}{n} \Big( var(u)+ \bar{X}^2 \frac{var(X)var(u)+\mu_x^2var(u)}{var(x)^2} \Big)$

$var(\hat{\beta}_0) = \frac{1}{n} var(u) \Big( 1+ \bar{X}^2 \frac{var(X)+\mu_x^2}{var(x)^2} \Big)$

Then one might to say that $\bar{X}$ is a consistent estimator of $\mu_x$ given that n is large and $X$ is iid.

$var(\hat{\beta}_0) = \frac{1}{n} var(u) \Big( 1+ \mu_x^2 \frac{var(X)+\mu_x^2}{var(x)^2} \Big)$

$var(\hat{\beta}_0) = \frac{1}{n} var(u) \Big( 1+ E(X)^2 \frac{E(X^2)}{var(x)^2} \Big)$

$var(\hat{\beta}_0) = \frac{1}{n} var(u) \Big( 1+ \frac{E(X)^2E(X^2)}{(E(X^2)-E(X))^2} \Big)$

The final expression looks already very familiar with the $H_i$ in the textbook. But I am not there yet.

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These are not the usual assumptions for linear regression. $\bar{X}$ is not usually assumed to have a normal distribution. The $X_i$ are assumed fixed. $Var[Y_i]$ is assumed to be $\sigma^2$. I don't know what $u_i$ means. Also, I don't know what it means to use subscripts $i$ in the formula for the variances without a summation symbol.

$$E[\hat{\beta_0}-\beta_0]=E[\bar{Y}-\hat{\beta_1}\bar{X}-\beta_0]=E[\bar{Y}]-\beta_1\bar{X}-\beta_0$$
$$=\frac{1}{n}\sum_{i=1}^n{\left(\beta_1 x_i+\beta_0\right)}-\beta_1\bar{X}-\beta_0=0$$

To find $Var[\hat{\beta_0}]$, I don't think you don't have enough information from what is shown there to calculate it.

The easiest way to derive the variance of both $\beta_0$ and $\beta_1$ and the correlation between them is shown here.

I can't find the variance from the information provided there. But, if I assume the formula for $\hat{\beta_1}$ is known, then I can find it this way. All sums are from $i=1,...,n$.

$$Var[\hat{\beta_0}]=Var[\bar{Y}-\hat{\beta_1}\bar{X}]$$
$$=Var[\bar{Y}]+(\bar{X})^2 Var[\hat{\beta_1}]-2 \bar{X} Cov[\bar{Y},\hat{\beta_1}]$$
$$=Var[\bar{Y}]+(\bar{X})^2 Var[\hat{\beta_1}]-2 \bar{X} Cov[\frac{1}{n} \sum{Y_i},\frac{\sum(X_i-\bar{X})(Y_i-\bar{Y})}{\sum{(X_i-\bar{X})^2}}]$$
$$=Var[\bar{Y}]+(\bar{X})^2 Var[\hat{\beta_1}]-\frac{2 \bar{X}}{n \sum{(X_i-\bar{X})^2}} Cov[ \sum{Y_i},\sum(X_i-\bar{X})(Y_i-\bar{Y})]$$

Now, since $\sum(X_i-\bar{X})]=0$, we have
$$Cov[ \sum{Y_i},\sum(X_i-\bar{X})(Y_i-\bar{Y})]$$ $$=Cov[ \sum{Y_i},\sum Y_i(X_i-\bar{X})-\bar{Y} \sum(X_i-\bar{X})]$$ $$=Cov[ \sum{Y_i},\sum Y_i(X_i-\bar{X})]$$ $$=\sum (X_i-\bar{X}) Var[Y_i]=\sum (X_i-\bar{X}) ] \sigma^2=0$$

So, all together,
$$Var[\hat{\beta_0}]=Var[\bar{Y}]+(\bar{X})^2 Var[\hat{\beta_1}]=\frac{\sigma^2}{n}+(\bar{X})^2 Var[\hat{\beta_1}]$$

From here, again, I do not see how to simplify this from what you are given. But, if you know that $Var[\hat{\beta_1}]=\frac{\sigma^2}{\sum(X_i-\bar{X})^2}$, then you can plug that in here and simplify it to get $Var[\hat{\beta_0}]=\frac{\sigma^2 \sum {X_i}^2}{n \sum(X_i-\bar{X})^2}=\frac{\sum {X_i}^2}{n } Var[\hat{\beta_1}]$.

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  • $\begingroup$ Thanks for attempting to answer the question. I intentionally wanted to derive the variance of $\beta_0$ in scalar format, not in matrices. As of the potential information that is missing, I doubt that is the case. Some identities that I could further think of are given in the edit. $\endgroup$ Feb 11 at 17:59
  • $\begingroup$ I believe you made a mistake, the variance of a constant should be squared. The $\bar{X}$ should be $\bar{X}^2$. I managed to make a few others transformations in order to find the expression in the textbook for the variance of $\hat{\beta_0}$. However, still failed to obtain the final expression. $\endgroup$ Feb 15 at 9:59
  • $\begingroup$ @NadiaMerquez fixed now $\endgroup$
    – John L
    Feb 15 at 15:35
  • $\begingroup$ Thanks. Any idea how $H_i$ may be constructed based on the latest edit? $\endgroup$ Feb 15 at 20:22
  • $\begingroup$ @NadiaMerquez I don't know what the formula means because I don't understand how there can be a subscript $i$. Since there are many different values of $X_i$, there would have to be many different values of $H_i$. Then, I would get a different value of the variance depending on which one I picked. Unless it means the variance will be the same regardless of which $i$ I pick. But, that doesn't seem possible to me. $\endgroup$
    – John L
    Feb 15 at 22:02

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