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Given the regression eqn: $y_0= \beta_0 +\beta_1 x_i + \epsilon_i$

I am having difficulty in calculating the variance of $\beta_0$

Here is how I proceeded:-

$\operatorname{Var}(b_0)= \operatorname{Var}(\bar Y -b_1\bar X)$, where $b_0,b_1 \text{are parameters estimator} $;

\begin{align} \operatorname{Var}(b_0)& =\operatorname{Var}(\bar Y)+\operatorname{Var}(b_1\bar X) -2\operatorname{Cov}(\bar Y,b_1\bar X)\\[10pt] &= \operatorname{Var}(\bar Y)+(\bar X)^2\operatorname{Var}(b_1) -2\bar X\operatorname{Cov}(\bar Y,b_1)\\ \end{align}

I have already got the value of $Var(b_1)$,but i cannot prove $\operatorname{Cov}(\bar Y,b_1)=0$.

Thanks!

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1 Answer 1

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Recall that \begin{align} b_1 = \frac{\sum (x_i - \bar{x})(Y_i - \bar{Y}_n) }{\sum (x_i - \bar{x}_n ) ^2} = \frac{ \sum(x_i - \bar{x}_n)}{ \sum(x_i - \bar{x}_n) ^ 2 }Y_i. \end{align} and that $Y_1,..., Y_n$ are i.i.d with $\mathrm{cov}(Y_i, Y_i) = \sigma^2$, hence \begin{align} \operatorname{cov} (\bar{Y}_n , b_1) &= \operatorname{cov}\left( \bar{Y}_n , \frac{ \sum(x_i - \bar{x}_n)}{ \sum(x_i - \bar{x}_n) ^ 2 }Y_i \right)\\ &= \frac{ 1}{ n \sum(x_i - \bar{x}_n) ^ 2 } \sum_{i=1}^n \operatorname{cov}\left( Y_i , (x_i - \bar{x}_n) Y_i \right) \\ &= \frac{ \sum(x_i - \bar{x}_n) \sigma^2}{ n \sum(x_i - \bar{x}_n) ^ 2 } \\ & = 0. \end{align} The last equation stems from $$ \sum (x_i - \bar{x}_n) = n\bar{x}_n - n \bar{x}_n = 0. $$

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