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So far I have

$b=\frac{\sum_{i=1}^{n} x_{i}y_{i}}{\sum_{i=1}^{n} x_{i}^{2}}$

So I substituted for $Y_i$ and got

$Var(b)=Var(\beta + \frac{\sum_{i=1}^{n} x_{i} \epsilon_{i}}{\sum_{i=1}^{n} x_{i}^{2}})$

But I got stuck after this and not sure how to get $Var(b)=\frac{\sigma^{2}}{\sum_{i=1}^{n} x_{i}^{2}}$

Gamecocks99
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1 Answers1

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Lets prove this for a more general case $Y=X\beta+\varepsilon$, where $Y$ is a vector of observations $X$ is the model or design matrix, $\beta$ is a vector of regression coefficients and $\varepsilon\sim N(0,\sigma^{2}I_{n})$ is a multivariate normal. Then for a single observation $Y_{i}=X_{1}\beta_{1}+X_{2}\beta_{2}...+\varepsilon_{i}$. The OLS which also happens to be the MLE in the normal error case is given by

$\hat\beta=(X'X)^{-1}X'Y$

$Var[\hat\beta]=Var[(X'X)^{-1}X'Y]=(X'X)^{-1}X'Var[Y]X(X'X)^{-1}=\sigma^{2}(X'X)^{-1}X'I_{n}X(X'X)^{-1}=\sigma^{2}(X'X)^{-1}$

so for a single regression coefficient, $Var[\hat\beta_{i}]=\sigma^{2}(X'X)^{-1}_{ii}$.

That is a more general proof. In your case you only have one column in your model/design matrix, so $Var[\hat\beta_{i}]=\frac{\sigma_{2}}{X'X}=\frac{\sigma_{2}}{\sum_{}x_{i}^{2}}$

Mael
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