Let $$y_i=B_0+B_1X_i+\varepsilon_i$$ where $\varepsilon_i\sim N(0,\sigma^2)$. Find the least squares estimator of $B_0$ and show that it is unbiased and has minimum variance.

I will not write in detail all the steps I went through, but $$\hat{B_1}=\frac{\sum((X_i-\overline{X})(Y_i-\overline{Y})}{\sum(X_i-\overline{X})^2}$$ and $$\hat{B_0}=\overline{Y}-\hat{B_1}\overline{X} \,.$$ Taking the expectation: $$\mathbb{E}[\hat{B_0}]=\mathbb{E}[\overline{Y}-\hat{B_1}\overline{X}]=\mathbb{E}[\overline{Y}]-\overline{X}\mathbb{E}[\hat{B_1}]=B_0+B_1\overline{X}-B_1\overline{X}=B_0$$ then this is unbiased.

But how can I show that the estimator has minimum variance in this case?

EDIT: Since I already proved that $B_0$ is unbiased, and since the distribution of $B_0$ belongs to an exponential family, it's a complete and sufficient statistic. Thus this estimator has minimum variance by the Lehmann–Scheffé theorem.

  • 1
    $\begingroup$ Have you calculated the Cramer-Rao Bound for this? $\endgroup$ Oct 3, 2015 at 21:45
  • 1
    $\begingroup$ @rightskewed No, it does not seem an appropriate way, but maybe it is. $\endgroup$
    – user72621
    Oct 3, 2015 at 21:49
  • 1
    $\begingroup$ Cramer-Rao rule gives you a lower bound on the variance of the estimator. Think about the case when the equality holds $\endgroup$ Oct 3, 2015 at 21:53
  • 1
    $\begingroup$ @rightskewed Your point is that doesn't exist a unbiased estimator that attains the CRLB? $\endgroup$
    – user72621
    Oct 3, 2015 at 22:01
  • 1
    $\begingroup$ I think you can also use the Lehman-Scheffe theorem $\endgroup$
    – JohnRos
    Oct 4, 2015 at 10:13

1 Answer 1


For establishing a more general result, I am referring to the lecture notes here.

Suppose we have the multiple linear regression model

$$y=X\beta + \varepsilon$$

, where the design matrix $X$ (with non-random entries) of order $n\times k$ has full column rank and $\beta=(\beta_1,\ldots,\beta_k)^T$ is the vector of regression coefficients.

Further assume that $\varepsilon \sim N_n(0,\sigma^2 I_n)$ with $\sigma^2$ unknown, so that $y\sim N_n(X\beta,\sigma^2 I_n)$.

Under this setting, we know that the OLS estimator of $\beta$ is $$\hat\beta=(X^T X)^{-1}X^T y\sim N_k\left(\beta,\sigma^2(X^T X)^{-1}\right)$$

So, $$(y-X\hat\beta)^T X(\hat\beta-\beta)=(y^TX-y^T X)(\hat\beta-\beta)=0$$


\begin{align} \left\| y-X\beta \right\|^2 &=\left\|y-X\hat\beta+X\hat\beta-X\beta\right\|^2 \\&=\left\|y-X\hat\beta\right\|^2+\left\|X\hat\beta-X\beta\right\|^2 \\&=\left\|y-X\hat\beta\right\|^2+\left\|X\hat\beta\right\|^2+\left\|X\beta\right\|^2-2\beta^T X^T y \end{align}

The pdf of $y$ now looks like

\begin{align} f(y;\beta,\sigma^2)&=\frac{1}{(2\pi\sigma^2)^{n/2}}\exp\left[-\frac{1}{2\sigma^2}\left\| y-X\beta \right\|^2\right] \\\\&=\frac{1}{(2\pi\sigma^2)^{n/2}}\exp\left[\frac{1}{\sigma^2}\beta^T X^T y-\frac{1}{2\sigma^2}\left(\left\|y-X\hat\beta\right\|^2+\left\|X\hat\beta\right\|^2\right)-\frac{1}{2\sigma^2}\left\|X\beta\right\|^2\right] \end{align}

Noting that $X\hat\beta$ is a function of $X^T y$ and that the above density is a member of the exponential family, a complete sufficient statistic for $(\beta,\sigma^2)$ is given by $$T=\left(X^Ty,\left\|y-X\hat\beta\right\|^2\right)$$

Again $\hat\beta$, a function of $X^T y$, is also a function of $T$ and it is unbiased for $\beta$.

So by Lehmann-Scheffe theorem $\hat\beta$ is the UMVUE of $\beta$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.