如何计算估计的OLS的方差

我知道

\hat{β_{0}} = \bar{y} - \hat{β_{1}} \bar{x}

$\hat{\beta_0}=\bar{y}-\hat{\beta_1}\bar{x}$ ，这是我得到多远，当我计算方差：

\begin{aligned} V a r (\hat{β_{0}}) & = V a r (\bar{y} - \hat{β_{1}} \bar{x}) \\ = V a r ((- \bar{x}) \hat{β_{1}} + \bar{y}) \\ = V a r ((- \bar{x}) \hat{β_{1}}) + V a r (\bar{y}) \\ = (- \bar{x})^{2} V a r (\hat{β_{1}}) + 0 \\ = (\bar{x})^{2} V a r (\hat{β_{1}}) + 0 \\ = \frac{σ^{2} (\bar{x})^{2}}{\sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}} \end{aligned}

$\begin{align*} Var(\hat{\beta_0}) &= Var(\bar{y} - \hat{\beta_1}\bar{x}) \\ &= Var((-\bar{x})\hat{\beta_1}+\bar{y}) \\ &= Var((-\bar{x})\hat{\beta_1})+Var(\bar{y}) \\ &= (-\bar{x})^2 Var(\hat{\beta_1}) + 0 \\ &= (\bar{x})^2 Var(\hat{\beta_1}) + 0 \\ &= \frac{\sigma^2 (\bar{x})^2}{\displaystyle\sum\limits_{i=1}^n (x_i - \bar{x})^2} \end{align*}$

但这距离我还很远。我想要计算的最终公式是

\begin{aligned} V a r (\hat{β_{0}}) & = \frac{σ^{2} n^{- 1} \sum_{i = 1}^{n} x_{i}^{2}}{\sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}} \end{aligned}

$\begin{align*} Var(\hat{\beta_0}) &= \frac{\sigma^2 n^{-1}\displaystyle\sum\limits_{i=1}^n x_i^2}{\displaystyle\sum\limits_{i=1}^n (x_i - \bar{x})^2} \end{align*}$

我不确定如何获得假设我的数学到那里是正确的）。

(\bar{x})^{2} = \frac{1}{n} \sum_{i = 1}^{n} x_{i}^{2}

$(\bar{x})^2 = \frac{1}{n}\displaystyle\sum\limits_{i=1}^n x_i^2$

这是正确的道路吗？

\begin{aligned} (\bar{x})^{2} & = {(\frac{1}{n} \sum_{i = 1}^{n} x_{i})}^{2} \\ = \frac{1}{n^{2}} {(\sum_{i = 1}^{n} x_{i})}^{2} \end{aligned}

$\begin{align} (\bar{x})^2 &= \left(\frac{1}{n}\displaystyle\sum\limits_{i=1}^n x_i\right)^2 \\ &= \frac{1}{n^2} \left(\displaystyle\sum\limits_{i=1}^n x_i\right)^2 \end{align}$

我敢肯定，这很简单，因此，如果有人暗示将我推向正确的方向，答案可能会稍等。

regression self-study

— 公吨
source

这不是正确的道路。第四个方程式不成立。例如，对于

x_{1} = - 1

$x_1=−1$ ，

x_{2} = 0

$x_2=0$ ，和

x_{3} = 1

$x_3=1$ ，左项为零，而正确的术语是

2 / 3

$2/3$ 。问题来自拆分方差的步骤（第二个方程的第三行）。明白为什么吗？

— QuantIbex

对Quantlbex点的提示：方差不是线性函数。它违反了可加性和标量乘法。

— 大卫·马克思

@DavidMarx即步骤应该是

= V a r ((- \bar{x}) \hat{β_{1}} + \bar{y}) = (\bar{x})^{2} V a r (\hat{β_{1}}) + \bar{y}

$=Var((-\bar{x})\hat{\beta_1}+\bar{y})=(\bar{x})^2Var(\hat{\beta_1})+\bar{y}$ ，我想，然后一旦我在替代对于

\hat{β_{1}}

$\hat{\beta_1}$ 和

\bar{y}

$\bar{y}$ （不知道该怎么为这个做的，但我会更多的考虑一下），是把我在正确的道路，我希望。

— MT

这是不正确的。考虑一下求和等于方差之和的条件。

— QuantIbex

否，

\bar{y}

$\bar{y}$ 是随机的，因为

y_{i} = β_{0} + β_{1} x_{i} + ϵ

$y_i = \beta_0 + \beta_1 x_i + \epsilon$ ，其中，

ϵ

$\epsilon$ 表示（随机）的噪声。但是，好的，我之前的评论可能会引起误解。另外，如果

和

表示常数，则

。

V a r (a X + b) = a^{2} V a r (X)

${\rm Var}(aX + b)= a^2{\rm Var}(X)$

a

$a$

b

$b$

— QuantIbex

Answers:

这是一个自学型问题，因此，我提供了一些提示，希望可以帮助找到解决方案，并且将根据您的反馈/进度来编辑答案。

参数估计值，最大限度地减少平方的总和是要获得的方差，从它的表达启动和替代的表达，并执行代数

\begin{aligned} {\hat{β}}_{0} & = \bar{y} - {\hat{β}}_{1} \bar{x}, \\ {\hat{β}}_{1} & = \frac{\sum_{i = 1}^{n} (x_{i} - \bar{x}) y_{i}}{\sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}} . \end{aligned}

$\begin{align} \hat{\beta}_0 &= \bar{y} - \hat{\beta}_1 \bar{x} , \\ \hat{\beta}_1 &= \frac{ \sum_{i = 1}^n(x_i - \bar{x})y_i }{ \sum_{i = 1}^n(x_i - \bar{x})^2 } . \end{align}$

{\hat{β}}_{0}

$\hat{\beta}_0$

{\hat{β}}_{1}

$\hat{\beta}_1$

V a r ({\hat{β}}_{0}) = V a r (\bar{Y} - {\hat{β}}_{1} \bar{x}) = \dots

${\rm Var}(\hat{\beta}_0) = {\rm Var} (\bar{Y} - \hat{\beta}_1 \bar{x}) = \ldots$

编辑：
我们有两个方差术语是

\begin{aligned} V a r ({\hat{β}}_{0}) & = V a r (\bar{Y} - {\hat{β}}_{1} \bar{x}) \\ = V a r (\bar{Y}) + (\bar{x})^{2} V a r ({\hat{β}}_{1}) - 2 \bar{x} C o v (\bar{Y}, {\hat{β}}_{1}) . \end{aligned}

$\begin{align} {\rm Var}(\hat{\beta}_0) &= {\rm Var} (\bar{Y} - \hat{\beta}_1 \bar{x}) \\ &= {\rm Var} (\bar{Y}) + (\bar{x})^2 {\rm Var} (\hat{\beta}_1) - 2 \bar{x} {\rm Cov} (\bar{Y}, \hat{\beta}_1). \end{align}$

和

V a r (\bar{Y}) = V a r (\frac{1}{n} \sum_{i = 1}^{n} Y_{i}) = \frac{1}{n^{2}} \sum_{i = 1}^{n} V a r (Y_{i}) = \frac{σ^{2}}{n},

${\rm Var} (\bar{Y}) = {\rm Var} \left(\frac{1}{n} \sum_{i = 1}^n Y_i \right) = \frac{1}{n^2} \sum_{i = 1}^n {\rm Var} (Y_i) = \frac{\sigma^2}{n},$

并且协方差项是

\begin{aligned} V a r ({\hat{β}}_{1}) & = \frac{1}{{[\sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}]}^{2}} \sum_{i = 1}^{n} (x_{i} - \bar{x})^{2} V a r (Y_{i}) \\ = \frac{σ^{2}}{\sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}}, \end{aligned}

$\begin{align} {\rm Var} (\hat{\beta}_1) &= \frac{ 1 }{ \left[\sum_{i = 1}^n(x_i - \bar{x})^2 \right]^2 } \sum_{i = 1}^n(x_i - \bar{x})^2 {\rm Var} (Y_i) \\ &= \frac{ \sigma^2 }{ \sum_{i = 1}^n(x_i - \bar{x})^2 } , \end{align}$

\begin{aligned} C o v (\bar{Y}, {\hat{β}}_{1}) & = C o v {\frac{1}{n} \sum_{i = 1}^{n} Y_{i}, \frac{\sum_{j = 1}^{n} (x_{j} - \bar{x}) Y_{j}}{\sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}}} \\ = \frac{1}{n} \frac{1}{\sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}} C o v {\sum_{i = 1}^{n} Y_{i}, \sum_{j = 1}^{n} (x_{j} - \bar{x}) Y_{j}} \\ = \frac{1}{n \sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}} \sum_{i = 1}^{n} (x_{j} - \bar{x}) \sum_{j = 1}^{n} C o v (Y_{i}, Y_{j}) \\ = \frac{1}{n \sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}} \sum_{i = 1}^{n} (x_{j} - \bar{x}) σ^{2} \\ = 0 \end{aligned}

$\begin{align} {\rm Cov} (\bar{Y}, \hat{\beta}_1) &= {\rm Cov} \left\{ \frac{1}{n} \sum_{i = 1}^n Y_i, \frac{ \sum_{j = 1}^n(x_j - \bar{x})Y_j }{ \sum_{i = 1}^n(x_i - \bar{x})^2 } \right \} \\ &= \frac{1}{n} \frac{ 1 }{ \sum_{i = 1}^n(x_i - \bar{x})^2 } {\rm Cov} \left\{ \sum_{i = 1}^n Y_i, \sum_{j = 1}^n(x_j - \bar{x})Y_j \right\} \\ &= \frac{ 1 }{ n \sum_{i = 1}^n(x_i - \bar{x})^2 } \sum_{i = 1}^n (x_j - \bar{x}) \sum_{j = 1}^n {\rm Cov}(Y_i, Y_j) \\ &= \frac{ 1 }{ n \sum_{i = 1}^n(x_i - \bar{x})^2 } \sum_{i = 1}^n (x_j - \bar{x}) \sigma^2 \\ &= 0 \end{align}$ since

\sum_{i = 1}^{n} (x_{j} - \bar{x}) = 0

$\sum_{i = 1}^n (x_j - \bar{x})=0$ .
And since

\sum_{i = 1}^{n} (x_{i} - \bar{x})^{2} = \sum_{i = 1}^{n} x_{i}^{2} - 2 \bar{x} \sum_{i = 1}^{n} x_{i} + \sum_{i = 1}^{n} {\bar{x}}^{2} = \sum_{i = 1}^{n} x_{i}^{2} - n {\bar{x}}^{2},

$\sum_{i = 1}^n(x_i - \bar{x})^2 = \sum_{i = 1}^n x_i^2 - 2 \bar{x} \sum_{i = 1}^n x_i + \sum_{i = 1}^n \bar{x}^2 = \sum_{i = 1}^n x_i^2 - n \bar{x}^2,$ we have

\begin{aligned} V a r ({\hat{β}}_{0}) & = \frac{σ^{2}}{n} + \frac{σ^{2} {\bar{x}}^{2}}{\sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}} \\ = \frac{σ^{2}}{n \sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}} {\sum_{i = 1}^{n} (x_{i} - \bar{x})^{2} + n {\bar{x}}^{2}} \\ = \frac{σ^{2} \sum_{i = 1}^{n} x_{i}^{2}}{n \sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}} . \end{aligned}

$\begin{align} {\rm Var}(\hat{\beta}_0) &= \frac{\sigma^2}{n} + \frac{ \sigma^2 \bar{x}^2}{ \sum_{i = 1}^n(x_i - \bar{x})^2 } \\ &= \frac{\sigma^2 }{ n \sum_{i = 1}^n(x_i - \bar{x})^2 } \left\{ \sum_{i = 1}^n(x_i - \bar{x})^2 + n \bar{x}^2 \right\} \\ &= \frac{\sigma^2 \sum_{i = 1}^n x_i^2}{ n \sum_{i = 1}^n(x_i - \bar{x})^2 }. \end{align}$

Edit 2

Why do we have ${\rm var} ( \sum_{i = 1}^n Y_i) = \sum_{i = 1}^n {\rm Var} (Y_i)$ ?

The assumed model is $Y_i = \beta_0 + \beta_1 X_i + \epsilon_i$ , where the $\epsilon_i$ are independant and identically distributed random variables with ${\rm E}(\epsilon_i) = 0$ and ${\rm var}(\epsilon_i) = \sigma^2$ .

Once we have a sample, the $X_i$ are known, the only random terms are the $\epsilon_i$ . Recalling that for a random variable $Z$ and a constant $a$ , we have ${\rm var}(a+Z) = {\rm var}(Z)$ . Thus,

\begin{aligned} v a r (\sum_{i = 1}^{n} Y_{i}) & = v a r (\sum_{i = 1}^{n} β_{0} + β_{1} X_{i} + ϵ_{i}) \\ = v a r (\sum_{i = 1}^{n} ϵ_{i}) = \sum_{i = 1}^{n} \sum_{j = 1}^{n} c o v (ϵ_{i}, ϵ_{j}) \\ = \sum_{i = 1}^{n} c o v (ϵ_{i}, ϵ_{i}) = \sum_{i = 1}^{n} v a r (ϵ_{i}) \\ = \sum_{i = 1}^{n} v a r (β_{0} + β_{1} X_{i} + ϵ_{i}) = \sum_{i = 1}^{n} v a r (Y_{i}) . \end{aligned}

$\begin{align} {\rm var} \left( \sum_{i = 1}^n Y_i \right) &= {\rm var} \left( \sum_{i = 1}^n \beta_0 + \beta_1 X_i + \epsilon_i \right)\\ &= {\rm var} \left( \sum_{i = 1}^n \epsilon_i \right) = \sum_{i = 1}^n \sum_{j = 1}^n {\rm cov} (\epsilon_i, \epsilon_j)\\ &= \sum_{i = 1}^n {\rm cov} (\epsilon_i, \epsilon_i) = \sum_{i = 1}^n {\rm var} (\epsilon_i)\\ &= \sum_{i = 1}^n {\rm var} (\beta_0 + \beta_1 X_i + \epsilon_i) = \sum_{i = 1}^n {\rm var} (Y_i).\\ \end{align}$ The 4th equality holds as

c o v (ϵ_{i}, ϵ_{j}) = 0

${\rm cov} (\epsilon_i, \epsilon_j) = 0$ for

i \neq j

$i \neq j$ by the independence of the

ϵ_{i}

$\epsilon_i$ .

— QuantIbex
source

I think I got it! The book has suggested steps, and I was able to prove each step separately (I think). It's not as satisfying as just sitting down and grinding it out from this step, since I had to prove intermediate conclusions for it to help, but I think everything looks good.

— M T

See edit for the development of the suggested approach.

— QuantIbex

The variance of the sum equals the sum of the variances in this step:

V a r (\bar{Y}) = V a r (\frac{1}{n} \sum_{i = 1}^{n} Y_{i}) = \frac{1}{n^{2}} \sum_{i = 1}^{n} V a r (Y_{i})

${\rm Var} (\bar{Y}) = {\rm Var} \left(\frac{1}{n} \sum_{i = 1}^n Y_i \right) = \frac{1}{n^2} \sum_{i = 1}^n {\rm Var} (Y_i)$ because since the

X_{i}

$X_i$ are independent, this implies that the

Y_{i}

$Y_i$ are independent as well, right?

— M T

Also, you can factor out a constant from the covariance in this step:

\frac{1}{n} \frac{1}{\sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}} C o v {\sum_{i = 1}^{n} Y_{i}, \sum_{j = 1}^{n} (x_{j} - \bar{x}) Y_{j}}

$\frac{1}{n} \frac{ 1 }{ \sum_{i = 1}^n(x_i - \bar{x})^2 } {\rm Cov} \left\{ \sum_{i = 1}^n Y_i, \sum_{j = 1}^n(x_j - \bar{x})Y_j \right\}$ even though it's not in both elements because the formula for covariance is multiplicative, right?

— M T

@oort, in the numerator you have the sum of

n

$n$ terms that are identical (and equal to

σ^{2}

$\sigma^2$ ), so the numerator is

n σ^{2}

$n \sigma^2$ .

— QuantIbex

I got it! Well, with help. I found the part of the book that gives steps to work through when proving the $Var \left( \hat{\beta}_0 \right)$ formula (thankfully it doesn't actually work them out, otherwise I'd be tempted to not actually do the proof). I proved each separate step, and I think it worked.

I'm using the book's notation, which is:

S S T_{x} = \sum_{i = 1}^{n} (x_{i} - \bar{x})^{2},

$SST_x = \displaystyle\sum\limits_{i=1}^n (x_i - \bar{x})^2,$ and

u_{i}

$u_i$ is the error term.

1) Show that $\hat{\beta}_1$ can be written as $\hat{\beta}_1 = \beta_1 + \displaystyle\sum\limits_{i=1}^n w_i u_i$ where $w_i = \frac{d_i}{SST_x}$ and $d_i = x_i - \bar{x}$ .

This was easy because we know that

\begin{aligned} {\hat{β}}_{1} & = β_{1} + \frac{\sum_{i = 1}^{n} (x_{i} - \bar{x}) u_{i}}{S S T_{x}} \\ = β_{1} + \sum_{i = 1}^{n} \frac{d_{i}}{S S T_{x}} u_{i} \\ = β_{1} + \sum_{i = 1}^{n} w_{i} u_{i} \end{aligned}

$\begin{align} \hat{\beta}_1 &= \beta_1 + \frac{\displaystyle\sum\limits_{i=1}^n (x_i - \bar{x}) u_i}{SST_x} \\ &= \beta_1 + \displaystyle\sum\limits_{i=1}^n \frac{d_i}{SST_x} u_i \\ &= \beta_1 + \displaystyle\sum\limits_{i=1}^n w_i u_i \end{align}$

2) Use part 1, along with $\displaystyle\sum\limits_{i=1}^n w_i = 0$ to show that $\hat{\beta_1}$ and $\bar{u}$ are uncorrelated, i.e. show that $E[(\hat{\beta_1}-\beta_1) \bar{u}] = 0$ .

\begin{aligned} E [(\hat{β_{1}} - β_{1}) \bar{u}] & = E [\bar{u} \sum_{i = 1}^{n} w_{i} u_{i}] \\ = \sum_{i = 1}^{n} E [w_{i} \bar{u} u_{i}] \\ = \sum_{i = 1}^{n} w_{i} E [\bar{u} u_{i}] \\ = \frac{1}{n} \sum_{i = 1}^{n} w_{i} E (u_{i} \sum_{j = 1}^{n} u_{j}) \\ = \frac{1}{n} \sum_{i = 1}^{n} w_{i} [E (u_{i} u_{1}) + \dots + E (u_{i} u_{j}) + \dots + E (u_{i} u_{n})] \end{aligned}

$\begin{align} E[(\hat{\beta_1}-\beta_1) \bar{u}] &= E[\bar{u}\displaystyle\sum\limits_{i=1}^n w_i u_i] \\ &=\displaystyle\sum\limits_{i=1}^n E[w_i \bar{u} u_i] \\ &=\displaystyle\sum\limits_{i=1}^n w_i E[\bar{u} u_i] \\ &= \frac{1}{n}\displaystyle\sum\limits_{i=1}^n w_i E\left(u_i\displaystyle\sum\limits_{j=1}^n u_j\right) \\ &= \frac{1}{n}\displaystyle\sum\limits_{i=1}^n w_i \left[E\left(u_i u_1\right) +\cdots + E(u_i u_j) + \cdots+ E\left(u_i u_n \right)\right] \\ \end{align}$

and because the $u$ are i.i.d., $E(u_i u_j) = E(u_i) E(u_j)$ when $j \neq i$ .

When $j = i$ , $E(u_i u_j) = E(u_i^2)$ , so we have:

\begin{aligned} = \frac{1}{n} \sum_{i = 1}^{n} w_{i} [E (u_{i}) E (u_{1}) + \dots + E (u_{i}^{2}) + \dots + E (u_{i}) E (u_{n})] \\ = \frac{1}{n} \sum_{i = 1}^{n} w_{i} E (u_{i}^{2}) \\ = \frac{1}{n} \sum_{i = 1}^{n} w_{i} [V a r (u_{i}) + E (u_{i}) E (u_{i})] \\ = \frac{1}{n} \sum_{i = 1}^{n} w_{i} σ^{2} \\ = \frac{σ^{2}}{n} \sum_{i = 1}^{n} w_{i} \\ = \frac{σ^{2}}{n \cdot S S T_{x}} \sum_{i = 1}^{n} (x_{i} - \bar{x}) \\ = \frac{σ^{2}}{n \cdot S S T_{x}} (0) & = 0 \end{aligned}

$\begin{align} &= \frac{1}{n}\displaystyle\sum\limits_{i=1}^n w_i \left[E(u_i) E(u_1) +\cdots + E(u_i^2) + \cdots + E(u_i) E(u_n)\right] \\ &= \frac{1}{n}\displaystyle\sum\limits_{i=1}^n w_i E(u_i^2) \\ &= \frac{1}{n}\displaystyle\sum\limits_{i=1}^n w_i \left[Var(u_i) + E(u_i) E(u_i)\right] \\ &= \frac{1}{n}\displaystyle\sum\limits_{i=1}^n w_i \sigma^2 \\ &= \frac{\sigma^2}{n}\displaystyle\sum\limits_{i=1}^n w_i \\ &= \frac{\sigma^2}{n \cdot SST_x}\displaystyle\sum\limits_{i=1}^n (x_i - \bar{x}) \\ &= \frac{\sigma^2}{n \cdot SST_x} \left(0\right) &= 0 \end{align}$

3) Show that $\hat{\beta_0}$ can be written as $\hat{\beta_0} = \beta_0 + \bar{u} - \bar{x}(\hat{\beta_1} - \beta_1)$ . This seemed pretty easy too:

\begin{aligned} \hat{β_{0}} & = \bar{y} - \hat{β_{1}} \bar{x} \\ = (β_{0} + β_{1} \bar{x} + \bar{u}) - \hat{β_{1}} \bar{x} \\ = β_{0} + \bar{u} - \bar{x} (\hat{β_{1}} - β_{1}) . \end{aligned}

$\begin{align} \hat{\beta_0} &= \bar{y} - \hat{\beta_1} \bar{x} \\ &= (\beta_0 + \beta_1 \bar{x} + \bar{u}) - \hat{\beta_1} \bar{x} \\ &= \beta_0 + \bar{u} - \bar{x}(\hat{\beta_1} - \beta_1). \end{align}$

4) Use parts 2 and 3 to show that $Var(\hat{\beta_0}) = \frac{\sigma^2}{n} + \frac{\sigma^2 (\bar{x}) ^2} {SST_x}$ :

\begin{aligned} V a r (\hat{β_{0}}) & = V a r (β_{0} + \bar{u} - \bar{x} (\hat{β_{1}} - β_{1})) \\ = V a r (\bar{u}) + (- \bar{x})^{2} V a r (\hat{β_{1}} - β_{1}) \\ = \frac{σ^{2}}{n} + (\bar{x})^{2} V a r (\hat{β_{1}}) \\ = \frac{σ^{2}}{n} + \frac{σ^{2} (\bar{x})^{2}}{S S T_{x}} . \end{aligned}

$\begin{align} Var(\hat{\beta_0}) &= Var(\beta_0 + \bar{u} - \bar{x}(\hat{\beta_1} - \beta_1)) \\ &= Var(\bar{u}) + (-\bar{x})^2 Var(\hat{\beta_1} - \beta_1) \\ &= \frac{\sigma^2}{n} + (\bar{x})^2 Var(\hat{\beta_1}) \\ &= \frac{\sigma^2}{n} + \frac{\sigma^2 (\bar{x}) ^2} {SST_x}. \end{align}$

I believe this all works because since we provided that $\bar{u}$ and $\hat{\beta_1} - \beta_1$ are uncorrelated, the covariance between them is zero, so the variance of the sum is the sum of the variance. $\beta_0$ is just a constant, so it drops out, as does $\beta_1$ later in the calculations.

5) Use algebra and the fact that $\frac{SST_x}{n} = \frac{1}{n} \displaystyle\sum\limits_{i=1}^n x_i^2 - (\bar{x})^2$ :

\begin{aligned} V a r (\hat{β_{0}}) & = \frac{σ^{2}}{n} + \frac{σ^{2} (\bar{x})^{2}}{S S T_{x}} \\ = \frac{σ^{2} S S T_{x}}{S S T_{x} n} + \frac{σ^{2} (\bar{x})^{2}}{S S T_{x}} \\ = \frac{σ^{2}}{S S T_{x}} (\frac{1}{n} \sum_{i = 1}^{n} x_{i}^{2} - (\bar{x})^{2}) + \frac{σ^{2} (\bar{x})^{2}}{S S T_{x}} \\ = \frac{σ^{2} n^{- 1} \sum_{i = 1}^{n} x_{i}^{2}}{S S T_{x}} \end{aligned}

$\begin{align} Var(\hat{\beta_0}) &= \frac{\sigma^2}{n} + \frac{\sigma^2 (\bar{x}) ^2} {SST_x} \\ &= \frac{\sigma^2 SST_x}{SST_x n} + \frac{\sigma^2 (\bar{x})^2}{SST_x} \\ &= \frac{\sigma^2}{SST_x} \left( \frac{1}{n} \displaystyle\sum\limits_{i=1}^n x_i^2 - (\bar{x})^2 \right) + \frac{\sigma^2 (\bar{x})^2}{SST_x} \\ &= \frac{\sigma^2 n^{-1} \displaystyle\sum\limits_{i=1}^n x_i^2}{SST_x} \end{align}$

— M T
source

There might be a typo in point 1; I think

v a r (\hat{β})

${\rm var(\hat{\beta})}$ should read

\hat{β}

$\hat{\beta}$ .

— QuantIbex

You might want to clarify notations, and specify what

u_{i}

$u_i$ and

{S S T}_{x}

${\rm SST}_x$ are.

— QuantIbex

u_{i}

$u_i$ is the error term and

S S T_{x}

$SST_x$ is the total sum of squares for

x

$x$ (defined in the edit).

— M T

In point 1, the term

β_{1}

$\beta_1$ is missing in the last two lines.

— QuantIbex

In point 2, you can't take

\bar{u}

$\bar{u}$ out of the expectation, it's not a constant.

— QuantIbex