E(b1)=beta1 regression

Question

In the prove of $E[b_1]=\beta_1$ I saw this steps : $$ E[b_1]=E[\frac{S_{xy}}{S_{xx}}]=E[\sum[\frac{[(x_i - \bar{X})*y_i]}{\sum(x_i-\bar{X})}]]=\frac{\sum[(x_i-\bar{x})*E(y_i)]}{\sum(x_i-\bar{X})} ...$$ why the last move is legal?

Form: $E[\sum\frac{[(x_i - \bar{X})*y_i]}{[\sum(x_i-\bar{X})]}$

To this : $\frac{\sum[x_i-\bar{x})*E(y_i)]}{\sum(x_i-\bar{X})}$?

Why $E[(x_i-\bar{X})*y_i]=(x_i-(\bar X))*E[y_i]$?

Answer 1

$$ E[b_1]=E[\frac{S_{xy}}{S_{xx}}]=E[\sum[\frac{[(x_i - \bar{X})*y_i]}{\sum(x_i-\bar{X})}]]=\frac{\sum[(x_i-\bar{x})*E(y_i)]}{\sum(x_i-\bar{X})} ...$$ why the last move is legal? Because $y_i$ are variables and $(x_i - \bar{X})$ are constants (remember, we want to estimate $y_i$, like the height of a tree, wich we don't know (assume we are too small), which is a variable, by something we know, such as their circumference, which we know(I'm 1m63, I'm okay with calculating trees'circumferences)).

Therefore, by applying the properties of the expectation on $E[(x_i-\bar{X})*y_i]=$

$$E(aY)=aE(Y)$$

we got: I let you deduce...