MSE(b) = E[(b - b)²]

Expand the quadratic expression

E[(b - b)²] = E[b² - 2bb + b²]

= E[b²] - 2E[b]b + b²

What am I trying to get?

Bias(b)² = (E[b] - b)² = (E[b])² - 2E[b]b + b²

Var[b] = E[(b- E[b])²] = E[b²] - (E[b])² (show this)

So, to get what I want I just add and subtract (E[b])²

E[(b - b)²] = E[b²] + {-(E[b])² + (E[b])²}- 2E[b]b + b²

⁼ {E[b²] - (E[b])²} + {(E[b])² - 2E[b]b + b²}

= Var[b] + Bias(b)²

Sc_i = (Sx_i)/Sx_i²

is the sum of the deviations from the mean divided by a constant; but the sum of the deviations from the mean is zero.

The numerator of Sc_iX_i is Sx_iX_i

This is the same as Sx_i(X_i - Xbar), since subtracting XbarSx_i is the same as subtracting zero.

This is the same as Sx_i², which is the same as the denominator of Sc_iX_i

Hence Sc_iX_i = 1

Sure, move all constants outside the expected value operator

E[(b - b)/s_b] = E[b]/s_b - b/s_b

= b/s_b - b/s_b= 0

The key is to note that the term after the minus sign is a constant; constants don't vary.

var[(b - b)/s_b] = var[(1/s_b)*b]

Recall that for a random variable X, var[aX] = a²var[X]. Treat (1/s_b) as the constant a:

var[(1/s_b)*b] = var[b]*(1/s_b)²

= s_b²*(1/s_b)² = 1

Pr(Z>1.29) = .098

Pr(Z< -1.1) = .136

Pr(b>1) = Pr(Z > (1-(-2))/3 ) = Pr(Z>1) = .159