# Changing units in linear regression

A linear model:

$$y_{i}=\alpha +\beta x_{i}+u_{i}$$

was estimated using least squares and the following estimate was calculated:

$$\widehat{\beta }=\frac{\sum (x_{i}-\overline{x})y_{i}}{\sum (x_{i}-\overline{x})x_{i}}$$

Now we want to change the units of the independent variable from Celsius to Fahrenheit:

$$z_{i}=32+1.8x_{i}$$

The OLS is now:

$$\widehat{\gamma }=\frac{\sum (z_{i}-\overline{z})y_{i}}{\sum (z_{i}-\overline{z})z_{i}}$$

I need to show that:

$$\widehat{\gamma }=\frac{5}{9}\widehat{\beta }$$

And I can't do it, trying for a couple of hours. Can you please assist ? Once I prove this, I already know how to prove the equation for the new intercept.

Thank you !

## 1 Answer

Insert $$z_i$$ into $$\hat \gamma$$. For the numerator you obtain $$\sum(1.8(x_i-\bar x_i))$$, as the 32 in each $$z_i$$ cancels with that in $$\bar z$$. The same happens in the denominator, leaving you with $$\sum(1.8(x_i-\bar x_i)) (32+1.8x_i)$$ But as $$\sum(x_i-\bar x_i)$$=0, this equals $$\sum(1.8(x_i-\bar x_i)) 1.8x_i$$. Thus $$\hat \gamma=\frac{1}{1.8}\hat\beta$$.

• How come you use the fact that the sum of xi-xbar is 0 if this term is multiplied by another term in the sum ? – user3275222 Mar 28 '19 at 23:23
• I use this only with the 'other term' being 32 (which can be taken out of the sum). The term involving the $1.8x_i$ is kept as it is and leads to the 1/1.8. – user1587692 Mar 28 '19 at 23:34