# Finding the original weights after data normalisation

Suppose we have vectors $$x_1$$ and $$x_2$$, each has ($$n$$) samples. Both $$x_1$$ and $$x_2$$ are my independent variables.

Suppose we also have a vector $$y$$ which has ($$n$$) samples and is y my dependent variable.

I would like to perform a linear regression in the form: $$y=b_0 + b_1 x_1 + b_2 x_2$$

however, before performing the regression, all my variables ($$y,x_1,x_2$$) were normalised such that their mean is zero and std. deviation is $$1$$.

After the regression, we get $$a_0, a_1$$ and $$a_2$$ which are the weights. How can I get the original weights ($$b_0,b_1,b_2$$) given the weights obtained from the normalised data ($$a_0,a_1$$ and $$a_2$$)?

Suppose the original vectors are $$y, x_1, x_2$$ and we have

$$\hat{y}=\frac{y-\mu_y}{\sigma_y},\hat{x}_i=\frac{x_i-\mu_{x_i}}{\sigma_{x_i}}$$

and we have

$$\hat{y}=a_0+a_1\hat{x}_1+a_2\hat{x}_2$$

then we have

$$\frac{y-\mu_y}{\sigma_y}=a_0+\sum_{i=1}^2a_i\frac{x_i-\mu_{x_i}}{\sigma_{x_i}}$$

Hence, $$b_0=\mu_y+\sigma_y\left(a_0-\sum_{i=1}^2 \frac{a_i\mu_{x_i}}{\sigma_{x_i}}\right)$$ For $$i\ge 1$$, $$b_i=\frac{\sigma_ya_i}{\sigma_{x_i}}$$

• Can you please clarify the math right after "Hence"? How did we get b0 and bi given the equations before? – HaneenSu Feb 21 '19 at 5:21
• I just perform some algebra, that is first I multiply both sides by $\sigma_y$, then I move $\mu_y$ over to the right hand side and collect coefficients of $x_i$ and the constant term. – Siong Thye Goh Feb 21 '19 at 5:29