How to obtain original coefficients after performing linear regression on normalized data?
Coefficients of the linear regression for unnormalized features

If parameters in the normalized space are denoted as $(\theta_0', \theta_1', \theta_2')$, parameters in the original space $(\theta_0, \theta_1, \theta_2)$ can be derived as follows
$$\begin{align*}
y' &amp;= \theta_2'x_2'+\theta_1'x_1'+\theta_0'\\
\frac{y-\mu_Y}{\sigma_Y} &amp;= \theta_2'\frac{x_2 - \mu_{X_2}}{\sigma_{X_2}} + \theta_1'\frac{x_1 - \mu_{X_1}}{\sigma_{X_1}} +\theta_0' \\
 y &amp;= \overbrace{\left(\frac{\sigma_{Y}}{\sigma_{X_2}}\theta_2'\right)}^{\theta_2}x_2+ \overbrace{\left(\frac{\sigma_{Y}}{\sigma_{X_1}}\theta_1'\right)}^{\theta_1}x_1
 + \overbrace{\sigma_Y\left(-\theta_2'\frac{\mu_{X_2}}{\sigma_{X_2}}-\theta_1'\frac{\mu_{X_1}}{\sigma_{X_1}} + \theta_0'\right) + \mu_Y}^{\theta_0}
\end{align*}$$

Generalization to D features

$$\begin{align*}
\theta_d = \left\{\begin{matrix}
 \sigma_Y \left(\theta_0' - \sum_{i=1}^{D}\theta_i'\frac{\mu_{X_i}}{\sigma_{X_i}} \right) + \mu_Y&amp; d=0\\ 
\frac{\sigma_{Y}}{\sigma_{X_d}}\theta_d' &amp; d &gt; 0
\end{matrix}\right.
\end{align*}$$

A trick

For visualization, we can plot the plane in original, un-normalized space without changing the parameters (Thetas). We only need to re-label (re-scale) the plot axes.

For example, a point $(x_1', x_2', y')$ in the normalized space corresponds to point $$(x_1, x_2, y)=(\sigma_{X_1}x_1'+\mu_{X_1}, \sigma_{X_2}x_2'+\mu_{X_2}, \sigma_{Y}y'+\mu_{Y})$$

in the original space. So you just need to rename the plot axes from $(x_1', x_2', y')$ to $(x_1, x_2, y)$.

Note that $y'=\theta_2'x_2'+\theta_1'x_1'+\theta_0'$ is still calculated using normalized features.