Partial Derivative of Loss Function for Linear Regression

Question

Consider the loss function of linear regression given by $J(\theta_0, \theta_1)$ . Given $(\theta_0, \theta_1)$ = 0, 0.5. Estimate $\frac{\partial J}{\partial \theta_1}$ using the data points below:

$\begin{tabular}{|c|c|c|c|c|c|} \hline x & 2 & 4 & 7 & 8 & 10\\ \hline y & 1 & 2 & 2.5 & 3.5 & 5.5\\\hline \end{tabular}$

Solution

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We know that $(\theta_0, \theta_1)$ = 0, 0.5 means that $y = ax + b \Rightarrow y = 0.5x$

$\begin{align*} \left\frac{\partial J}{\partial \theta_1} = \frac{1}{n} \ \[ \sum_{i=1}^{n} (h_{\theta}(x^{(i)})) - y^{(i)}) \ x_j^{(i)} \] \end{align*}$

From the table, we get $n=5$ .

$\begin{align*} \left\frac{\partial J}{\partial \theta_1} = \frac{1}{5} \ [ [ ((0.5 \ * \ 2) \ - \ 1) * 2 ] + [ ((0.5 \ * \ 4) \ - \ 2) * 4 ] + [ ((0.5 \ * \ 7) \ - \ 2.5) * 7 ] + [ ((0.5 \ * \ 8) \ - \ 3.5) * 8 ] + [ ((0.5 \ * \ 10) \ - \ 5.5) * 10 ] ] \end{align*}$

$\begin{align*} \left = \frac{1}{5} \ [ 0 \ + \ 0 \ + \ 7 \ + \ 4 \ + \ (-5)] = 1.2 \end{align*}$

Thus, $\frac{\partial J}{\partial \theta_1} = 1.2$

Question

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