This problem involves updating the parameters of a multivariate linear regression model using gradient descent. The model is defined by the hypothesis function \( h_{\theta}(x) = \theta_0 + \theta_1 x_1 + \theta_2 x_2 \), and the cost function is the mean squared error. Given the initial parameters \( \theta = [\theta_0, \theta_1, \theta_2] = [0, 0.5, 1] \) and a learning rate \( \alpha = 0.8 \) for the first iteration and \( \alpha = 0.4 \) for the second iteration, we need to perform the updates for two iterations. The update rule for gradient descent is: \[ \theta_j := \theta_j - \alpha \frac{1}{m} \sum_{i=1}^{m} (h_{\theta}(x^{(i)}) - y^{(i)}) \cdot x_j^{(i)} \] where \( m \) is the number of training examples, \( x^{(i)} \) is the input features of the \( i \)-th training example, \( y^{(i)} \) is the actual output of the \( i \)-th training example, and \( x_j^{(i)} \) is the \( j \)-th feature of the \( i \)-th training example. Let's calculate the updates for each \( \theta_j \) for the first iteration: First, we need to compute the hypothesis for each instance: - For instance 1: \( h_{\theta}(x^{(1)}) = 0 + 0.5 \cdot (-1) + 1 \cdot 0.5 = 0 - 0.5 + 0.5 = 0 \) - For instance 2: \( h_{\theta}(x^{(2)}) = 0 + 0.5 \cdot (-0.5) + 1 \cdot 1 = 0 - 0.25 + 1 = 0.75 \) - For instance 3: \( h_{\theta}(x^{(3)}) = 0 + 0.5 \cdot 2 + 1 \cdot 0.5 = 0 + 1 + 0.5 = 1.5 \) Now, we calculate the gradient for each \( \theta_j \): - For \( \theta_0 \): \( \frac{1}{3} \sum_{i=1}^{3} (h_{\theta}(x^{(i)}) - y^{(i)}) \cdot x_0^{(i)} \), where \( x_0^{(i)} = 1 \) for all \( i \) (since \( x_0 \) is the bias term). - Gradient for \( \theta_0 \): \( \frac{1}{3} [(0 - 0) \cdot 1 + (0.75 - 1) \cdot 1 + (1.5 - 1) \cdot 1] = \frac{1}{3} [0 - 0.25 + 0.5] = \frac{1}{3} \cdot 0.25 = \frac{1}{12} \) - For \( \theta_1 \): - Gradient for \( \theta_1 \): \( \frac{1}{3} [(0 - 0) \cdot (-1) + (0.75 - 1) \cdot (-0.5) + (1.5 - 1) \cdot 2] = \frac{1}{3} [0 + 0.125 + 1] = \frac{1}{3} \cdot 1.125 = \frac{3.375}{12} \) - For \( \theta_2 \): - Gradient for \( \theta_2 \): \( \frac{1}{3} [(0 - 0) \cdot 0.5 + (0.75 - 1) \cdot 1 + (1.5 - 1) \cdot 0.5] = \frac{1}{3} [0 - 0.25 + 0.25] = 0 \) Now we update the parameters using the learning rate \( \alpha = 0.8 \): - \( \theta_0 := \theta_0 - 0.8 \cdot \frac{1}{12} = 0 - 0.8 \cdot \frac{1}{12} = 0 - \frac{1}{15} = -\frac{1}{15} \) - \( \theta_1 := \theta_1

This problem involves updating the parameters of a multivariate linear regression model using gradient descent. The model is defined by the hypothesis function $h_{\theta}(x) = \theta_0 + \theta_1 x_1 + \theta_2 x_2$, and the cost function is the mean squared error. Given the initial parameters $\theta = [\theta_0, \theta_1, \theta_2] = [0, 0.5, 1]$ and a learning rate $\alpha = 0.8$ for the first iteration and $\alpha = 0.4$ for the second iteration, we need to perform the updates for two iterations. The update rule for gradient descent is: $\theta_j := \theta_j - \alpha \frac{1}{m} \sum_{i=1}^{m} (h_{\theta}(x^{(i)}) - y^{(i)}) \cdot x_j^{(i)}$ where $m$ is the number of training examples, $x^{(i)}$ is the input features of the $i$-th training example, $y^{(i)}$ is the actual output of the $i$-th training example, and $x_j^{(i)}$ is the $j$-th feature of the $i$-th training example. Let's calculate the updates for each $\theta_j$ for the first iteration: First, we need to compute the hypothesis for each instance: - For instance 1: $h_{\theta}(x^{(1)}) = 0 + 0.5 \cdot (-1) + 1 \cdot 0.5 = 0 - 0.5 + 0.5 = 0$ - For instance 2: $h_{\theta}(x^{(2)}) = 0 + 0.5 \cdot (-0.5) + 1 \cdot 1 = 0 - 0.25 + 1 = 0.75$ - For instance 3: $h_{\theta}(x^{(3)}) = 0 + 0.5 \cdot 2 + 1 \cdot 0.5 = 0 + 1 + 0.5 = 1.5$ Now, we calculate the gradient for each $\theta_j$: - For $\theta_0$: $\frac{1}{3} \sum_{i=1}^{3} (h_{\theta}(x^{(i)}) - y^{(i)}) \cdot x_0^{(i)}$, where $x_0^{(i)} = 1$ for all $i$ (since $x_0$ is the bias term). - Gradient for $\theta_0$: $\frac{1}{3} [(0 - 0) \cdot 1 + (0.75 - 1) \cdot 1 + (1.5 - 1) \cdot 1] = \frac{1}{3} [0 - 0.25 + 0.5] = \frac{1}{3} \cdot 0.25 = \frac{1}{12}$ - For $\theta_1$: - Gradient for $\theta_1$: $\frac{1}{3} [(0 - 0) \cdot (-1) + (0.75 - 1) \cdot (-0.5) + (1.5 - 1) \cdot 2] = \frac{1}{3} [0 + 0.125 + 1] = \frac{1}{3} \cdot 1.125 = \frac{3.375}{12}$ - For $\theta_2$: - Gradient for $\theta_2$: $\frac{1}{3} [(0 - 0) \cdot 0.5 + (0.75 - 1) \cdot 1 + (1.5 - 1) \cdot 0.5] = \frac{1}{3} [0 - 0.25 + 0.25] = 0$ Now we update the parameters using the learning rate $\alpha = 0.8$: - $\theta_0 := \theta_0 - 0.8 \cdot \frac{1}{12} = 0 - 0.8 \cdot \frac{1}{12} = 0 - \frac{1}{15} = -\frac{1}{15}$ - ( \theta_1 := \theta_1