Question

Consider the Ridge LRR is being used to learn prediction function \( y_{\text{pred}} = w^T x \) where \( w, x \in \mathbb{R}^2 \) & mean absolute error (MAE) is used to measure the prediction error. A weight of 0.20 is associated with the regularizer. At an intermediate step of training process assume that the parameter \( w = [-3.00, 4.00]^T \). In the next step for the I/P \( x = [1.00, 2.00]^T \), the predicted value of \( y \) is noted. Let the relation b/w \( x = [x_1, x_2]^T \) & the true value of \( y \) be \( y_{\text{true}} = x_1 + x_2 \). The value of the overall regularized loss for instance is __________ (upto 2 decimal).

Hint:

Regularized loss functions are central to many machine learning models. Always break the calculation into two parts: first compute the data-dependent loss (like MSE, MAE, or cross-entropy), then compute the model-dependent regularization penalty (like L1 or L2 norm of weights), and finally combine them using the regularization strength parameter $\lambda$.

Answer 1

Correct Answer: 7

Step 1: Understanding the Question:
We need to calculate the value of a regularized loss function for a single data instance. The loss function has two components: a prediction error term (MAE) and a regularization term (Ridge, or L2 norm).
Step 2: Key Formula or Approach:
The overall regularized loss is defined as: \[ \text{Loss} = \text{Prediction Error} + \lambda \times \text{Regularization Term} \] Given the problem specifics: - Prediction Error = MAE = $|y_{\text{true}} - y_{\text{pred}}|$ - Regularization Term = L2 norm of weights = $||w||^2_2 = w_1^2 + w_2^2$ - $\lambda$ (weight of regularizer) = 0.20 So, the formula is: \[ \text{Loss} = |y_{\text{true}} - y_{\text{pred}}| + 0.20 \times (w_1^2 + w_2^2) \] Step 3: Detailed Explanation:
Let's calculate each part of the formula.
- Given values: - $w = [-3.00, 4.00]^T$ - $x = [1.00, 2.00]^T$ - $\lambda = 0.20$ - Calculate $y_{\text{true}$:} The true value is given by the relation $y_{\text{true}} = x_1 + x_2$. \[ y_{\text{true}} = 1.00 + 2.00 = 3.00 \] - Calculate $y_{\text{pred}$:} The predicted value is given by $y_{\text{pred}} = w^T x$. \[ y_{\text{pred}} = [-3.00, 4.00] \begin{pmatrix} 1.00
2.00 \end{pmatrix} = (-3.00 \times 1.00) + (4.00 \times 2.00) = -3.00 + 8.00 = 5.00 \] - Calculate the MAE term: \[ \text{MAE} = |y_{\text{true}} - y_{\text{pred}}| = |3.00 - 5.00| = |-2.00| = 2.00 \] - Calculate the Regularization term: \[ ||w||^2_2 = (-3.00)^2 + (4.00)^2 = 9.00 + 16.00 = 25.00 \] The weighted regularization term is $\lambda ||w||^2_2$. \[ 0.20 \times 25.00 = 5.00 \] - Calculate the Overall Regularized Loss: \[ \text{Loss} = \text{MAE} + \lambda ||w||^2_2 = 2.00 + 5.00 = 7.00 \] Step 4: Final Answer:
The value of the overall regularized loss for the given instance is 7.00.