Question
Compute the Linear Discriminant projection for the following two-dimensional dataset:
X1 =(x1, x2) = (4, 2), (2, 2), (3, 2), (3, 5), (3, 4)
X2 = (x1, x2) = (8, 7), (9, 6), (7, 7), (9, 8), (10, 9)
Answer :
Word Count : 657
Linear Discriminant Analysis (LDA) finds a projection vector \( w \) that maximizes the separation between two classes by maximizing the between-class scatter while minimizing the within-class scatter. Below are the step-by-step computations. --- ### Step 1: Compute the Mean Vectors The mean vector for each class is calculated as: #### Class \( X_1 \) Mean: \[ \mu_1 = \frac{1}{N_1} \sum_{i=1}^{N_1} X_1^i \] \[ \mu_1 = \frac{1}{5} \left( \begin{bmatrix} 4 \\ 2 \end{bmatrix} + \begin{bmatrix} 2 \\ 2 \end{bmatrix} + \begin{bmatrix} 3 \\ 2 \end{bmatrix} + \begin{bmatrix} 3 \\ 5 \end{bmatrix} + \begin{bmatrix} 3 \\ 4 \end{bmatrix} \right) \] \[ \mu_1 = \frac{1}{5} \begin{bmatrix} 4+2+3+3+3 \\ 2+2+2+5+4 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 15 \\ 15 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \end{bmatrix} \] #### Class \( X_2 \) Mean: \[ \mu_2 = \frac{1}{5} \left( \begin{bmatrix} 8 \\ 7 \end{bmatrix} + \begin{bmatrix} 9 \\ 6 \end{bmatrix} + \begin{bmatrix} 7 \\ 7 \end{bmatrix} + \begin{bmatrix} 9 \\ 8 \end{bmatrix} + \begin{bmatrix} 10 \\ 9 \end{bmatrix} \right) \] ____ ____ __________ ________ ________ _______ _____ ______ ________ ________ ________.
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Linear Discriminant Analysis (LDA) finds a projection vector \( w \) that maximizes the separation between two classes by maximizing the between-class scatter while minimizing the within-class scatter. Below are the step-by-step computations. --- ### Step 1: Compute the Mean Vectors The mean vector for each class is calculated as: #### Class \( X_1 \) Mean: \[ \mu_1 = \frac{1}{N_1} \sum_{i=1}^{N_1} X_1^i \] \[ \mu_1 = \frac{1}{5} \left( \begin{bmatrix} 4 \\ 2 \end{bmatrix} + \begin{bmatrix} 2 \\ 2 \end{bmatrix} + \begin{bmatrix} 3 \\ 2 \end{bmatrix} + \begin{bmatrix} 3 \\ 5 \end{bmatrix} + \begin{bmatrix} 3 \\ 4 \end{bmatrix} \right) \] \[ \mu_1 = \frac{1}{5} \begin{bmatrix} 4+2+3+3+3 \\ 2+2+2+5+4 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 15 \\ 15 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \end{bmatrix} \] #### Class \( X_2 \) Mean: \[ \mu_2 = \frac{1}{5} \left( \begin{bmatrix} 8 \\ 7 \end{bmatrix} + \begin{bmatrix} 9 \\ 6 \end{bmatrix} + \begin{bmatrix} 7 \\ 7 \end{bmatrix} + \begin{bmatrix} 9 \\ 8 \end{bmatrix} + \begin{bmatrix} 10 \\ 9 \end{bmatrix} \right) \] ____ ____ __________ ________ ________ _______ _____ ______ ________ ________ ________.
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