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Unsupervised Learning: Difference between revisions
→EM Algorithm
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The lagrangian for this is <math>\max_{\phi_1,...,\phi_k} \min_{\beta} \left[ \sum_{i=1}^{m} \sum_{j=1}^{k}Q^{(i)}_{(j)} \log( \phi_j) + \beta(\sum_{j=1}^{k}\phi_j - 1) \right]</math><br> | The lagrangian for this is <math>\max_{\phi_1,...,\phi_k} \min_{\beta} \left[ \sum_{i=1}^{m} \sum_{j=1}^{k}Q^{(i)}_{(j)} \log( \phi_j) + \beta(\sum_{j=1}^{k}\phi_j - 1) \right]</math><br> | ||
This can be rewritten as <math>\max_{\phi_1,...,\phi_k} \min_{\beta} \sum_{j=1}^{k}\left[\log( \phi_j) \sum_{i=1}^{m} Q^{(i)}_{(j)} + \beta(\phi_j - 1/k) \right]</math><br> | This can be rewritten as <math>\max_{\phi_1,...,\phi_k} \min_{\beta} \sum_{j=1}^{k}\left[\log( \phi_j) \sum_{i=1}^{m} Q^{(i)}_{(j)} + \beta(\phi_j - 1/k) \right]</math><br> | ||
The dual of this problem is <math> \min_{\beta | The dual of this problem is <math> \min_{\beta} \max_{\phi_1,...,\phi_k} \sum_{j=1}^{k} \left[\log( \phi_j) \sum_{i=1}^{m} Q^{(i)}_{(j)} + \beta(\phi_j - 1/k) \right]</math><br> | ||
Taking the gradient w.r.t <math>\phi</math>, we get <math>\frac{1}{\phi_j}\sum_{i}Q^{(i)}_{(j)} + \beta = 0 \implies \phi_j = \frac{-1}{\beta}(\sum_{i}Q^{(i)}_{(j)})</math><br> | Taking the gradient w.r.t <math>\phi</math>, we get <math>\frac{1}{\phi_j}\sum_{i}Q^{(i)}_{(j)} + \beta = 0 \implies \phi_j = \frac{-1}{\beta}(\sum_{i}Q^{(i)}_{(j)})</math><br> | ||
Plugging this into our dual problem we get:<br> | Plugging this into our dual problem we get:<br> | ||