a. The value of \displaystyle

  1. Consider the general case where the two classes have different means and possibly different variances:\displaystyle
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  2. Given two data points in 2 dimensions:\displaystyle \displaystyle \mathbf{x}^{(1)} \displaystyle = \displaystyle (x^{(1)},
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  3. Given two data points in 2 dimensions:\displaystyle \displaystyle \mathbf{x}^{(1)} \displaystyle = \displaystyle (x^{(1)},
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  4. Recall from the slides that the Gamma distribution can be reparameterized using the two parameters a, the shape parameter, and
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  5. Consider the same statistical set-up as above. Suppose we observe a data set consisting of 1000 observations as described in the
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  6. For the two following pmfs with one parameter \theta that are written in the form\displaystyle \displaystyle f_\theta (y)
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  7. Setting these equal to zero and isolating terms with a and b to one side, we obtain a system of linear equations\displaystyle
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  8. Let\displaystyle \psi : \mathbb {R} \times (0, \infty ) \displaystyle \to \mathbb {R}^2 \displaystyle (\mu , \sigma )
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  9. The normal distribution \mathcal{N}(\theta ,1) with with mean \theta and known variance \sigma ^2=1 has pdf\displaystyle
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  10. We will now work through an example where the principal components cannot easily determined by inspection.Given 4 data points in
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