Question

(14 points) We have a dataset D = {x_n, t_n} (n = 1,..., N) and each data point x_n has a weighting factor p_n ? 0. Thus, the sum-of-squares error function becomes $qquad E_D(mathbf{w}) = frac{1}{2} sum_{n=1}^N p_n {t_n - mathbf{w}^T Phi(mathbf{x}_n)}^2.$ Derive step by step an expression for the solution mathbf{w}^* that minimizes this error function (10 points). Give two interpretations of the weighted sum-of-squares error function (i.e., what the error function measures) in terms of (i) data dependent noise variance (2 points) and (ii) replicated data points (2 points).

          (14 points) We have a dataset D = {x_n, t_n} (n = 1,..., N) and each data point x_n has
a weighting factor p_n ? 0. Thus, the sum-of-squares error function becomes
$qquad E_D(mathbf{w}) = frac{1}{2} sum_{n=1}^N p_n {t_n - mathbf{w}^T Phi(mathbf{x}_n)}^2.$ 
Derive step by step an expression for the solution mathbf{w}^* that minimizes this error function
(10 points). Give two interpretations of the weighted sum-of-squares error function (i.e.,
what the error function measures) in terms of (i) data dependent noise variance (2 points)
and (ii) replicated data points (2 points).

$(14 points) We have a dataset D = xn, tn (n = 1,..., N) and each data point xn has a weighting factor pn ? 0. Thus, the sum-of-squares error function becomes qquad ED(mathbfw) = frac12 sumn=1^N pn tn - mathbfw^T Phi(mathbfxn)^2. Derive step by step an expression for the solution mathbfw^* that minimizes this error function (10 points). Give two interpretations of the weighted sum-of-squares error function (i.e., what the error function measures) in terms of (i) data dependent noise variance (2 points) and (ii) replicated data points (2 points).$

Added by Donald H.

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