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10 | 10.5 | medium | - 10.5 (\*\*) Consider a model in which the set of all hidden stochastic variables, denoted collectively by $\mathbf{Z}$ , comprises some latent variables $\mathbf{z}$ together with some model parameters $\boldsymbol{\theta}$ . Suppose we use a variational distribution that factorizes between latent variables and ... | We introduce a property of Dirac function:
$$\int \delta(\boldsymbol{\theta} - \boldsymbol{\theta}_0) f(\boldsymbol{\theta}) d\boldsymbol{\theta} = f(\boldsymbol{\theta}_0)$$
We first calculate the optimal $q(\mathbf{z}, \boldsymbol{\theta})$ by fixing $q_{\boldsymbol{\theta}}(\boldsymbol{\theta})$ . This is achie... | 5,129 |
10 | 10.6 | medium | The alpha family of divergences is defined by $D_{\alpha}(p||q) = \frac{4}{1 - \alpha^2} \left( 1 - \int p(x)^{(1+\alpha)/2} q(x)^{(1-\alpha)/2} dx \right)$. Show that the Kullback-Leibler divergence $\mathrm{KL}(p\|q)$ corresponds to $\alpha \to 1$ . This can be done by writing $p^{\epsilon} = \exp(\epsilon \ln p)... | Let's use the hint by first enforcing $\alpha \to 1$ .
$$\begin{split} D_{\alpha}(p||q) &= \frac{4}{1-\alpha^2} \Big\{ 1 - \int p^{(1+\alpha)/2} q^{(1-\alpha)/2} \, dx \Big\} \\ &= \frac{4}{1-\alpha^2} \Big\{ 1 - \int \frac{p}{p^{(1-\alpha)/2}} \left[ 1 + \frac{1-\alpha}{2} \ln q + O(\frac{1-\alpha}{2})^2 \right] dx ... | 4,965 |
10 | 10.8 | easy | Consider the variational posterior distribution for the precision of a univariate Gaussian whose parameters are given by $a_N = a_0 + \frac{N}{2}$ and $b_N = b_0 + \frac{1}{2} \mathbb{E}_{\mu} \left[ \sum_{n=1}^{N} (x_n - \mu)^2 + \lambda_0 (\mu - \mu_0)^2 \right].$. By using the standard results for the mean and varia... | According to Eq (B.27), we have:
$$\mathbb{E}[\tau] = \frac{a_0 + (N+1)/2}{b_0 + \frac{1}{2} \mathbb{E}_{\mu} \left[ \sum_{n=1}^{N} (x_n - \mu)^2 + \lambda_0 (\mu - \mu_0)^2 \right]}$$
$$\approx \frac{N/2}{\frac{1}{2} \mathbb{E}_{\mu} \left[ \sum_{n=1}^{N} (x_n - \mu)^2 \right]}$$
$$= \frac{N}{\mathbb{E}_{\mu} \left... | 831 |
10 | 10.9 | medium | By making use of the standard result $\mathbb{E}[\tau] = a_N/b_N$ for the mean of a gamma distribution, together with $\mu_N = \frac{\lambda_0 \mu_0 + N\overline{x}}{\lambda_0 + N}$, $\lambda_N = (\lambda_0 + N) \mathbb{E}[\tau].$, $a_N = a_0 + \frac{N}{2}$, and $b_N = b_0 + \frac{1}{2} \mathbb{E}_{\mu} \left[ \sum_{... | The underlying assumption of this problem is $a_0 = b_0 = \lambda_0 = 0$ . According to Eq $\mu_N = \frac{\lambda_0 \mu_0 + N\overline{x}}{\lambda_0 + N}$, Eq $\lambda_N = (\lambda_0 + N) \mathbb{E}[\tau].$ and the definition of variance, we can obtain:
$$\begin{split} \mathbb{E}[\mu^2] &= \lambda_N^{-1} + \mathbb{E}... | 1,875 |
11 | 11.1 | easy | Show that the finite sample estimator $\hat{f}$ defined by $\widehat{f} = \frac{1}{L} \sum_{l=1}^{L} f(\mathbf{z}^{(l)}).$ has mean equal to $\mathbb{E}[f]$ and variance given by $\operatorname{var}[\widehat{f}] = \frac{1}{L} \mathbb{E}\left[ (f - \mathbb{E}[f])^2 \right]$. | Based on definition, we can write down:
$$\begin{split} \mathbb{E}[\widehat{f}] &= \mathbb{E}[\frac{1}{L}\sum_{l=1}^{L}f(\mathbf{z}^{(l)})] \\ &= \frac{1}{L}\sum_{l=1}^{L}\mathbb{E}[f(\mathbf{z}^{(l)})] \\ &= \frac{1}{L}\cdot L\cdot \mathbb{E}[f] = \mathbb{E}[f] \end{split}$$
Where we have used the fact that the expe... | 1,560 |
11 | 11.10 | easy | Show that the simple random walk over the integers defined by $p(z^{(\tau+1)} = z^{(\tau)}) = 0.5$, $p(z^{(\tau+1)} = z^{(\tau)} + 1) = 0.25$, and $p(z^{(\tau+1)} = z^{(\tau)} - 1) = 0.25$ has the property that $\mathbb{E}[(z^{(\tau)})^2] = \mathbb{E}[(z^{(\tau-1)})^2] + 1/2$ and hence by induction that $\mathbb{E}[... | Based on definition and Eq (11.34)-(11.36), we can write down:
$$\mathbb{E}_{\tau}[(z^{(\tau)})^{2}] = 0.5 \cdot \mathbb{E}_{\tau-1}[(z^{\tau-1})^{2}] + 0.25 \cdot \mathbb{E}_{\tau-1}[(z^{\tau-1}+1)^{2}] + 0.25 \cdot \mathbb{E}_{\tau-1}[(z^{\tau-1}-1)^{2}]$$
$$= \mathbb{E}_{\tau-1}[(z^{\tau-1})^{2}] + 0.5$$
If the i... | 521 |
11 | 11.11 | medium | Show that the Gibbs sampling algorithm, discussed in Section 11.3, satisfies detailed balance as defined by $p^{}(\mathbf{z})T(\mathbf{z}, \mathbf{z}') = p^{}(\mathbf{z}')T(\mathbf{z}', \mathbf{z})$. | This problem requires you to know the definition of detailed balance, i.e., Eq (11.40):
$$p^{}(\mathbf{z})T(\mathbf{z},\mathbf{z}') = p^{}(\mathbf{z}')T(\mathbf{z}',\mathbf{z})$$
Note that here **z** and **z**' are the sampled values of $[z_1, z_2, ..., z_M]^T$ in two consecutive Gibbs Sampling step. Without loss o... | 1,513 |
11 | 11.12 | easy | Consider the distribution shown in Figure 11.15. Discuss whether the standard Gibbs sampling procedure for this distribution is ergodic, and therefore whether it would sample correctly from this distribution | Obviously, Gibbs Sampling is not ergodic for this specific distribution, and the quick reason is that neither the projection of the two shaded region on $z_1$ axis nor $z_2$ axis overlaps. For instance, we denote the left down shaded region as region 1. If the initial sample falls into this region, no matter how ma... | 1,568 |
11 | 11.14 | easy | Verify that the over-relaxation update $z_i' = \mu_i + \alpha(z_i - \mu_i) + \sigma_i(1 - \alpha_i^2)^{1/2}\nu$, in which $z_i$ has mean $\mu_i$ and variance $\sigma_i$ , and where $\nu$ has zero mean and unit variance, gives a value $z_i'$ with mean $\mu_i$ and variance $\sigma_i^2$ . | Based on definition, we can write down:
$$\begin{split} \mathbb{E}[z_i'] &= \mathbb{E}[\mu_i + \alpha(z_i - \mu_i) + \sigma_i (1 - \alpha_i^2)^{1/2} v] \\ &= \mu_i + \mathbb{E}[\alpha(z_i - \mu_i)] + \mathbb{E}[\sigma_i (1 - \alpha_i^2)^{1/2} v] \\ &= \mu_i + \alpha \cdot \mathbb{E}[z_i - \mu_i] + [\sigma_i (1 - \alph... | 1,535 |
11 | 11.15 | easy | Using $K(\mathbf{r}) = \frac{1}{2} \|\mathbf{r}\|^2 = \frac{1}{2} \sum_{i} r_i^2.$ and $H(\mathbf{z}, \mathbf{r}) = E(\mathbf{z}) + K(\mathbf{r})$, show that the Hamiltonian equation $\frac{\mathrm{d}z_i}{\mathrm{d}\tau} = \frac{\partial H}{\partial r_i}$ is equivalent to $r_i = \frac{\mathrm{d}z_i}{\mathrm{d}\tau}$. S... | Using Eq $H(\mathbf{z}, \mathbf{r}) = E(\mathbf{z}) + K(\mathbf{r})$, we can write down:
$$\frac{\partial H}{\partial r_i} = \frac{\partial K}{\partial r_i} = r_i$$
Comparing this with Eq $r_i = \frac{\mathrm{d}z_i}{\mathrm{d}\tau}$, we obtain Eq $\frac{\mathrm{d}z_i}{\mathrm{d}\tau} = \frac{\partial H}{\partial r_i}... | 750 |
11 | 11.16 | easy | By making use of $K(\mathbf{r}) = \frac{1}{2} \|\mathbf{r}\|^2 = \frac{1}{2} \sum_{i} r_i^2.$, $H(\mathbf{z}, \mathbf{r}) = E(\mathbf{z}) + K(\mathbf{r})$, and $p(\mathbf{z}, \mathbf{r}) = \frac{1}{Z_H} \exp(-H(\mathbf{z}, \mathbf{r})).$, show that the conditional distribution $p(\mathbf{r}|\mathbf{z})$ is a Gaussian... | According to Bayes' Theorem and Eq $p(\mathbf{z}) = \frac{1}{Z_p} \exp\left(-E(\mathbf{z})\right)$, $p(\mathbf{z}, \mathbf{r}) = \frac{1}{Z_H} \exp(-H(\mathbf{z}, \mathbf{r})).$, we have:
$$p(\mathbf{r}|\mathbf{z}) = \frac{p(\mathbf{z}, \mathbf{r})}{p(\mathbf{z})} = \frac{1/Z_H \cdot \exp(-H(\mathbf{z}, \mathbf{r}))}{... | 709 |
11 | 11.17 | easy | Verify that the two probabilities $\frac{1}{Z_H} \exp(-H(\mathcal{R})) \delta V_{\frac{1}{2}} \min \{1, \exp(-H(\mathcal{R}) + H(\mathcal{R}'))\}.$ and $\frac{1}{Z_H} \exp(-H(\mathcal{R}')) \delta V_{\frac{1}{2}} \min \{1, \exp(-H(\mathcal{R}') + H(\mathcal{R}))\}.$ are equal, and hence that detailed balance holds for ... | There are typos in Eq $\frac{1}{Z_H} \exp(-H(\mathcal{R})) \delta V_{\frac{1}{2}} \min \{1, \exp(-H(\mathcal{R}) + H(\mathcal{R}'))\}.$ and $\frac{1}{Z_H} \exp(-H(\mathcal{R}')) \delta V_{\frac{1}{2}} \min \{1, \exp(-H(\mathcal{R}') + H(\mathcal{R}))\}.$. The signs in the exponential of the second argument of the min f... | 1,280 |
11 | 11.2 | easy | Suppose that z is a random variable with uniform distribution over (0,1) and that we transform z using $y = h^{-1}(z)$ where h(y) is given by $z = h(y) \equiv \int_{-\infty}^{y} p(\widehat{y}) \,\mathrm{d}\widehat{y}$. Show that y has the distribution p(y). | What this problem wants us to prove is that if we use $y = h^{-1}(z)$ to transform the value of z to y, where z satisfies a uniform distribution over [0,1] and $h(\cdot)$ is defined by Eq(11.6), we can enforce y to satisfy a specific desired distribution p(y). Let's prove it beginning by Eq (11.1):
$$p^{}(y) = p(z... | 467 |
11 | 11.3 | easy | Given a random variable z that is uniformly distributed over (0, 1), find a transformation y = f(z) such that y has a Cauchy distribution given by $p(y) = \frac{1}{\pi} \frac{1}{1 + y^2}.$. | We use what we have obtained in the previous problem.
$$h(y) = \int_{-\infty}^{y} p(\hat{y}) d\hat{y}$$
$$= \int_{-\infty}^{y} \frac{1}{\pi} \frac{1}{1 + \hat{y}^2} d\hat{y}$$
$$= \tan^{-1}(y)$$
Therefore, since we know that $z = h(y) = \tan^{-1}(y)$ , we can obtain the transformation from z to y: $y = \tan(z)$ . | 318 |
11 | 11.4 | medium | Suppose that $z_1$ and $z_2$ are uniformly distributed over the unit circle, as shown in Figure 11.3, and that we make the change of variables given by $y_1 = z_1 \left(\frac{-2\ln z_1}{r^2}\right)^{1/2}$ and $y_2 = z_2 \left(\frac{-2\ln z_2}{r^2}\right)^{1/2}$. Show that $(y_1, y_2)$ will be distributed accordin... | First, I believe there is a typo in Eq $y_1 = z_1 \left(\frac{-2\ln z_1}{r^2}\right)^{1/2}$ and $y_2 = z_2 \left(\frac{-2\ln z_2}{r^2}\right)^{1/2}$. Both $\ln z_1$ and $\ln z_2$ should be $\ln(z_1^2 + z_2^2)$ . In the following, we will solve the problem under this assumption.
We only need to calculate the Jacob... | 3,089 |
11 | 11.5 | easy | - 11.5 (\*) Let z be a D-dimensional random variable having a Gaussian distribution with zero mean and unit covariance matrix, and suppose that the positive definite symmetric matrix $\Sigma$ has the Cholesky decomposition $\Sigma = \mathbf{L}\mathbf{L}^T$ where $\mathbf{L}$ is a lower-triangular matrix (i.e., o... | This is a linear transformation of $\mathbf{z}$ , we still obtain a Gaussian random variable $\mathbf{y}$ . We only need to match its moments (mean and variance). We know that $\mathbf{z} \sim \mathcal{N}(\mathbf{0}, \mathbf{I})$ , $\mathbf{\Sigma} = \mathbf{L}\mathbf{L}^T$ , and $\mathbf{y} = \boldsymbol{\mu} + \... | 1,529 |
11 | 11.6 | medium | - 11.6 (\*\*) In this exercise, we show more carefully that rejection sampling does indeed draw samples from the desired distribution $p(\mathbf{z})$ . Suppose the proposal distribution is $q(\mathbf{z})$ and show that the probability of a sample value $\mathbf{z}$ being accepted is given by $\widetilde{p}(\math... | This problem is all about definition. According to the description of rejection sampling, we know that: for a specific value $\mathbf{z}_0$ (drawn from $q(\mathbf{z})$ ), we will generate a random variable $u_0$ , which satisfies a uniform distribution in the interval $[0, kq(\mathbf{z}_0)]$ , and if the generated... | 2,556 |
11 | 11.7 | easy | Suppose that z has a uniform distribution over the interval [0,1]. Show that the variable $y = b \tan z + c$ has a Cauchy distribution given by $q(z) = \frac{k}{1 + (z - c)^2 / b^2}.$. | Notice that the symbols used in the main text is different from those in the problem description. in the following, we will use those in the main text. Namely, y satisfies a uniform distribution on interval [0,1], and $z = b \tan y + c$ . Then we aims to prove Eq $q(z) = \frac{k}{1 + (z - c)^2 / b^2}.$. Since we know ... | 1,112 |
11 | 11.8 | medium | Determine expressions for the coefficients $k_i$ in the envelope distribution $q(z) = k_i \lambda_i \exp\{-\lambda_i (z - z_{i-1})\} \qquad z_{i-1} < z \leqslant z_i.$ for adaptive rejection sampling using the requirements of continuity and normalization. | There is a typo in Eq $q(z) = k_i \lambda_i \exp\{-\lambda_i (z - z_{i-1})\} \qquad z_{i-1} < z \leqslant z_i.$, which is not difficult to observe, if we carefully examine Fig.11.6. The correct form should be:
$$q_i(z) = k_i \lambda_i \exp\{-\lambda_i (z - z_i)\}, \quad \tilde{z}_{i-1,i} < z \le \tilde{z}_{i,i+1}, \qu... | 1,990 |
11 | 11.9 | medium | By making use of the technique discussed in Section 11.1.1 for sampling from a single exponential distribution, devise an algorithm for sampling from the piecewise exponential distribution defined by $q(z) = k_i \lambda_i \exp\{-\lambda_i (z - z_{i-1})\} \qquad z_{i-1} < z \leqslant z_i.$. | In this problem, we will still use the same notation as in the previous one. First, we need to know the probability of sampling from each segment. Notice that Eq $q(z) = k_i \lambda_i \exp\{-\lambda_i (z - z_{i-1})\} \qquad z_{i-1} < z \leqslant z_i.$ is not correctly normalized, we first calculate its normalization co... | 3,763 |
12 | 12.11 | medium | Show that in the limit $\sigma^2 \to 0$ , the posterior mean for the probabilistic PCA model becomes an orthogonal projection onto the principal subspace, as in conventional PCA. | Taking $\sigma^2 \to 0$ in $\mathbf{M} = \mathbf{W}^{\mathrm{T}} \mathbf{W} + \sigma^{2} \mathbf{I}.$ and substituting into $\mathbb{E}[\mathbf{z}|\mathbf{x}] = \mathbf{M}^{-1}\mathbf{W}_{\mathrm{ML}}^{\mathrm{T}}(\mathbf{x} - \overline{\mathbf{x}})$ we obtain the posterior mean for probabilistic PCA in the form
$$(... | 1,139 |
12 | 12.28 | medium | **** Use the transformation property $= p_{x}(g(y)) |g'(y)|.$ of a probability density under a change of variable to show that any density p(y) can be obtained from a fixed density q(x) that is everywhere nonzero by making a nonlinear change of variable y = f(x) in which f(x) is a monotonic function so that $0 \le f'(... | If we assume that the function y = f(x) is *strictly* monotonic, which is necessary to exclude the possibility for spikes of infinite density in p(y), we are guaranteed that the inverse function $x = f^{-1}(y)$ exists. We can then use $= p_{x}(g(y)) |g'(y)|.$ to write
$$p(y) = q(f^{-1}(y)) \left| \frac{\mathrm{d}f^{... | 617 |
12 | 12.3 | easy | Verify that the eigenvectors defined by $\mathbf{u}_i = \frac{1}{(N\lambda_i)^{1/2}} \mathbf{X}^{\mathrm{T}} \mathbf{v}_i.$ are normalized to unit length, assuming that the eigenvectors $\mathbf{v}_i$ have unit length. | According to Eq $\mathbf{u}_i = \frac{1}{(N\lambda_i)^{1/2}} \mathbf{X}^{\mathrm{T}} \mathbf{v}_i.$, we can obtain:
$$\mathbf{u}_i^T \mathbf{u}_i = \frac{1}{N\lambda_i} \mathbf{v}_i^T \mathbf{X} \mathbf{X}^T \mathbf{v}_i$$
We left multiply $\mathbf{v}_i^T$ on both sides of Eq $\frac{1}{N} \mathbf{X} \mathbf{X}^{\ma... | 745 |
12 | 12.4 | easy | - 12.4 (\*) Suppose we replace the zero-mean, unit-covariance latent space distribution $p(\mathbf{z}) = \mathcal{N}(\mathbf{z}|\mathbf{0}, \mathbf{I}).$ in the probabilistic PCA model by a general Gaussian distribution of the form $\mathcal{N}(\mathbf{z}|\mathbf{m}, \boldsymbol{\Sigma})$ . By redefining the paramete... | We know $p(\mathbf{z}) = \mathcal{N}(\mathbf{z}|\mathbf{m}, \boldsymbol{\Sigma})$ , and $p(\mathbf{x}|\mathbf{z}) = \mathcal{N}(\mathbf{x}|\mathbf{W}\mathbf{z} + \boldsymbol{\mu}, \sigma^2\mathbf{I})$ . According to Eq (2.113)-(2.115), we have:
$$p(\mathbf{x}) = \mathcal{N}(\mathbf{x}|\mathbf{W}\mathbf{m} + \boldsym... | 872 |
12 | 12.6 | easy | Draw a directed probabilistic graph for the probabilistic PCA model described in Section 12.2 in which the components of the observed variable x are shown explicitly as separate nodes. Hence verify that the probabilistic PCA model has the same independence structure as the naive Bayes model discussed in Section 8.2.2. | Omitting the parameters, W, $\mu$ and $\sigma$ , leaving only the stochastic variables z and x, the graphical model for probabilistic PCA is identical with the the 'naive Bayes' model24 in Section 8.2.2. Hence these two models exhibit the same independence structure. | 270 |
13 | 13.1 | easy | Use the technique of d-separation, discussed in Section 8.2, to verify that the Markov model shown in Figure 13.3 having N nodes in total satisfies the conditional independence properties $p(\mathbf{x}_n|\mathbf{x}_1,\dots,\mathbf{x}_{n-1}) = p(\mathbf{x}_n|\mathbf{x}_{n-1})$ for n = 2, ..., N. Similarly, show that a m... | Since the arrows on the path from $x_m$ to $x_n$ , with m < n - 1, will meet head-to-tail at $x_{n-1}$ , which is in the conditioning set, all such paths are blocked by $x_{n-1}$ and hence $p(\mathbf{x}_n|\mathbf{x}_1,\dots,\mathbf{x}_{n-1}) = p(\mathbf{x}_n|\mathbf{x}_{n-1})$ holds.
The same argument applies in... | 443 |
13 | 13.13 | medium | - 13.13 (\*\*) Use the definition $\mu_{f_s \to x}(x) \equiv \sum_{X_s} F_s(x, X_s)$ of the messages passed from a factor node to a variable node in a factor graph, together with the expression $p(\mathbf{x}_1, \dots, \mathbf{x}_N, \mathbf{z}_1, \dots, \mathbf{z}_N) = p(\mathbf{z}_1) \left[ \prod_{n=2}^N p(\mathbf{z}_... | Using $\mu_{f_s \to x}(x) \equiv \sum_{X_s} F_s(x, X_s)$, we can rewrite $\alpha(\mathbf{z}_n) = \mu_{f_n \to \mathbf{z}_n}(\mathbf{z}_n)$ as
$$\alpha(\mathbf{z}_n) = \sum_{\mathbf{z}_{n-1}} F_n(\mathbf{z}_n, \{\mathbf{z}_1, \dots, \mathbf{z}_{n-1}\}), \tag{148}$$
where $F_n(\cdot)$ is the product of all factors co... | 1,314 |
13 | 13.17 | easy | Show that the directed graph for the input-output hidden Markov model, given in Figure 13.18, can be expressed as a tree-structured factor graph of the form shown in Figure 13.15 and write down expressions for the initial factor $h(\mathbf{z}_1)$ and for the general factor $f_n(\mathbf{z}_{n-1}, \mathbf{z}_n)$ wher... | The emission probabilities over observed variables $\mathbf{x}_n$ are absorbed into the corresponding factors, $f_n$ , analogously to the way in which14 was transformed into15. The factors then take the form
$$h(\mathbf{z}_1) = p(\mathbf{z}_1|\mathbf{u}_1)p(\mathbf{x}_1|\mathbf{z}_1,\mathbf{u}_1)$$
(150)
$$f_n(\m... | 455 |
13 | 13.22 | medium | Using $c_1\widehat{\alpha}(\mathbf{z}_1) = p(\mathbf{z}_1)p(\mathbf{x}_1|\mathbf{z}_1).$, together with the definitions $p(\mathbf{x}_n|\mathbf{z}_n) = \mathcal{N}(\mathbf{x}_n|\mathbf{C}\mathbf{z}_n, \mathbf{\Sigma}).$ and $p(\mathbf{z}_1) = \mathcal{N}(\mathbf{z}_1 | \boldsymbol{\mu}_0, \mathbf{V}_0).$ and the result... | Using $p(\mathbf{x}_n|\mathbf{z}_n) = \mathcal{N}(\mathbf{x}_n|\mathbf{C}\mathbf{z}_n, \mathbf{\Sigma}).$, $p(\mathbf{z}_1) = \mathcal{N}(\mathbf{z}_1 | \boldsymbol{\mu}_0, \mathbf{V}_0).$ and $\widehat{\alpha}(\mathbf{z}_n) = \mathcal{N}(\mathbf{z}_n | \boldsymbol{\mu}_n, \mathbf{V}_n).$, we can write $c_1\widehat{\al... | 3,668 |
13 | 13.4 | medium | Consider a hidden Markov model in which the emission densities are represented by a parametric model $p(\mathbf{x}|\mathbf{z},\mathbf{w})$ , such as a linear regression model or a neural network, in which $\mathbf{w}$ is a vector of adaptive parameters. Describe how the parameters $\mathbf{w}$ can be learned from ... | The learning of w would follow the scheme for maximum learning described in Section 13.2.1, with w replacing $\phi$ . As discussed towards the end of Section 13.2.1, the precise update formulae would depend on the form of regression model used and how it is being used.
The most obvious situation where this would occu... | 717 |
13 | 13.8 | medium | - 13.8 (\*\*) For a hidden Markov model having discrete observations governed by a multinomial distribution, show that the conditional distribution of the observations given the hidden variables is given by $p(\mathbf{x}|\mathbf{z}) = \prod_{i=1}^{D} \prod_{k=1}^{K} \mu_{ik}^{x_i z_k}$ and the corresponding M step equ... | Only the final term of $Q(\theta, \theta^{\text{old}})$ given by $Q(\boldsymbol{\theta}, \boldsymbol{\theta}^{\text{old}}) = \sum_{k=1}^{K} \gamma(z_{1k}) \ln \pi_k + \sum_{n=2}^{N} \sum_{j=1}^{K} \sum_{k=1}^{K} \xi(z_{n-1,j}, z_{nk}) \ln A_{jk} + \sum_{n=1}^{N} \sum_{k=1}^{K} \gamma(z_{nk}) \ln p(\mathbf{x}_n | \bol... | 2,368 |
13 | 13.9 | medium | Use the d-separation criterion to verify that the conditional independence properties (13.24)–(13.31) are satisfied by the joint distribution for the hidden Markov model defined by $p(\mathbf{x}_1, \dots, \mathbf{x}_N, \mathbf{z}_1, \dots, \mathbf{z}_N) = p(\mathbf{z}_1) \left[ \prod_{n=2}^N p(\mathbf{z}_n | \mathbf{z}... | We can verify all these independence properties using d-separation by refering to5.
$p(\mathbf{x}_{n+1}, \dots, \mathbf{x}_{N}|\mathbf{z}_{n}) \qquad$ follows from the fact that arrows on paths from any of $\mathbf{x}_1, \dots, \mathbf{x}_n$ to any of $\mathbf{x}_{n+1}, \dots, \mathbf{x}_N$ meet head-to-tail or ta... | 2,825 |
14 | 14.1 | medium | - 14.1 (\*\*) Consider a set models of the form $p(\mathbf{t}|\mathbf{x}, \mathbf{z}_h, \boldsymbol{\theta}_h, h)$ in which $\mathbf{x}$ is the input vector, $\mathbf{t}$ is the target vector, h indexes the different models, $\mathbf{z}_h$ is a latent variable for model h, and $\boldsymbol{\theta}_h$ is the ... | The required predictive distribution is given by
$$p(\mathbf{t}|\mathbf{x}, \mathbf{X}, \mathbf{T}) = \sum_{h} p(h) \sum_{\mathbf{z}_{h}} p(\mathbf{z}_{h}) \int p(\mathbf{t}|\mathbf{x}, \boldsymbol{\theta}_{h}, \mathbf{z}_{h}, h) p(\boldsymbol{\theta}_{h}|\mathbf{X}, \mathbf{T}, h) d\boldsymbol{\theta}_{h}, \quad (154... | 1,289 |
14 | 14.13 | easy | Verify that the complete-data log likelihood function for the mixture of linear regression models is given by $\ln p(\mathbf{t}, \mathbf{Z}|\boldsymbol{\theta}) = \sum_{n=1}^{N} \sum_{k=1}^{K} z_{nk} \ln \left\{ \pi_k \mathcal{N}(t_n | \mathbf{w}_k^{\mathrm{T}} \boldsymbol{\phi}_n, \beta^{-1}) \right\}.$. | Starting from the mixture distribution in $p(t|\boldsymbol{\theta}) = \sum_{k=1}^{K} \pi_k \mathcal{N}(t|\mathbf{w}_k^{\mathrm{T}} \boldsymbol{\phi}, \beta^{-1})$, we follow the same steps as for mixtures of Gaussians, presented in Section 9.2. We introduce a *K*-nomial latent variable, **z**, such that the joint distr... | 1,118 |
14 | 14.15 | easy | - 14.15 (\*) We have already noted that if we use a squared loss function in a regression problem, the corresponding optimal prediction of the target variable for a new input vector is given by the conditional mean of the predictive distribution. Show that the conditional mean for the mixture of linear regression mode... | The predictive distribution from the mixture of linear regression models for a new input feature vector, $\hat{\phi}$ , is obtained from $p(t|\boldsymbol{\theta}) = \sum_{k=1}^{K} \pi_k \mathcal{N}(t|\mathbf{w}_k^{\mathrm{T}} \boldsymbol{\phi}, \beta^{-1})$, with $\phi$ replaced by $\hat{\phi}$ . Calculating the ex... | 910 |
14 | 14.17 | medium | Consider a mixture model for a conditional distribution $p(t|\mathbf{x})$ of the form
$$p(t|\mathbf{x}) = \sum_{k=1}^{K} \pi_k \psi_k(t|\mathbf{x})$$
$p(t|\mathbf{x}) = \sum_{k=1}^{K} \pi_k \psi_k(t|\mathbf{x})$
in which each mixture component $\psi_k(t|\mathbf{x})$ is itself a mixture model. Show that this two-l... | If we define $\psi_k(t|\mathbf{x})$ in $p(t|\mathbf{x}) = \sum_{k=1}^{K} \pi_k \psi_k(t|\mathbf{x})$ as
$$\psi_k(t|\mathbf{x}) = \sum_{m=1}^{M} \lambda_{mk} \phi_{mk}(t|\mathbf{x}),$$
we can rewrite $p(t|\mathbf{x}) = \sum_{k=1}^{K} \pi_k \psi_k(t|\mathbf{x})$ as
$$p(t|\mathbf{x}) = \sum_{k=1}^{K} \pi_k \sum_{m=1}... | 3,368 |
14 | 14.3 | easy | By making use of Jensen's inequality $f\left(\sum_{i=1}^{M} \lambda_i x_i\right) \leqslant \sum_{i=1}^{M} \lambda_i f(x_i)$, for the special case of the convex function $f(x) = x^2$ , show that the average expected sum-of-squares error $E_{AV}$ of the members of a simple committee model, given by $E_{\text{AV}} = \f... | We start by rearranging the r.h.s. of $E_{\text{AV}} = \frac{1}{M} \sum_{m=1}^{M} \mathbb{E}_{\mathbf{x}} \left[ \epsilon_m(\mathbf{x})^2 \right].$, by moving the factor 1/M inside the sum and the expectation operator outside the sum, yielding
$$\mathbb{E}_{\mathbf{x}} \left[ \sum_{m=1}^{M} \frac{1}{M} \epsilon_m(\mat... | 966 |
14 | 14.5 | medium | $ Consider a committee in which we allow unequal weighting of the constituent models, so that
$$y_{\text{COM}}(\mathbf{x}) = \sum_{m=1}^{M} \alpha_m y_m(\mathbf{x}). \tag{14.55}$$
In order to ensure that the predictions $y_{\text{COM}}(\mathbf{x})$ remain within sensible limits, suppose that we require that they ... | To prove that $\alpha_m \geqslant 0, \qquad \sum_{m=1}^{M} \alpha_m = 1.$ is a sufficient condition for $y_{\min}(\mathbf{x}) \leqslant y_{\text{COM}}(\mathbf{x}) \leqslant y_{\max}(\mathbf{x}).$ we have to show that $y_{\min}(\mathbf{x}) \leqslant y_{\text{COM}}(\mathbf{x}) \leqslant y_{\max}(\mathbf{x}).$ follows fro... | 2,788 |
14 | 14.6 | easy | By differentiating the error function $= (e^{\alpha_m/2} - e^{-\alpha_m/2}) \sum_{n=1}^N w_n^{(m)} I(y_m(\mathbf{x}_n) \neq t_n) + e^{-\alpha_m/2} \sum_{n=1}^N w_n^{(m)}.$ with respect to $\alpha_m$ , show that the parameters $\alpha_m$ in the AdaBoost algorithm are updated using $\alpha_m = \ln \left\{ \frac{1 - \e... | If we differentiate $= (e^{\alpha_m/2} - e^{-\alpha_m/2}) \sum_{n=1}^N w_n^{(m)} I(y_m(\mathbf{x}_n) \neq t_n) + e^{-\alpha_m/2} \sum_{n=1}^N w_n^{(m)}.$ w.r.t. $\alpha_m$ we obtain
$$\frac{\partial E}{\partial \alpha_m} = \frac{1}{2} \left( (e^{\alpha_m/2} + e^{-\alpha_m/2}) \sum_{n=1}^{N} w_n^{(m)} I(y_m(\mathbf{x... | 1,018 |
14 | 14.9 | easy | Show that the sequential minimization of the sum-of-squares error function for an additive model of the form $f_m(\mathbf{x}) = \frac{1}{2} \sum_{l=1}^{m} \alpha_l y_l(\mathbf{x})$ in the style of boosting simply involves fitting each new base classifier to the residual errors $t_n f_{m-1}(\mathbf{x}_n)$ from the pre... | The sum-of-squares error for the additive model of $f_m(\mathbf{x}) = \frac{1}{2} \sum_{l=1}^{m} \alpha_l y_l(\mathbf{x})$ is defined as
$$E = \frac{1}{2} \sum_{n=1}^{N} (t_n - f_m(\mathbf{x}_n))^2.$$
Using $f_m(\mathbf{x}) = \frac{1}{2} \sum_{l=1}^{m} \alpha_l y_l(\mathbf{x})$, we can rewrite this as
$$\frac{1}{2} ... | 651 |
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