# Explain the expected value of an indicator variable in section 9.3.1 of PRML

The posterior distribution in Bishop's "Pattern Recognition and Machine Learning" book is written as: $$p(\textbf{Z} \mid \textbf{X}, \mu, \pi) \approx \Pi_n \Pi_k [\pi_k \mathcal{N}(x_n \mid \mu_k, \Sigma_k)]^{z_{nk}}$$

where $$z_n$$ is a 1-of-K binary rperesentation in which a particular element $$z_{nk} \in \{0,1 \}$$ and $$\sum_k z_{nk} = 1$$ and $$z_{nk}$$ denotes the indicator for a data point $$x_n$$.

Then the author states that: The expected value of the indicator variable $$z_{nk}$$ under this posterior distribution is then given by, \begin{align} \mathbb{E}[z_{nk}] &= \frac{\sum_{z_{nk}} z_{nk} [\pi_k \mathcal{N}(x_n \mid \mu_k, \Sigma_k)]^{z_{nk}}}{\sum_{z_{nj}} [\pi_j \mathcal{N}(x_n \mid \mu_j, \Sigma_j)]^{z_{nj}}} \\[1em] &= \frac{\pi_k \mathcal{N}(x_n \mid \mu_k, \Sigma_k)}{\sum^K_j \pi_j \mathcal{N}(x_n \mid \mu_j, \Sigma_j)} \\[1em] &= \gamma(z_{nk}) \end{align}

I am aware of similar question, but the answer assumed the posterior distribution is $$p(z_n | x_n)$$, which is unlikely the assumption of Bishop. At first I agreed because that explained nicely the expression, but when Bishop used the expression to explain the expected value of the log likelihood below, I think the answer may not be correct,

$$\ln p(X, Z | \mu, \Sigma, \pi) = \sum^N_n \sum^K_k z_{nk} \Big\{ \ln \pi_k + \ln \mathcal{N}(x_n | \mu_k, \Sigma_k) \Big\}$$ $$E_Z [\ln p(X, Z | \mu, \Sigma, \pi)] = \sum^N_n \sum^K_k \gamma(z_{nk}) \Big\{ \ln \pi_k + \ln \mathcal{N}(x_n | \mu_k, \Sigma_k) \Big\}$$

So, how can I obtain the expression of $$\mathbb{E}[z_{nk}]$$ under the posterior distribution $$p(Z | X, \mu, \Sigma, \pi)$$?

Update: the "corrected version" of $$\mathbb{E}[z_{nk}]$$, by Bishop,

\begin{align} \mathbb{E}[z_{nk}] &= \frac{\sum_{\mathbf{z}_n} z_{nk} \prod_{k'} \left[\pi_{k'} \mathcal{N}(\mathbf{x}_n | \boldsymbol{\mu}_{k'}, \boldsymbol{\Sigma}_{k'}) \right]^{z_{nk'}}}{\sum_{\mathbf{z}_n} \prod_{j} \left[\pi_{j} \mathcal{N}(\mathbf{x}_n | \boldsymbol{\mu}_{j}, \boldsymbol{\Sigma}_{j}) \right]^{z_{nj}}} \end{align}

I don't replace the version with the original because I don't see} the version will make my question easier to answer, and they are both equivalent if we assume the posterior distribution of interest is p(z_n | x_n) (The original version is not correct because the nominator in the first equation is not equal to that in the second)

Typo.

There is a typo in the 2006 version of Bishop's PRML, which has been corrected in later versions.

Instead, equation (9.39) on p443 should read "the expectation of the indicator variable $$z_{nk}$$ under this posterior distribution is then given by:

\begin{align} \mathbb{E}[z_{nk}] &= \frac{\sum_{\mathbf{z}_n} z_{nk} \prod_{k'} \left[\pi_{k'} \mathcal{N}(\mathbf{x}_n | \boldsymbol{\mu}_{k'}, \boldsymbol{\Sigma}_{k'}) \right]^{z_{nk'}}}{\sum_{\mathbf{z}_n} \prod_{j} \left[\pi_{j} \mathcal{N}(\mathbf{x}_n | \boldsymbol{\mu}_{j}, \boldsymbol{\Sigma}_{j}) \right]^{z_{nj}}} \\ &= \frac{\pi_{k} \mathcal{N}(\mathbf{x}_n | \boldsymbol{\mu}_k, \boldsymbol{\Sigma}_k)}{\sum^K_{j=1} \pi_j \mathcal{N}(\mathbf{x}_n | \boldsymbol{\mu}_j, \boldsymbol{\Sigma}_j)} = \gamma(z_{nk}) \end{align}

If you want further confirmation that this correction is indeed the case, see p19 of the publisher errata on the 2006 version of Bishop here.

Derivation.

When I worked through this I found it tricky getting my head around the indicators and the selection of the appropriate dimensionality of the posterior for computing expectations. Here is one way of deriving the result.

The graphical model for the Gaussian mixture model, from Bishops's PRML is the following: Notice that if we are interested in computing the posterior distribution $$p(\mathbf{z}_n | \mathbf{x}_n, \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi})$$, this amounts to reversing the direction of the arrow so that it goes from $$\mathbf{x}_n$$ to $$\mathbf{z}_n$$. The conditional independencies in the diagram also imply that we can factorise the posterior over all latent variables $$\mathbf{Z}$$ into a product of $$n$$ independent posterior distributions:

$$p(\mathbf{Z} | \mathbf{X}, \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi})= \prod^N_{n=1} p(\mathbf{z}_n | \mathbf{x}_n, \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi})$$

We are now interested in computing the expectation of the indicator variable $$z_{nk}$$, that is, $$\mathbb{E}[z_{nk}]$$, where the expectation is with respect to the posterior $$p(\mathbf{Z} | \mathbf{X}, \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi})$$.

Now for clarity, I am going to explicitly choose to work at the dimensionality of vectors that is, rewriting everything in terms of $$\mathbf{z}_n \in \mathbb{R}^K$$, rather than matrices $$\mathbf{Z}$$, or scalars $$z_{nk}$$.

The indicator variable $$z_{nk}$$ takes the value 1 when the $$n$$th latent mixture assignment $$\mathbf{z}_n$$ is component $$k$$, and $$0$$ otherwise. Explicitly writing the indicator variable in terms of an indicator function on $$\mathbf{z}_n$$, we have that $$z_{nk} = \mathbb{I}(\mathbf{z}_n = k)$$. Due to the factorisation of the posterior over the latent variable matrix $$\mathbf{Z}$$ into $$n$$ independent posterior distributions over vectors $$\mathbf{z}_n$$, the expectation simplifies so that we only need to consider it with respect to the posterior $$p(\mathbf{z}_n | \mathbf{x}_n, \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi})$$:

\begin{align} \mathbb{E}_{p(\mathbf{Z} | \mathbf{X}, \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi})}[z_{nk}] &= \mathbb{E}_{p(\mathbf{Z} | \mathbf{X}, \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi})}[\mathbb{I}(\mathbf{z}_n = k)] \\ &= \mathbb{E}_{\prod^N_{n=1} p(\mathbf{z}_n | \mathbf{x}_n, \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi})}[\mathbb{I}(\mathbf{z}_n = k)] \\ &= \mathbb{E}_{p(\mathbf{z}_n | \mathbf{x}_n, \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi})}[\mathbb{I}(\mathbf{z}_n = k)] \\ &= p(\mathbf{z}_n = k | \mathbf{x}_n, \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi}) \\ \tag{1} \end{align}

Where in the last line we have used the fact that the expectation of an indicator function is a probability. As you have sought further clarity on the steps behind this derivation, I have put further exposition at the bottom of the post.

Using standard probability formulae, the posterior probability is:

$$p(\mathbf{z}_n | \mathbf{x}_n, \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi}) = \frac{p(\mathbf{x}_n, \mathbf{z}_n)}{p(\mathbf{x}_n)} = \frac{p(\mathbf{x}_n, \mathbf{z}_n)}{\sum_{\mathbf{z}_n} p(\mathbf{x}_n, \mathbf{z}_n)}$$

Using the factorisation induced by the graphical rerpesentation of the Gaussian mixture model, and also the distributions specified by the model in equation $$(9.10)$$ and $$(9.11)$$, the joint distribution is:

\begin{align} p(\mathbf{x}_n, \mathbf{z}_n) &= p(\mathbf{z}_n) \cdot p(\mathbf{x}_n | \mathbf{z}_n) \\ &= \prod^K_{k'=1} \pi_k^{z_{nk'}} \cdot \prod^K_{k'=1} \mathcal{N}(\mathbf{x}_n | \boldsymbol{\mu}_{k'}, \boldsymbol{\Sigma}_{k'})^{z_{nk'}} \\ &= \prod^K_{k'=1} [\pi_{k'} \mathcal{N}(\mathbf{x}_n | \boldsymbol{\mu}_{k'}, \boldsymbol{\Sigma}_{k'})]^{z_{nk'}} \end{align}

The marginal distribution is:

$$p(\mathbf{x}_n) = \sum_{\mathbf{z}_n} p(\mathbf{x}_n, \mathbf{z}_n) = \sum_{\mathbf{z}_n} \prod^K_{j=1} [\pi_k \mathcal{N}(\mathbf{x}_n | \boldsymbol{\mu}_j, \boldsymbol{\Sigma}_j)]^{z_{nj}} = \sum^K_{j=1} \pi_j \mathcal{N}(\mathbf{x}_n | \boldsymbol{\mu}_j, \boldsymbol{\Sigma}_j)$$

Hence the posterior probability distribution is:

$$p(\mathbf{z}_n | \mathbf{x}_n, \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi}) = \frac{\prod^K_{k'=1} [\pi_{k'} \mathcal{N}(\mathbf{x}_n | \boldsymbol{\mu}_{k'}, \boldsymbol{\Sigma}_{k'})]^{z_{nk'}}}{\sum^K_{j=1} \pi_j \mathcal{N}(\mathbf{x}_n | \boldsymbol{\mu}_j, \boldsymbol{\Sigma}_j)}$$

Putting this all together, and using the above distribution to query the probability that the $$n$$th latent mixture assignment is component $$k$$, we have:

$$\mathbb{E}[z_{nk}] = p(\mathbf{z}_n = k | \mathbf{x}_n, \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi}) = \frac{\pi_k \mathcal{N}(\mathbf{x}_n | \boldsymbol{\mu}_{k}, \boldsymbol{\Sigma}_{k})}{\sum^K_{j=1} \pi_j \mathcal{N}(\mathbf{x}_n | \boldsymbol{\mu}_j, \boldsymbol{\Sigma}_j)} = \gamma(z_{nk})$$

Further derivation of $$(1)$$ in response to comments.

This derivation relies on two results that are quite useful, which I will not prove formally with any rigour, but state.

a. Computing expectations with respect to a factorisable joint distribution.

If a joint distribution over a random vector $$\mathbf{x} \in \mathbb{R}^N$$ can be factorised into a product of marginal distributions over each of its $$N$$ components, that is, $$p(\mathbf{x}) = p(X_1, ..., X_N) = \prod^N_{i=1} p(X_i)$$, and we wish to compute the expectation of a deterministic function of one of the components with respect to the joint distribution, then we only need to compute the expectation with respect to the relevant marginal distribution. To see this:

\begin{align} \mathbb{E}_{p(\mathbf{x})}[f(X_i)] &= \int_{\mathbb{R}^N} f(x_i) p(\mathbf{x}) d\mathbf{x} \\ &= \int ... \int f(x_i) p(x_1, ..., x_N) dx_1 ... dx_N \\ &= \int ... \int f(x_i) \prod^N_{i=1} p(x_i) dx_1 ... dx_N \\ &= \int ... \int f(x_i) p(x_i) \prod_{-i} p(x_{-i}) dx_i dx_{-i} \\ &= \int ... \int \prod_{-i} p(x_{-i}) \int f(x_i) p(x_i) dx_i dx_{-i} \\ &= \mathbb{E}_{p(x_i)}[f(X_i)]\int ... \int \prod_{-i} p(x_{-i}) dx_{-i} \\ &= \mathbb{E}_{p(x_i)}[f(X_i)] \end{align}

Where $$\int ... \int d_{x_{-i}}$$ refers to the multiple integrals with respect to all components other than component $$i$$, forming a total of $$(N-1)$$ integrals. And where I have assumed that all distributions are appropriately normalised - meaning that the rightmost term in the penultimate equality is equal to 1.

This argument is being generalised to the case where the posterior distribution over a matrix $$\mathbf{Z}$$, $$p(\mathbf{Z} | \mathbf{X}, \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi})$$ can be factorised into $$N$$ independent posterior distributions over vectors $$\mathbf{z}_n$$, $$p(\mathbf{z}_n | \mathbf{x}_n, \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi})$$. Hence

$$\mathbb{E}_{p(\mathbf{Z} | \mathbf{X}, \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi})}[f(\mathbf{z}_n)] = \mathbb{E}_{\prod^N_{n=1} p(\mathbf{z}_n | \mathbf{x}_n, \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi})}[f(\mathbf{z}_n)] = \mathbb{E}_{p(\mathbf{z}_n | \mathbf{x}_n, \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi})}[f(\mathbf{z}_n)]$$

Normally you would use the unconscious/lazy statistician rule to evaluate this, but in the case that $$f(\cdot)$$ is an indicator function, then we use the next result.

b. Expectation of an indicator function.

There is a fairly standard result that the expectation of an indicator function of a random variable is its probability. For a set $$A$$ and random variable $$X$$:

$$\mathbb{E}[\mathbb{I}(X \in A)] = 1 \cdot P(X \in A) + 0 \cdot P(X \notin A) = P(X \in A)$$

Applying this result, we have that:

\begin{align}\mathbb{E}_{p(\mathbf{z}_n | \mathbf{x}_n, \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi})}[z_{nk}] &= \mathbb{E}_{p(\mathbf{z}_n | \mathbf{x}_n, \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi})}[\mathbb{I}(\mathbf{z}_n = k)] \\ &= 1 \cdot p(\mathbf{z}_n = k | \mathbf{x}_n, \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi}) + 0 \cdot p(\mathbf{z}_n \neq k | \mathbf{x}_n, \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi})\\ &= p(\mathbf{z}_n = k | \mathbf{x}_n, \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi}) \end{align}

Lastly, if you find this confusing, you can think of $$z_{nk}$$ as a Bernoulli random variable. The mean of a Bernoulli random variable is the parameter $$p$$, which in this case is the posterior probability that $$\mathbf{z}_n$$ is component $$k$$.

Hopefully that addresses the difficulties you are experiencing.

• I know the version of equation 9.39, but could you explain how the first equation is equal to the expectation under the posterior distribution (which is the first expression in my question)?
– Hiep
Feb 25, 2021 at 13:52
• No problem, I will edit and update. Feb 25, 2021 at 15:36
• Sorry I only saw the edit you made in the post after I wrote my answer. Does the answer clear up your confusion at all? Feb 25, 2021 at 16:33
• Could you explain the conclusion "the expectation simplifies so that we only need to consider it with respect to the posterior ..." (or the third equation of the expression immediately below it)? The expectation can be expressed as summation of two terms, which are product of values of the indicator function I(z_n) and p(Z| X). Because z_nk can only 0 or 1, so expectation is equal to the posterior p(Z|X). I cannot figure out how the (n-1) factors in the posterior p(Z| X) can be canceled out.
– Hiep
Feb 26, 2021 at 3:26
• I do think the notation in that part of Bishop is somewhat confusing. In particular when specifying $\mathbb{E}[z_{nk}]$, Bishop uses $\sum_{\mathbf{z}_n}$, implying that we are summing over all $K$ distinct states of $\mathbf{z}_n$, but then the summand is the product of $z_{nk}$ (and the posterior probability) which can range over only 2 states. Hence the reason for my choosing to present the indicator variable $z_{nk}$ as an indicator function on $\mathbf{z}_n$, and to gloss over the problematic notation altogether. Feb 27, 2021 at 13:17