# Derivation of maximum likelihood for a Gaussian mixture model

I'm working my way through the derivation of EM in Bishop (p. 435).

I'm stuck trying to derive to MLE for $$\mu_k$$ for the gaussian mixture model.

Basically I get an extra sum in the numerator.

For those that don't have the book:

The log likelihood for the gaussian mixture model is:

$$ln\; p(X|\pi,\mu,\Sigma) = \sum_{n=1}^{N} ln \left\{ \sum_{k=1}^K \pi_k N(x_n|\mu_k,\Sigma_k) \right\}$$

When I take derivatives wrt $$\mu_k$$:

1. recognise that we're dealing with $$ln(f(x))$$ and the derivative is $$\frac{f'(x)}{f(x)}$$

2. This gives us:

$$\sum_{n=1}^{N} \frac{1}{\sum_{k=1}^K \pi_k N(x_n|\mu_k,\Sigma_k)} \times \frac{\partial \sum_{k=1}^K \pi_k N(x_n|\mu_k,\Sigma_k) }{\partial \mu_k}$$

1. Now only have to solve the differential in the right most term:

$$\frac{ \partial \sum_{k=1}^K \pi_k N(x_n|\mu_k,\Sigma_k) }{\partial \mu_k} = \sum_{k=1}^K -0.5(2\Sigma^{-1}(x-\mu_k)\times \pi_k N(x_n|\mu_k,\Sigma_k)$$

1. This leaves me with:

$$\sum_{n=1}^{N} \frac{ \sum_{k=1}^K \pi_k N(x_n|\mu_k,\Sigma_k) }{\sum_{k=1}^K \pi_k N(x_n|\mu_k,\Sigma_k)} \times -0.5(2\Sigma^{-1}(x-\mu_k))$$

1. The solution in the book is:

$$\sum_{n=1}^{N} \frac{ \pi_k N(x_n|\mu_k,\Sigma_k) }{\sum_{j} \pi_j N(x_n|\mu_j,\Sigma_j)} \times 0.5(2\Sigma^{-1}(x-\mu_k))$$

How is it that

1. There's no summation in their numerator?

2. They've changed what they're summing over (k -> j) ?

3. They have a positive final term, whereas I have a negative?

Thanks

To avoid any confusion, the summation index and the index of the $$\mu$$ that you differentiate with should be different. From the beginning, assume the likelihood is written with index $$j$$ and you want to differentiate it with $$\mu_k$$:
$$\frac{\partial \sum_{j=1}^K \pi_j N(x_n|\mu_j,\Sigma_j) }{\partial \mu_k}=\frac{ \pi_k\partial N(x_n|\mu_k,\Sigma_k)}{\partial \mu_k}$$ which explains why the answer doesn't have a summation in the numerator.
You'll have a minus in $$(x-\mu_k)$$, i.e. differentiating wrt $$\mu_k$$ gives $$-1$$, and also another minus in $$\exp(-(\ldots))$$ expression in normal PDF. They'll cancel out each other.
• The summation index can be any letter, but we want to differentiate with respect to a specific $k$. In the summation, if $j\neq k$, the derivative will be $0$, so the only term remaining there will be when $j=k$. – gunes Mar 10 '20 at 15:37
• exactly, e.g. $\mu_1=a,\mu_2=b$, and you'll differentiate wrt $a$. – gunes Mar 10 '20 at 15:40