# Maximum Likelihood Example

I have the problem to understand the following simple Maximum (Log) Likelihood example. Let $X$ be a discrete variable with domain $\{1,\dots,K\}$ and the discrete distribution is parametrized

$$P(X=k;\pi) = \pi_k$$

With parameters $\pi = (\pi_1,\dots,\pi_K)$ that are constrainted to fulfill $\sum_k \pi_k = 1$ and there is some data $D = \{x_i\}_{i = 1}^n$

What is the log likelihood $\mathcal{L}(\pi)$ of the data under the model?

I have applied the definition which gives me:

$$\mathcal{L}(\pi) = \log P(x_{1:n};\pi) = \sum_{i = 1}^n \log P(x_i;\pi)$$

At first I thought that it must sum up to 1 and the liklihood is 0, but this does not make sense, this the sum is over the data set which can have different occurences of different $X=k$ values, also the $\log$ is applied every time. The only thing I can think of is that $$\mathcal{L}(\pi) \leq 0$$ since there is no value which is $\geq$ 1.

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Your likelihood is correct to begin with, but you must now write it explicitly. Hint: when you are at a loss, try with some concrete example. Say, take K=2, an n=5, and write down the likelihood. – leonbloy May 31 '12 at 20:17
@leonbloy Does this involve making use of the Lagrange multiplier? – Mahoni May 31 '12 at 21:28
Solving some concrete examples just leads me to likelihood with a negative result, but I lack the idea to generalize this. – Mahoni May 31 '12 at 21:49
It is not the likelihood that is negative, it is the log likelihood that is working out to be negative. – Dilip Sarwate May 31 '12 at 22:46
It does not matter at all that the log-likelihood is negative, the log-lilelihood is not a probability density (not even the likelihood is), you just want to consider it as a function with the parameter/s as variable, and find its maximum – leonbloy May 31 '12 at 23:07

Suppose $K=3$ and you data consists of 5 samples ${\mathbb X} =\{ 1, 3, 1, 1, 2 \}$ The likelihood of this realization would be $P(X=1)P(X=3)P(X=1)P(X=1) P(X=2) = \pi_1 \pi_3\pi_1\pi_1\pi_2 = \pi_1^3 \pi_2\pi_3$ Calling $n_i$ the number of samples with value $i$, this can be written in general as $\pi_1^{n_1} \pi_2^{n_2} \pi_3^{n_3}$

In general, then $$\mathcal{L}(\pi) = \sum_{i = 1}^n \log \pi_{x_i} = \sum_{j = 1}^K \log \pi_{j}^{n_j} = \sum_{j = 1}^K {n_j} \log \pi_{j}$$

Now, you must consider this as a function of $\pi=\{\pi_j\}$ ($n_j$ are given by the realization) and find the value of $\pi$ that maximizes this - subjected to the restriction that $\sum \pi_j=1$ and $\pi_j \ge 0$.

This is now a typical problem of multivariate Calculus (maximize a differentiable funcion of several variables subjected to a restriction given as another function) Lagrange multipliers is the standard method . Can you go on from here?

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Thanks a lot these were many hints, but they helped me to understand the problem at least in terms of the notation. I applied Lagrange multiplier and my result is $\pi_{k}^{ML} = \frac{m_k}{N}$, where $N$ is the size of the given data. – Mahoni Jun 1 '12 at 11:03
Anyway, can be said, that $\mathcal{L}(\pi)$ is the same as $\pi_k^{ML}$ ? – Mahoni Jun 1 '12 at 11:12
I meant is $\pi_k^{ML}$ the result for maximizing $\mathcal{L}(\pi)$? I forgot to say, that my $m_k$ is your $n_i$ – Mahoni Jun 1 '12 at 11:22
Yes, that's the definition of Maximum Likelihood (the (log)likelihood is a function that has the parameter as variable; the ML estimator is the value of the variable that maximizes the function). BTW, your result is correct, and it's good to understand that it is intuitively satisfactory. – leonbloy Jun 1 '12 at 11:42