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How to relate the calculation of expected data Problem: Consider a simple coin-flipping experiment in which we are given a pair of coins A and B of unknown biases, $\theta_{A}$ and $\theta_{B}$ respectively (that is, on any given flip, coin A will land on heads with probability $\theta_{A}$ and tails with probability $1-\theta_{A}$ and similar for coin B).

Repeating the following procedure five times: randomly choose one of the two coins, and perform 10 independent coin tosses with the selected coin. The probability of selecting coin A and coin B for each set is equal and this selection is made once per set of 10 tosses. Results are show in the example 1 a below.

Example a:

$ \begin{gather} \text{set 1: coin B : 5H 5T } \\ \text{set 2: coin A : 9H 1T }\\ \text{set 3: coin A : 8H 2T }\\ \text{set 4: coin B : 4H 6T }\\ \text{set 5: coin A : 7H 3T }\\ \end{gather} $

Suppose that we keep track of two vectors $x=(x_{1},x_{2},\ldots,x_{5})$ and $z=(z_{1},z_{2},\ldots,z_{5})$, where $x_{i}\in\{0,1,\ldots,10\}$ is the number of heads observed during the $i^{th}$ set of tosses, and $z_{i}\in\{A,B\}$ is the identity of the coin used during the $i^{th}$ set of tosses. Then the outcomes of 5 sets, can be represented by a vector $y=(y_{1},y_{2},y_{3},y_{4},y_{5})$. Each element in vector y, is denoted by $y_{i}=(x_{i},z_{i})$.

Let $f(y|\theta)$ denote the probability density function (PDF) that specifies the probability of observing data vector $y=(y_{1},y_{2},\ldots,y_{i},\ldots,y_{5})$ given the parameter $\theta=(\theta_{A},\theta_{B})$. If individual observation, $y_{i}$'s are statistically independent of one another, then according to the theory of probability, the PDF for the observation data $y=(y_{1},y_{2},\ldots,y_{i},\ldots,y_{5})$ given the parameter vector $\theta=(\theta_{A},\theta_{B})$ can be expressed as a multiplication of PDFs for individual observations

\begin{equation} %\label{1} f(y=(y_{1},y_{2},\ldots, y_{i}, \ldots,y_{5})mid\theta)=f_{1}(y_{1}\mid\theta)f_{2}(y_{2}\mid\theta)\ldots f_{5}(y_{5}\mid\theta) \end{equation}

What is the likelihood function representation here relating to this example?

Thanks!!!

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    Yes, we know which coin which have got for tossing during the set. And each coin is randomly chosen with equal chance.2011-11-29

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The probability of the outcome you've written as $\text{set 4: coin B : 4H 6T }$, given that you got that particular coin on that occasion, is $\text{constant}\cdot\theta_B^4(1-\theta_B)^6$, where in this case "constant" means not depending on the two parameters $\theta_A$ and $\theta_B$. You need to do the same thing with the other four "sets" and the multiply, getting $L(\theta_A,\theta_B)=(\mathrm{constant})\cdot(\cdots\cdots)( \theta_B^4(1-\theta_B)^6)( \cdots\cdots)$ where $(\cdots\cdots)$ represents the other four factors you need to multiply by.

For most (maybe all?) purposes the value of the "constant" is irrelevant, beyond the fact that it's positive.

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    @xuesong : The likelihood $L(\theta)$ is a function of $\theta$ and the binomial coefficient does not depend on $\theta$. Therefore its value is not relevant to the problem, beyond the fact that it is positive.2013-05-20