# What is the likelihood function of a two coin tossing?

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

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

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

Thanks!!!

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Just to be crystal-clear: You're saying in each set we know which coin we've got. If we didn't, we'd get a different likelihood function. And when you say we randomly choose a coin, does that mean they have equal chances of being chosen? (That won't actually affect the likelihood function, beyond a constant multiple, if we know which coin we've got in each set. But it will if we don't know.) –  Michael Hardy Nov 29 '11 at 6:55
Yes, we know which coin which have got for tossing during the set. And each coin is randomly chosen with equal chance. –  xuesong Nov 29 '11 at 8:51
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.
In fact, since $x_i$ is the number of heads on the $i$-th trial and $z_i \in \{A,B\}$ tells us which coin was used on the $i$-th trial, it is possible to state more succinctly that the likelihood is proportional to $$\prod_{i=1}^5\theta_{z_i}^{x_i}(1-\theta_{z_i})^{10-x_i}$$ –  Dilip Sarwate Nov 29 '11 at 18:40
Thanks Michael and Dilip for your answer. but I notice that the binomial coefficients are not mentioned in neither of your formula. Is $L(\theta_{z_i})=\prod_{i=1}^5\binom{10}{x_i}\theta_{z_i}^{x_i}(1-\theta_{z_i})^‌​{10-x_i}\;$.correct or binomial coefficients are not necessary here ? –  xuesong Nov 29 '11 at 23:01
@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. –  Michael Hardy May 20 at 1:02