Final answer:
The value of p that maximizes the likelihood for the given geometric distribution samples with x_1=1, x_2=2 and x_3=1 is p=3/4. This is found by taking the derivative of the likelihood function and setting it to zero, confirming the value is a maximum by examining the trend of the function.
Step-by-step explanation:
To find the value of p that maximizes the likelihood for independent samples drawn from a geometric distribution with outcomes x_1=1, x_2=2, x_3=1, we must first write down the likelihood function of the geometric distribution. The probability mass function (PMF) for a geometric distribution is P(X=x) = p(1-p)^(x-1), where x is the number of trials up to and including the first success, and p is the probability of success on any individual trial.
The likelihood function L is the product of the individual probabilities for each observed sample:
L(p) = p(1-p)^(x_1-1) * p(1-p)^(x_2-1) * p(1-p)^(x_3-1)
Substituting the values we have, this becomes:
L(p) = p(1-p)^(1-1) * p(1-p)^(2-1) * p(1-p)^(1-1) which simplifies to L(p) = p^3(1-p).
To maximize this function, we take the derivative with respect to p and set it equal to zero:
L'(p) = 3p^2 - 4p^3
Setting the derivative equal to zero gives p = 3/4, which is the critical point. To show this maximizes the likelihood, we can examine the second derivative or observe that the likelihood function increases as p increases from 0 to 3/4 and then decreases as p goes from 3/4 to 1. Therefore, p = 3/4 maximizes the likelihood.