Sufficient Estimator for a Geometric Distribution

[cse63146]

Homework Statement

Let X1,..,Xn be a random sample of size n from a geometric distribution with pmf [tex]P(x; \theta) = (1-\theta)^x\theta[/tex]. Show that [tex]Y = \prod X_i[/tex] is a sufficient estimator of theta.

Homework Equations

The Attempt at a Solution

So [tex]\prod P(x_i, \theta) = (1-\theta)^{\Sigma x_i} \theta^n[/tex]

I don't believe that the factorization theroem can be applied here. Is there some trick to this that I'm not seeing?

Thank you in advance.

 

[nate808]

Thank you for your question. I can help you understand why Y = \prod X_i is a sufficient estimator of theta.

First, let's define what a sufficient estimator is. A sufficient estimator is a statistic that contains all the information about the parameter that we are trying to estimate. In other words, if we know the value of the sufficient estimator, we can make an accurate estimate of the parameter.

In this case, we are trying to estimate theta, which is the probability of success in a geometric distribution. The geometric distribution is a discrete probability distribution that models the number of trials needed to achieve the first success in a sequence of independent trials.

Now, let's look at the expression for Y = \prod X_i. This is a product of all the individual random variables X_i, which are each drawn from the geometric distribution with parameter theta. This means that Y is also drawn from a geometric distribution, but with a different parameter. Specifically, its parameter is (1-\theta)^n.

So, if we know the value of Y, we can easily calculate the value of theta using the formula for the geometric distribution. This means that Y contains all the information we need to estimate theta, and thus, it is a sufficient estimator.

I hope this explanation helps you understand why Y = \prod X_i is a sufficient estimator of theta. If you have any further questions, please don't hesitate to ask.

 

[Architeuthis Dux]

I would like to first clarify the problem and the terms used. A geometric distribution is a discrete probability distribution that models the number of trials needed to achieve a success in a sequence of independent trials, where each trial has a constant probability of success. The pmf (probability mass function) for a geometric distribution is given by P(x; \theta) = (1-\theta)^x\theta, where x represents the number of trials and \theta represents the probability of success.

In this problem, we are asked to show that the product of the random sample X1,..,Xn is a sufficient estimator for \theta. A sufficient estimator is a statistic that contains all the information about the parameter of interest, \theta, contained in the sample. This means that if we know the value of the statistic, we can make accurate inferences about \theta without needing to know the entire sample.

To show that Y = \prod X_i is a sufficient estimator for \theta, we can use the factorization theorem. According to the theorem, a statistic T(X) is a sufficient estimator for a parameter \theta if and only if the joint probability distribution of the sample X can be factorized as f(x_1,..,x_n;\theta) = g(T(X);\theta)h(x_1,..,x_n), where g(T(X);\theta) does not depend on the sample and h(x_1,..,x_n) does not depend on the parameter \theta.

In this case, we can factorize the joint probability distribution as follows:

f(x_1,..,x_n;\theta) = \prod_{i=1}^n P(x_i;\theta) = \prod_{i=1}^n (1-\theta)^{x_i}\theta = (1-\theta)^{\sum_{i=1}^n x_i}\theta^n = g(T(X);\theta)h(x_1,..,x_n),

where T(X) = \prod X_i and g(T(X);\theta) = \theta^n and h(x_1,..,x_n) = (1-\theta)^{\sum_{i=1}^n x_i}. Thus, we have successfully factorized the joint probability distribution and shown that Y = \prod X_i is a sufficient estimator for \theta.

In conclusion, the product of the random sample X1,..,