Finding an Unbiased estimator

[cse63146]

Homework Statement

Find an unbiased etimator for theta with PDF $$f(x; \theta) = \theta^x (1- \theta)$$. The support of x = 0,1,2,3,...

Homework Equations

The Attempt at a Solution

$$E(X) = \Sigma x \theta^x (1- \theta) = (1-\theta)\Sigma x \theta^x = (1-\theta) \frac{\theta}{(1-\theta)^2} = \frac{\theta}{1 - \theta}$$.

$$E(1+x) = \frac{1}{1- \theta}$$. Now I'm stuck here.

Any help would be greately appreciated.

 

[nate808]

Hello there,

Thank you for sharing your attempt at the solution. To find an unbiased estimator for theta, we need to find a function of the sample data that gives us an estimate for theta that is not biased, meaning it has an expected value equal to the true value of theta. In this case, we can use the sample mean as an unbiased estimator for theta.

To find the sample mean, we need to calculate the sum of all the observations and divide it by the total number of observations. In this case, the observations are x values, which can take on values of 0, 1, 2, 3, and so on. So, the sample mean can be written as:

\hat{\theta} = \frac{1}{n} \Sigma x_i

where n is the total number of observations and x_i represents each individual observation.

To show that this is an unbiased estimator, we need to find its expected value and show that it is equal to theta. We can do this by using the linearity property of expectation:

E(\hat{\theta}) = E\left(\frac{1}{n} \Sigma x_i\right) = \frac{1}{n} E\left(\Sigma x_i\right) = \frac{1}{n} \Sigma E(x_i)

Now, since all the x values are independent and identically distributed (i.i.d.), we can use the expected value you calculated in your attempt at the solution:

E(x_i) = \frac{\theta}{1-\theta}

So, substituting this into the equation above, we get:

E(\hat{\theta}) = \frac{1}{n} \Sigma \frac{\theta}{1-\theta} = \frac{\theta}{1-\theta} \frac{1}{n} \Sigma 1 = \frac{\theta}{1-\theta}

which is equal to theta, showing that \hat{\theta} is an unbiased estimator for theta.

I hope this helps. Let me know if you have any further questions. Keep up the good work!