- #1

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I have a lot of difficulty figuring out if it is an unbiased estimator or not. How do I determine the expected value of theta-hat?

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- Thread starter daoshay
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- #1

- 14

- 0

I have a lot of difficulty figuring out if it is an unbiased estimator or not. How do I determine the expected value of theta-hat?

- #2

- 2,161

- 79

I have a lot of difficulty figuring out if it is an unbiased estimator or not. How do I determine the expected value of theta-hat?

You're saying MLE in the text and MSE in the title. For the Mean Squared Error, you can get a Bayesian minimum. (see Gamma Distribution in the Wiki). You make a Maximum Likelihood Estimate of a parameter, not a distribution. You can write a likelihood function for a distribution.

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- #3

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Based on the gamma family, the mean of this distribution should be 4theta --> theta = mu/4

Bias is E(theta-hat)-theta, right?

Now, I'm supposed to find the MSE of theta-hat E[(theta-hat -theta)^2] right?

So, am I supposed to use the value of theta based on the distribution? I'll check my work on the expansion and check back later. Thanks for your patience. =)

- #4

- 2,161

- 79

Based on the gamma family, the mean of this distribution should be 4theta --> theta = mu/4

Bias is E(theta-hat)-theta, right?

Now, I'm supposed to find the MSE of theta-hat E[(theta-hat -theta)^2] right?

So, am I supposed to use the value of theta based on the distribution? I'll check my work on the expansion and check back later. Thanks for your patience. =)

Your equation has only one parameter so it's a simple exponential distribution (or gamma with k=1). [tex]\theta[/tex] is the reciprocal of the rate parameter which is often written as [tex]\lambda[/tex]. So [tex]E|X|=\theta=\frac{1}{\lambda}[/tex]. [tex]Var|X|=\frac{1}{\lambda^2}[/tex]. For MSE use the Baysian minimum that I referred to earlier. Are you using any dummy (or real) data here?

EDIT: I'm using [tex]\theta[/tex] above as the mean of the distribution [tex]F(x;\theta)[/tex]. Otherwise, this is not making any sense to me.

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- #5

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I have given some serious effort to working out and understanding the MLE of a distribution. From the distribution f(x;[tex]\theta[/tex])= [tex]x^{3}[/tex][tex]e^{-x/\theta}[/tex]/(6[tex]\theta[/tex][tex]^{4}[/tex]),

Are you sure that shouldn't be [tex] f(x|\theta)[/tex]? ie [tex]L(\theta|x)=f(x|\theta)[/tex].

EDIT: OK. I see that k=4. Now for some N you can estimate [tex]\theta[/tex] for the gamma distribution. As far as I know the ML estimate of [tex] \theta[/tex] is unbiased, assuming an unbiased sample. [tex]\hat{\theta} =\frac{1}{kN}\sum_{i=1}^{N}x_{i}[/tex].

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- #6

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I looked further into the issue of estimator bias for the gamma distribution. Numerical analysis for moderate sized samples indicate a small "upward" or positive bias for ML estimates of the mean and variance.

http://web.uvic.ca/econ/research/papers/ewp0908.pdf

The bias of [tex]\hat{\theta}[/tex] is shown on page 8, equation (14).

http://web.uvic.ca/econ/research/papers/ewp0908.pdf

The bias of [tex]\hat{\theta}[/tex] is shown on page 8, equation (14).

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