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    Unbiased Estimator Question




    Suppose that Y has a Poisson distribution with parameter lambda. How can I show that (-2)^lambda is an unbiased estimator of e^(-3*lambda)?

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    Re: Unbiased Estimator Question

    You can't? That isn't an estimator. I'm guessing you've copied something down wrong. But I also have to direct you here: http://www.talkstats.com/showthread....ho-show-effort

    and quite possibly here: http://www.talkstats.com/announcement.php?f=3

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    Re: Unbiased Estimator Question

    The question is written correctly. I am unsure how to find the expected value of these types of functions. Do you have any insight on how to find the expected value of these two functions?

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    Re: Unbiased Estimator Question

    Like I said they aren't estimators. You don't have any function of any data in there at all. Currently since you don't have any random values the expectation of (-2)^lambda is (-2)^lambda. Are you missing a hat or something?

    Like is it supposed to be (-2)^{\hat{\lambda}} where \hat{\lambda} = \frac{1}{n}\sum_{i=1}^n Y_i ?

    But as you have it written there is no random quantity so you're just taking the expectation of a constant.

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    Re: Unbiased Estimator Question

    Ah, yes. I made a typo. (-2)^Y. It has been a long day. I am not sure how to take the Expected value of this function where Y is poisson distributed. Do you have any suggestions as to how I can determine whether it is an unbiased estimator for e^(-3*lambda). thanks!

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    Re: Unbiased Estimator Question

    I would just find the expectation directly from the definition. Recall that if we let f(y| \lambda) be the poisson pmf at y then:

    E[g(y)] = \sum_{y = 0}^\infty g(y) f(y|\lambda)

    and in your case g(y) = (-2)^\lambda so

    = \sum_{y=0}^\infty (-2)^{y} \frac{\exp(-\lambda)\lambda^y}{y!}

    Working this out isn't too bad if you get things in the correct form. One nice little trick is that since you know what this should evaluate to you should be able to multiply by 1 in the form of \exp(-3\lambda)\exp(3\lambda) and pull the \exp(-3\lambda) out of the summation. That should simplify things a little bit.

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    Re: Unbiased Estimator Question

    Thanks, Dason.

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    Re: Unbiased Estimator Question

    I'm actually a bit stuck. My expression thus far looks like:

    e^(-3*lambda) summation of (-2)^lambda*e^(2*lambda)*lambda^(y)/y!

    How can I simplify this further? I notice that multiplying the exponential terms results in (-2)^lambda*(pmf for a poisson random variable). Shouldn't the final result be e^(-3*lambda) to prove that (-2)^Y is an unbiased estimator? Is there a way to simplify the expression inside of the summation?

    Thanks!

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    Re: Unbiased Estimator Question

    Note that if you shouldn't multiply by e^(-3*lambda). If anything you should multiply by 1. You can write 1 = e^(-3*lambda)*e^(3*lambda) and then pull the e^(-3*lambda) outside the summation.

    And then at that point the summation should give you 1. You want to rearrange things properly and then show that what you actually have is the pmf of a poisson (possibly for a different value of lambda - but it doesn't matter if it's different as long as it's a poisson pmf) and when you sum over the pmf of a poisson... you get 1.

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    Re: Unbiased Estimator Question

    I multiplied by that form of 1 to get the expression above. Because originally the expression is:

    summation of (-2)^lambda*e^(-lambda)*lambda^y/y!

    Multiplying by e^(-3*lambda)*e^(3*lambda) and pulling e^-3*lambda out, gives the following expression in the summation:

    (-2)^lambda*e^(-lambda)*e(3*lambda)*lambda^y/y!

    Simplifying the exponential terms in the summation expression, I get:

    (-2)^lambda*e^(2*lambda)*lambda^y/y!

    with e^(-3*lambda) outside of the summation.

    I'm just a little lost on how to convert and simplify the summation term to appear as the pmf of poisson rv. Or did I make a wrong turn somewhere?

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    Re: Unbiased Estimator Question

    Recall that x^cy^c = (xy)^c

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    Re: Unbiased Estimator Question

    Great. Got it. So my other question would be, how can I justify whether (-2)^Y would (or would not) be useful in practice as an estimator of e^(-3*lambda)? From what I know of unbiased estimators, they may not always be the optimal choice of estimator, as they may not have the minimum mean square error. But how can I approach this issue for the above function?

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    Re: Unbiased Estimator Question

    Hi,

    I'm also stuck at this question. I got to e^-3*lambda*summation((-2)^lambda*e^2lambda*lambda^y/y!

    I really can't figure out how to get it to be a poisson pmf. I could divide up the e^2lambda but I can't figure out how to get rid of the (-2e)^lambda...

    Thanks!

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    Re: Unbiased Estimator Question

    Huh - don't know how I didn't realize I had a mistake in my one post with the actual math written out but it was supposed to be (-2)^y not (-2)^lambda.

    You shouldn't have (-2)^lambda so you shouldn't get a (-2e)^lambda term anywhere.

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    Re: Unbiased Estimator Question


    Hmmm, ok, but that only confuses me more about how to get to that pmf...

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