Results 1 to 5 of 5

Math Help - Minimum variance unbiased estimator proof

  1. #1
    Member
    Joined
    Jun 2012
    From
    Georgia
    Posts
    193
    Thanks
    25

    Minimum variance unbiased estimator proof

    "Show that the mean of a random sample of size n is a minimum variance unbiased estimator of the parameter $\lambda$ of a Poisson distribution."

    Here are my steps. Somewhere along the way, I got lost.

    1. The distribution is Poisson, so $\mu = \sigma^2 = \lambda$. $E(\overline{X})$, so that the estimator is unbiased is clear.
    2. $f(x) = \frac{\lambda ^ x \exp(-\lambda)}{x!}$
    3. $\ln f(x) = x \ln \lambda - \ln x! - \lambda$
    4. $\{\frac{\partial}{\partial \lambda}[\ln f(x)] \}^2 = \frac{x^2}{\lambda^2} - \frac{2x}{\lambda} + 1$

    There's no way that can be right...What did I do wrong??
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,789
    Thanks
    1147

    Re: Minimum variance unbiased estimator proof

    a sample of n i.i.d poisson variates has distribution

    we know the maximum likelihood estimate is minimum variance, we need to see if it's unbiased.

    $\large p_{\vec{K}}(\vec{k})=\displaystyle{e^{-\lambda} \prod_{i=1}^n}\dfrac {\lambda^{k_i}}{k_i!}$

    $\large \ln(p_{\vec{K}}(\vec{k})=\displaystyle{\sum_{i=1}^ n}\left(k_i \ln(\lambda)-\ln(k_i!)-\lambda\right)$

    $\large \dfrac \partial {\partial \lambda}\ln(p_{\vec{K}}(\vec{k}))=\dfrac 1 \lambda \displaystyle{\sum_{i=1}^n}\left(k_i - 1\right)$

    setting this to zero we get

    $\dfrac 1 \lambda \displaystyle{\sum_{i=1}^n}k_i = n$

    $\dfrac 1 n \displaystyle{\sum_{i=1}^n}k_i = \lambda$

    as $E[k_i]=\lambda$ we see that it is unbiased and thus the mean is our unbiased minimum variance estimate.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2014
    From
    Little Neck, New York
    Posts
    6

    Re: Minimum variance unbiased estimator proof

    This is what I get but no guarantee it is correct:

    Step 4: x/λ - 1/λ - 1=0, Derivative wrt λ
    Step 5: (x^2 - λ) / λx = 1, Put under common denominator, move 1 to RHS
    Step 6: (x^2 - λ) = λx, Multiply both by denominator
    Step 7: x - λ = λ, Divide by x both sides
    Step 8: x= 2 λ, then square this function.

    I have the right solution somewhere, if I find it later and this incorrect let you know.
    Toni
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jun 2014
    From
    Little Neck, New York
    Posts
    6

    Re: Minimum variance unbiased estimator proof

    This is brilliant, I see my answer had a goof in it. But shouldn't the function be Sum of Ki * (Lambda -1)? Doesn't affect the final answer but I don't think we mean to multiply the 1 by 1/Lambda as you have there 1/L * ((SUM Ki-1), right? The -1 is just summed over 1 to n so becomes -n then goes to the other side as +n, not n/lambda.

    Looks like just minor parens mismatch if I am not mistaken (again! LOL).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jun 2012
    From
    Georgia
    Posts
    193
    Thanks
    25

    Re: Minimum variance unbiased estimator proof

    Quote Originally Posted by TSmarsco View Post
    This is what I get but no guarantee it is correct:

    Step 4: x/λ - 1/λ - 1=0, Derivative wrt λ
    Step 5: (x^2 - λ) / λx = 1, Put under common denominator, move 1 to RHS
    Step 6: (x^2 - λ) = λx, Multiply both by denominator
    Step 7: x - λ = λ, Divide by x both sides
    Step 8: x= 2 λ, then square this function.

    I have the right solution somewhere, if I find it later and this incorrect let you know.
    Toni
    OK I see that's not right, but really all I have to do is take my step 4 equal to zero and go from there? Note, that's a trinomial square, so I could just back it up and say:

    5. $\frac{x}{\lambda} - 1 = 0$
    6. $\Rightarrow x = \lambda$
    7. Observe, $E(\overline{X}) = \mu = \lambda,$ so $\overline{X}$ is an MVUE of $\lambda$.

    Right?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: April 1st 2011, 12:17 AM
  2. Calculating minimum variance unbiased estimator (MVUE)
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: March 31st 2010, 03:58 PM
  3. Finding minimum variance unbiased estimator (MVUE)
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: March 30th 2010, 06:49 PM
  4. Variance of Unbiased Estimator
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: February 16th 2010, 09:59 PM
  5. Minimum Variance Unbiased Estimator
    Posted in the Advanced Statistics Forum
    Replies: 6
    Last Post: August 24th 2009, 07:38 AM

Search Tags


/mathhelpforum @mathhelpforum