Get inspired by the success stories of our students in IIT JAM MS, ISI MStat, CMI MSc Data Science. Learn More

Content

[hide]

This is a very simple sample problem from ISI MStat PSB 2012 Problem 10. It's a very basic problem but very important and regular problem for statistics students, using one of the most beautiful theorem in Point Estimation. Try it!

Let \(X_1,X_2,.....X_{10}\) be i.i.d. Poisson random variables with unknown parameter \(\lambda >0\). Find the minimum variance unbiased estimator of exp{\(-2\lambda \)}.

Poisson Distribution

Minimum Variance Unbiased Estimators

Lehman-Scheffe's Theorem

Completeness and Sufficiency

Well, this is a very straight forward problem, where we just need to verify certain conditions, of sufficiency and completeness.

If, one is aware of the nature of Poisson Distribution, one knows that for a given sample \(X_1,X_2,.....X_{10}\), the sufficient statistics for the unknown parameter \(\lambda>0\), is \(\sum_{i=1}^{10} X_i \) , also by extension \(\sum_{i}X_i\) is also complete for \(\lambda\) (How??).

So, now first let us construct an unbiased estimator of \(e^{-2\lambda}\). Here, we need to observe patterns as usual. Let us define an Indicator Random variable,

\(I_X(x) = \begin{cases} 1 & X_1=0\ and\ X_2=0 \\ 0 & Otherwise \end{cases}\),

So, \(E(I_X(x))=P(X_1=0, X_2=0)=e^{-2\lambda}\), hence \(I_X(x)\) is an unbiased estimator of \(e^{-2\lambda}\). But is it a Minimum Variance ??

Well, Lehman-Scheffe answers that, Since we know that \(\sum X_i\) is complete and sufficient for \(\lambda \), By Lehman-Scheffe's theorem,

\(E(I_X(x)|\sum X_i=t)\) is the minimum variance unbiased estimator of \(e^{-2\lambda }\) for any \(t>0\). So, we need to find the following,

\(E(I_X(x)|\sum_{i=1}^{10}X_i=t)= \frac{P(X_1=0,X_2; \sum_{i}X_i=t)}{P(\sum_{i=3}^{10}X_i=t)}=\frac{e^{-2\lambda}e^{-8\lambda}\frac{(8\lambda)^t}{t!}}{e^{10\lambda}\frac{(10\lambda)^t}{t!}}=(\frac{8}{10})^t\).

So, the Minimum Variance Unbiased Estimator of exp{\(-2\lambda\)} is \((\frac{8}{10})^{\sum_{i=1}^{10}X_i}\)

Now can you generalize this for a sample of size n, again what if I defined \(I_X(x)\) as,

\(I_X(x) = \begin{cases} 1 & X_i=0\ &\ X_j=0 \\ 0 & Otherwise \end{cases}\), for some \(i \neq j\),

would it affected the end result ?? What do you think?

Let's not end our concern for Poisson, and think further, that for the given sample if the sample mean is \(\bar{X}\) and sample variance is \(S^2\). Can you show that \(E(S^2|\bar{X})=\bar{X}\), and further can you extend your deductions to \( Var(S^2) > Var(\bar{X}) \) ??

Finally can you generalize the above result ?? Give some thoughts to deepen your insights on MVUE.

Content

[hide]

This is a very simple sample problem from ISI MStat PSB 2012 Problem 10. It's a very basic problem but very important and regular problem for statistics students, using one of the most beautiful theorem in Point Estimation. Try it!

Let \(X_1,X_2,.....X_{10}\) be i.i.d. Poisson random variables with unknown parameter \(\lambda >0\). Find the minimum variance unbiased estimator of exp{\(-2\lambda \)}.

Poisson Distribution

Minimum Variance Unbiased Estimators

Lehman-Scheffe's Theorem

Completeness and Sufficiency

Well, this is a very straight forward problem, where we just need to verify certain conditions, of sufficiency and completeness.

If, one is aware of the nature of Poisson Distribution, one knows that for a given sample \(X_1,X_2,.....X_{10}\), the sufficient statistics for the unknown parameter \(\lambda>0\), is \(\sum_{i=1}^{10} X_i \) , also by extension \(\sum_{i}X_i\) is also complete for \(\lambda\) (How??).

So, now first let us construct an unbiased estimator of \(e^{-2\lambda}\). Here, we need to observe patterns as usual. Let us define an Indicator Random variable,

\(I_X(x) = \begin{cases} 1 & X_1=0\ and\ X_2=0 \\ 0 & Otherwise \end{cases}\),

So, \(E(I_X(x))=P(X_1=0, X_2=0)=e^{-2\lambda}\), hence \(I_X(x)\) is an unbiased estimator of \(e^{-2\lambda}\). But is it a Minimum Variance ??

Well, Lehman-Scheffe answers that, Since we know that \(\sum X_i\) is complete and sufficient for \(\lambda \), By Lehman-Scheffe's theorem,

\(E(I_X(x)|\sum X_i=t)\) is the minimum variance unbiased estimator of \(e^{-2\lambda }\) for any \(t>0\). So, we need to find the following,

\(E(I_X(x)|\sum_{i=1}^{10}X_i=t)= \frac{P(X_1=0,X_2; \sum_{i}X_i=t)}{P(\sum_{i=3}^{10}X_i=t)}=\frac{e^{-2\lambda}e^{-8\lambda}\frac{(8\lambda)^t}{t!}}{e^{10\lambda}\frac{(10\lambda)^t}{t!}}=(\frac{8}{10})^t\).

So, the Minimum Variance Unbiased Estimator of exp{\(-2\lambda\)} is \((\frac{8}{10})^{\sum_{i=1}^{10}X_i}\)

Now can you generalize this for a sample of size n, again what if I defined \(I_X(x)\) as,

\(I_X(x) = \begin{cases} 1 & X_i=0\ &\ X_j=0 \\ 0 & Otherwise \end{cases}\), for some \(i \neq j\),

would it affected the end result ?? What do you think?

Let's not end our concern for Poisson, and think further, that for the given sample if the sample mean is \(\bar{X}\) and sample variance is \(S^2\). Can you show that \(E(S^2|\bar{X})=\bar{X}\), and further can you extend your deductions to \( Var(S^2) > Var(\bar{X}) \) ??

Finally can you generalize the above result ?? Give some thoughts to deepen your insights on MVUE.

Cheenta is a knowledge partner of Aditya Birla Education Academy

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

JOIN TRIALAcademic Programs

Free Resources

Why Cheenta?

Google