A geometric distribution has a pdf given by P(X=x) = p(1-p)^x, where x = 0, 1, 2, …, and 0 < p=””>< 1.=”” this=”” form=”” of=”” the=”” geometric=”” starts=”” at=”” x=”0,” not=”” at=”” x=”1.” given=”” are=”” the=”” following=””>
E(X) = (1-p)/p, and Var(X) = (1-p)/p^2
A random sample of size n is drawn; the data are X1, X2, …, Xn.
A. Derive the Fisher information function for the parameter p.
B. Find the Cramér-Rao lower bound (CRLB) for the variance of an unbiased estimator for p.
C. Find a sufficient statistic for the parameter p
D. Show that the sample mean, xbar, is an unbiased estimate of E(X) = (1-p)/p Find the variance of xbar.
E. Argue whether or not the sample mean is a minimum variance unbiased estimate (MVUE) of population mean, mew.
The Correct Answer and Explanation is :
A. Fisher Information Function for Parameter ( p )
The Fisher information function ( I(p) ) is given by:
[
I(p) = – \mathbb{E} \left[ \frac{\partial^2}{\partial p^2} \log L(p) \right]
]
Where ( L(p) ) is the likelihood function of the sample. For a random sample of size ( n ), the likelihood function is:
[
L(p) = \prod_{i=1}^{n} P(X_i = x_i) = \prod_{i=1}^{n} p(1 – p)^{x_i}
]
Taking the logarithm of the likelihood:
[
\log L(p) = \sum_{i=1}^{n} \log \left[ p(1 – p)^{x_i} \right]
= n \log p + \sum_{i=1}^{n} x_i \log(1 – p)
]
Now, we compute the first and second derivatives with respect to ( p ):
- First derivative:
[
\frac{\partial}{\partial p} \log L(p) = \frac{n}{p} – \sum_{i=1}^{n} \frac{x_i}{1 – p}
]
- Second derivative:
[
\frac{\partial^2}{\partial p^2} \log L(p) = – \frac{n}{p^2} – \sum_{i=1}^{n} \frac{x_i}{(1 – p)^2}
]
The Fisher information is:
[
I(p) = – \mathbb{E} \left[ \frac{\partial^2}{\partial p^2} \log L(p) \right]
]
Since ( \mathbb{E}[X_i] = \frac{1 – p}{p} ), the Fisher information function is:
[
I(p) = \frac{n}{p^2} + \frac{n(1 – p)}{p^3}
]
B. Cramér-Rao Lower Bound (CRLB) for the Variance of an Unbiased Estimator of ( p )
The CRLB is given by:
[
\text{CRLB} = \frac{1}{I(p)}
]
From part A, we know the Fisher information ( I(p) ). Therefore, the CRLB is:
[
\text{CRLB} = \frac{p^2}{n(1 – p)}
]
This is the minimum variance for an unbiased estimator of ( p ).
C. Sufficient Statistic for ( p )
By the factorization theorem, a statistic is sufficient for ( p ) if the likelihood can be factored into a product of two terms, one of which depends only on the data through the statistic. The likelihood function is:
[
L(p) = p^n (1 – p)^{\sum_{i=1}^{n} x_i}
]
This can be factored as:
[
L(p) = \left( p^n \right) \left( (1 – p)^{\sum_{i=1}^{n} x_i} \right)
]
Thus, the statistic ( T(X) = \sum_{i=1}^{n} X_i ) is sufficient for ( p ).
D. Unbiased Estimate of ( E(X) = \frac{1 – p}{p} ) and Variance of ( \overline{X} )
The sample mean ( \overline{X} ) is an unbiased estimator of ( E(X) = \frac{1 – p}{p} ) because:
[
\mathbb{E}[\overline{X}] = \mathbb{E}[X_1] = \frac{1 – p}{p}
]
To find the variance of ( \overline{X} ), recall that:
[
\text{Var}(\overline{X}) = \frac{\text{Var}(X_1)}{n}
]
Since ( \text{Var}(X_1) = \frac{1 – p}{p^2} ), we have:
[
\text{Var}(\overline{X}) = \frac{1 – p}{n p^2}
]
E. Is the Sample Mean the Minimum Variance Unbiased Estimator (MVUE)?
To determine if ( \overline{X} ) is the MVUE, we need to check if it is unbiased and has the minimum variance among all unbiased estimators. From parts D and B, we know that ( \overline{X} ) is an unbiased estimator of ( \frac{1 – p}{p} ), and its variance is given by:
[
\text{Var}(\overline{X}) = \frac{1 – p}{n p^2}
]
From the CRLB, the minimum variance for an unbiased estimator of ( p ) is ( \frac{p^2}{n(1 – p)} ). Since the sample mean does not achieve this lower bound, it is not the MVUE.
Thus, while ( \overline{X} ) is an unbiased estimator, it is not the MVUE because it does not attain the Cramér-Rao lower bound. The MVUE for ( p ) could be a more complex estimator that incorporates the sufficient statistic ( T(X) = \sum_{i=1}^{n} X_i ).