Answer to Question #87366 in Statistics and Probability for Shivam Nishad

Question #87366
Let X1, X2, ..., Xn be a random sample from a Binomial distribution with parameters n
and p, both unkown. Obtain estimators of n and p, using method of moments.
1
Expert's answer
2019-04-09T10:40:44-0400

Since


"\\mu=EX_1=kp"

and


"EX_1^2=Var(X_1)-(EX_1)^2=\\sigma^2+\\mu^2"




"EX_1^2=kp(1-p)+k^2p^2"


Setting "\\widehat{\\mu}_1=\\mu" and "\\widehat{\\mu}_2=\\sigma^2+\\mu^2" we obtain the moment estimators

"\\widehat{\\theta}=(\\bar{X}, {1 \\over n}\\displaystyle\\sum_{i=1}^n(X_i-\\bar{X})^2)=(\\bar{X}, {n-1 \\over n}S^2)"


"np={1 \\over n}\\displaystyle\\sum_{i=1}^nX_i=\\bar{X}"

"np(1-p)={1 \\over n}\\displaystyle\\sum_{i=1}^n(X_i-\\bar{X})^2"




"S^2 ={1 \\over n-1}\\displaystyle\\sum_{i=1}^n(X_i-\\bar{X})^2"

"1-p={{1 \\over n}\\displaystyle\\sum_{i=1}^n(X_i-\\bar{X})^2 \\over \\bar{X}}={n-1 \\over n}S^2\/\\bar{X}"





"\\widehat{p}=(\\widehat{\\mu}_1+\\widehat{\\mu}_1^2-\\widehat{\\mu}_2)\/\\widehat{\\mu}_1=1-{n-1 \\over n}S^2\/\\bar{X}"

and


"\\widehat{k}\\widehat{p}=\\bar{X}\\Rightarrow\\widehat{k}={\\bar{X} \\over \\widehat{p}}"


"\\widehat{k}=\\widehat{\\mu}_1^2\/(\\widehat{\\mu}_1+\\widehat{\\mu}_1^2-\\widehat{\\mu}_2)=\\bar{X}\/(1-{n-1 \\over n}S^2\/\\bar{X})"

The estimator "\\widehat{p}" is in the range of (0, 1).

But "\\widehat{k}" may not be an integer.

It can be improved by an estimator that is "\\widehat{k}" rounded to the nearest positive integer.



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