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Statistics / Mathematical Statistics with Applications 7 / Chapter 3 / Problem 85E

Mathematical Statistics with Applications | 7th Edition | ISBN: 9780495110811 | Authors: Dennis Wackerly; William Mendenhall; Richard L. Scheaffer

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Find E[Y (Y ? 1)] for a geometric random variable Y by

Chapter 3, Problem 85E

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QUESTION:

Find \(E[Y(Y-1)]\) for a geometric random variable $Y$ by finding \(d^{2} / d q^{2}\left(\Sigma_{y=1}^{\infty} q^{y}\right)\). Use this result to find the variance of $Y$ .

Equation Transcription:

$E[Y(Y-1)]$

${d}^{2}/d{q}^{2}({\Sigma{}}_{y=1}^{\infty{}}{q}^{y})$

Text Transcription:

E[Y(Y-1)]

d^2/dq^2(Sigma_y=1^infinity q^y)

Questions & Answers

QUESTION:

Find \(E[Y(Y-1)]\) for a geometric random variable $Y$ by finding \(d^{2} / d q^{2}\left(\Sigma_{y=1}^{\infty} q^{y}\right)\). Use this result to find the variance of $Y$ .

Equation Transcription:

$E[Y(Y-1)]$

${d}^{2}/d{q}^{2}({\Sigma{}}_{y=1}^{\infty{}}{q}^{y})$

Text Transcription:

E[Y(Y-1)]

d^2/dq^2(Sigma_y=1^infinity q^y)

ANSWER:

Answer:

Step 1 of 1:

We need to find $E[Y(Y-1)]$ for a geometric random variable $Y$ by finding $\frac{{d}^{2}}{d{q}^{2}}\left({\sum\limits_{y\ =1}^{\infty{}}{q}^{y}}\right)$ . use this result to find the variance of $Y.$

The mean of a random variable with a geometric probability distribution can be written like this,

$E(Y)\ =\ \mu{}\ =\sum\limits_{y\ =1}^{\infty{}}{yq}^{y-1}p$ ……..(1)

$\ =p\sum\limits_{y\ =1}^{\infty{}}{yq}^{y-1}$

Hence we can write using equation (1)

$E[Y(Y-1)]\ =p\sum\limits_{y\ =1}^{\infty{}}{y(y-1)q}^{y-1}$ …….(2)

We need to solve the equation (2) by finding the value of,

$\frac{{d}^{2}}{d{q}^{2}}\left({\sum\limits_{y\ =1}^{\infty{}}{q}^{y}}\right)$

Lets solve it,

$\frac{{d}^{2}}{d{q}^{2}}({q}^{y})\ =y(y-1){q}^{y-2}\$

Thus,

$\frac{{d}^{2}}{d{q}^{2}}\left({\sum\limits_{y\ =1}^{\infty{}}{q}^{y}}\right)\ =\sum\limits_{y\ =1}^{\infty{}}y(y-1){q}^{y-2}\$ …….(3)

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