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Chapter 5: Q23E (page 326)

Let F be a continuous CDF satisfying F (0) = 0, and suppose that the distribution with CDF F has the memoryless property (5.7.18). Define f(x) = log[1 鈭� F (x)] for x > 0

a. Show that for all t, h > 0,

\({\bf{1 - F}}\left( {\bf{h}} \right){\bf{ = }}\frac{{{\bf{1 - F}}\left( {{\bf{t + h}}} \right)}}{{{\bf{1 - F}}\left( {\bf{t}} \right)}}\)

b. Prove that \(\ell \)(t + h) = \(\ell \)(t) + \(\ell \)(h) for all t, h > 0.

c. Prove that for all t > 0 and all positive integers k and m, \(\ell \)(kt/m) = (k/m)\(\ell \) (t).

d. Prove that for all t, c > 0, \(\ell \) (ct) = c\(\ell \) (t).

e. Prove that g(t) = \(\ell \) (t)/t is constant for t > 0

f. Prove that F must be the CDF of an exponential distribution.

Short Answer

Expert verified

a.\(1 - F\left( h \right) = \frac{{1 - F\left( {t + h} \right)}}{{1 - F\left( t \right)}}\)

b.\(\ell \) (t + h) = \(\ell \) (t) + \(\ell \) (h) for all t, h > 0.

c. for all t > 0 and all positive integers k and m, \(\ell \) (kt/m) = (k/m)\(\ell \) (t).

d. for all t, c > 0, \(\ell \) (ct) = c\(\ell \) (t).

e. g(t) = \(\ell \) (t)/t is constant for t > 0

f. F must be the CDF of an exponential distribution.

Step by step solution

01

Given information

F is a continuous CDF satisfying F (0) = 0 and the distribution with CDF F has the memoryless property (5.7.18) and f(x) = log[1 鈭� F (x)] for x > 0

02

Proof of part (a),(b),(c),(d),(e),(f)

a. Since F has memoryless property, it follows that

\(\begin{aligned}{}\frac{{1 - F\left( {t + h} \right)}}{{1 - F\left( t \right)}} &= \frac{{1 - P\left( {X < t + h} \right)}}{{1 - P\left( {X < t} \right)}}\\ &= \frac{{P\left( {X \ge t + h} \right)}}{{P\left( {X \ge t} \right)}}\\ &= \frac{{P\left( {X \ge t + h,X \ge t} \right)}}{{P\left( {X \ge t} \right)}}\\ &= P\left( {X \ge t + h|X \ge t} \right)\\ &= P\left( {X \ge h} \right)\\ &= 1 - P\left( {X < h} \right)\\& = 1 - F\left( h \right)\end{aligned}\)

Hence the relation is proved.

b. From part(a),

\(1 - F\left( h \right) = \frac{{1 - F\left( {t + h} \right)}}{{1 - F\left( t \right)}}\)

Taking logarithm on both sides, we have

\(\begin{aligned}{}{\rm{log}}\left( {1 - F\left( h \right)} \right)& = {\rm{log}}\left( {\frac{{1 - F\left( {t + h} \right)}}{{1 - F\left( t \right)}}} \right)\\{\rm{log}}\left( {1 - F\left( h \right)} \right) &= {\rm{log}}\left( {1 - F\left( {t + h} \right)} \right) - {\rm{log}}\left( {1 - F\left( t \right)} \right)\\\ell \left( h \right)& = \ell \left( {t + h} \right) - \ell \left( t \right)\\\ell \left( {t + h} \right) &= \ell \left( t \right) + \ell \left( h \right)\end{aligned}\)

Hence it is proved.

c.From part (b)

\(\begin{aligned}{}\ell \left( {\frac{t}{m} + \frac{t}{m}} \right) &= \ell \left( {\frac{t}{m}} \right) + \ell \left( {\frac{t}{m}} \right)\\\ell \left( {\frac{{2t}}{m}} \right)&= 2\ell \left( {\frac{t}{m}} \right)\,\,{\rm{and}}\\{\rm{similarly}}\,\ell \left( {\frac{{kt}}{m}} \right) &= k\ell \left( {\frac{t}{m}} \right)\\{\rm{Let}}\,{\rm{k = m}}\,{\rm{then}}\\\ell \left( {\frac{{mt}}{m}} \right) &= m\ell \left( {\frac{t}{m}} \right)\\\ell (t)& = m\ell \left( {\frac{t}{m}} \right)\\\ell \left( {\frac{t}{m}} \right)& = \frac{1}{m}\ell \left( t \right)\\Now,\\\ell \left( {\frac{{kt}}{m}} \right) &= k\ell \left( {\frac{t}{m}} \right)\\ &= k \times \frac{1}{m}\ell \left( t \right)\\ &= \frac{k}{m}\ell (t)\end{aligned}\)

Hence the relation is proved.

d. From part (c)

\(\begin{aligned}{}{\rm{if}}\frac{{\rm{k}}}{{\rm{m}}}{\rm{ = c}}\,{\rm{then}}\\\ell \left( {\frac{{kt}}{m}} \right) &= \frac{k}{m}\ell \left( t \right)\\\ell \left( {ct} \right)& = c\ell \left( t \right)\end{aligned}\)

e.From part (d)

\(\begin{aligned}{l}{\rm{ifc = }}\frac{{\rm{1}}}{{\rm{t}}}{\rm{then}}\\g\left( t \right) = \ell \left( {\frac{t}{t}} \right) = \frac{{\ell \left( t \right)}}{t}\\\frac{{\ell \left( t \right)}}{t} = \ell \left( 1 \right) = {\rm{constant}}\end{aligned}\)

f. From part(e)

\(\begin{aligned}{}\frac{{\ell \left( t \right)}}{t} &= \ell \left( 1 \right) = - \lambda \left( {{\rm{say}}} \right)\\\ell \left( t \right)& = - \lambda t\\{\rm{log}}\left( {1 - F\left( t \right)} \right) &= - \lambda t\\1 - F\left( t \right) &= {e^{ - \lambda t}}\\F\left( t \right) &= 1 - {e^{ - \lambda t}}\end{aligned}\)

Thus F must be CDF of an exponential distribution with parameter\(\lambda = - \ell \left( 1 \right)\)\(\)

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Most popular questions from this chapter

Suppose that the random variables \({{\bf{X}}_{\bf{1}}}{\bf{ \ldots }}{{\bf{X}}_{\bf{k}}}\) are independent and that \({{\bf{X}}_{\bf{i}}}\) has the Poisson distribution with mean \({{\bf{\lambda }}_{\bf{i}}}\left( {{\bf{i = 1, \ldots ,k}}} \right)\). Show that for each fixed positive integer n, the conditional distribution of the random Vector \({\bf{X = }}\left( {{{\bf{X}}_{\bf{1}}}{\bf{ \ldots }}{{\bf{X}}_{\bf{k}}}} \right)\), given that \(\sum\limits_{{\bf{i = 1}}}^{\bf{k}} {{{\bf{X}}_{\bf{i}}}{\bf{ = n}}} \) it is the multinomial distribution with parameters n and

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