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Chapter 4: Q19SE (page 273)

Suppose thatXis a random variable with m.g.f\(\psi \left( t \right)\),mean\(\mu \), and variance\({\sigma ^2}\); and let\(c\left( t \right) = \log \left( {\psi \left( t \right)} \right)\). Prove that\(c'\left( 0 \right) = \mu \;and\;c''\left( 0 \right) = {\sigma ^2}\).

Short Answer

Expert verified

It is proved that\(c'\left( 0 \right) = \mu \;\)and\(c''\left( 0 \right) = {\sigma ^2}\;\).

Step by step solution

01

Given information

A random variable X has a mean\(\mu \)and variance\({\sigma ^2}\). It has the moment generating function\(\psi \left( t \right)\).

Let\(c\left( t \right) = \left( {\log \psi \left( t \right)} \right)\).

02

Differentiate the given function

By differentiating c(t) we get,

\(\begin{aligned}{c}c'\left( t \right) = \frac{d}{{dt}}\left( {\log \psi \left( t \right)} \right)\\ = \frac{1}{{\psi \left( t \right)}} \times \frac{d}{{dt}}\psi \left( t \right)\\ = \frac{{\psi '\left( t \right)}}{{\psi \left( t \right)}}\end{aligned}\)

So, if t=0 then,

\(\begin{aligned}{c}c'\left( 0 \right) = \frac{{\psi '\left( 0 \right)}}{{\psi \left( 0 \right)}}\\ = \mu \end{aligned}\)

Thus, we proved that \(c'\left( 0 \right) = \mu \).

03

Double differentiate the given function

Differentiating\(c'\left( t \right)\)with respect to t,

\(\begin{aligned}{c}c''\left( t \right) = \frac{d}{{dt}}\left( {\frac{{\psi '\left( t \right)}}{{\psi \left( t \right)}}} \right)\\ = \frac{{\psi \left( t \right)\frac{d}{{dt}}\left( {\psi '\left( t \right)} \right) - \psi '\left( t \right)\frac{d}{{dt}}\psi \left( t \right)}}{{{{\left( {\psi \left( t \right)} \right)}^2}}}\\ = \frac{{\psi \left( t \right) \times \psi ''\left( t \right) - {{\left( {\psi '\left( t \right)} \right)}^2}}}{{{{\left( {\psi \left( t \right)} \right)}^2}}}\end{aligned}\)

So, if t=0 then,

\(c''\left( 0 \right) = \frac{{\psi \left( 0 \right) \times \psi ''\left( 0 \right) - {{\left( {\psi '\left( t \right)} \right)}^2}}}{{{{\left( {\psi \left( 0 \right)} \right)}^2}}}\)

Since\(\psi \left( t \right)\)is moment generating function of t,\(\psi \left( 0 \right) = 1,\psi '\left( 0 \right) = \mu \;and\;\psi ''\left( 0 \right) = E\left( {{X^2}} \right)\).

Now,

\(\begin{aligned}{c}{\sigma ^2} = E\left( {{X^2}} \right) - \left( {E{{\left( X \right)}^2}} \right)\\ = E\left( {{X^2}} \right) - {\mu ^2}\end{aligned}\)

So,\(E\left( {{X^2}} \right) = {\sigma ^2} + {\mu ^2}\)

Therefore, one get,

\(\begin{aligned}{c}c''\left( 0 \right) = \frac{{{\sigma ^2} + {\mu ^2} - {\mu ^2}}}{1}\\ = {\sigma ^2}\end{aligned}\)

Thus, it is proved that \(c''\left( 0 \right) = {\sigma ^2}\).

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Most popular questions from this chapter

Consider a utility function U for which\({\bf{U}}\left( {\bf{0}} \right){\bf{ = 5}}\), \({\bf{U}}\left( {\bf{1}} \right){\bf{ = 8}}\), and \({\bf{U}}\left( {\bf{2}} \right){\bf{ = 10}}\). Suppose that a person who has this utility function is indifferent to either of two gambles X and Y, for which the probability distributions of the gains are as follows:

\(\begin{align}{\bf{Pr}}\left( {{\bf{X = - 1}}} \right){\bf{ = 0}}{\bf{.6,}}\,{\bf{Pr}}\left( {{\bf{X = 0}}} \right){\bf{ = 0}}{\bf{.2,}}\,{\bf{Pr}}\left( {{\bf{X = 2}}} \right){\bf{ = 0}}{\bf{.2;}}\\{\bf{Pr}}\left( {{\bf{Y = 0}}} \right){\bf{ = 0}}{\bf{.9,}}\,\,\,\,{\bf{Pr}}\left( {{\bf{Y = 1}}} \right){\bf{ = 0}}{\bf{.1}}\end{align}\)

What is the value\({\bf{U}}\left( { - {\bf{1}}} \right)\)?

Suppose that a fair coin is tossed repeatedly until exactlykheads have been obtained. Determine the expected number of tosses that will be required.

Hint:Represent the total number of tossesXin the form \({\bf{X = }}{{\bf{X}}_{\bf{1}}}{\bf{ + }} \cdots {\bf{ + }}{{\bf{X}}_{\bf{k}}}\), where \({{\bf{X}}_{\bf{i}}}\)is the number of tosses required to obtain theith head afteri鈭�1 heads have been obtained.

Let X be a random variable. Suppose that there exists a number m such that \(\Pr \left( {X < m} \right) = \Pr \left( {X > m} \right)\).Prove that m is a median of the distribution ofX.

Suppose that the proportion of defective items in a large lot is p, and suppose that a random sample of n items is selected from the lot. Let X denote the number of defective items in the sample, and let Y denote the number of non-defective items. Find E (X 鈭� Y)

Suppose that the random variables \({{\bf{X}}_{\bf{1}}}{\bf{,}}{{\bf{X}}_{\bf{2}}}{\bf{,}}....{{\bf{X}}_{\bf{n}}}\)form a random sample of sizenfrom the uniform distribution on the interval [0,1].

Let \({{\bf{Y}}_{\bf{1}}}{\bf{ = min\{ }}{{\bf{X}}_{\bf{1}}}{\bf{,}}{{\bf{X}}_{\bf{2}}}{\bf{,}}....{{\bf{X}}_{\bf{n}}}{\bf{\} }}\), and let \({{\bf{Y}}_{\bf{n}}}{\bf{ = max\{ }}{{\bf{X}}_{\bf{1}}}{\bf{,}}{{\bf{X}}_{\bf{2}}}{\bf{,}}....{{\bf{X}}_{\bf{n}}}{\bf{\} }}\)

Find \({\bf{E(}}{{\bf{Y}}_{\bf{1}}}{\bf{)}}\)and \({\bf{E(}}{{\bf{Y}}_{\bf{n}}}{\bf{)}}\)

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