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Chapter 4: Q 15E (page 271)

Suppose that a person has a lottery ticket from which she will win X dollars, where X has the uniform distribution on the interval\(\left( {{\bf{0,4}}} \right)\). Suppose also that the person鈥檚 utility function is\({\bf{U}}\left( {\bf{x}} \right){\bf{ = }}{{\bf{x}}^{\bf{\alpha }}}\)for\({\bf{x}} \ge {\bf{0}}\), where 伪 is a given positive constant. For how many dollars\({{\bf{x}}_{\bf{0}}}\)would the person be willing to sell this lottery ticket?

Short Answer

Expert verified

The person will prefer to sell the lottery ticket for\({x_0}\)dollars if

\(x_0^\alpha > \frac{{{4^\alpha }}}{{\alpha + 1}}\)

Or if,

\({x_0} > \frac{4}{{{{\left( {\alpha + 1} \right)}^{{\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 \alpha }}\right.\ } \!\lower0.7ex\hbox{$\alpha $}}}}}}\)

Step by step solution

01

Given information 

The utility function is,

\(U\left( x \right) = {x^\alpha }\)

02

Necessary calculation

The expected utility for the lottery ticket is,

\(\begin{align}E\left( {U\left( X \right)} \right) &= \int_0^4 {\frac{1}{4}{x^\alpha }} dx\\ &= \frac{1}{4}\left( {\frac{{{x^{\alpha + 1}}}}{{\alpha + 1}}} \right)_0^4\\ &= \frac{1}{4}\left( {\frac{{{4^{\alpha + 1}}}}{{\alpha + 1}}} \right)\\ &= \frac{{{4^\alpha }}}{{\alpha + 1}}\end{align}\)

The utility of accepting\({x_0}\)dollars instead of the lottery ticket is,

\(U\left( {{x_0}} \right) = x_0^\alpha \)

Therefore, the person will prefer to sell the lottery ticket for\({x_0}\)dollars if

\(x_0^\alpha > \frac{{{4^\alpha }}}{{\alpha + 1}}\)

Or if,

\({x_0} > \frac{4}{{{{\left( {\alpha + 1} \right)}^{{\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 \alpha }}\right.\ } \!\lower0.7ex\hbox{$\alpha $}}}}}}\)

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Most popular questions from this chapter

Let X have the discrete uniform distribution on the integers \(1, \ldots \ldots ,n\).Compute the variance of X.

Hint: Youmay wish to use the formula \(\sum\limits_{k = 1}^n {{k^2}} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\)

Suppose that X and Y have a continuous joint distribution for which the joint p.d.f. is as follows:

\({\bf{f}}\left( {{\bf{x,y}}} \right){\bf{ = }}\left\{ {\begin{align}{}{{\bf{x + y}}}&{{\bf{for}}\,\,{\bf{0}} \le {\bf{x}} \le {\bf{1}}\,\,{\bf{and}}\,\,{\bf{0}} \le {\bf{y}} \le {\bf{1,}}}\\{\bf{0}}&{{\bf{otherwise}}{\bf{.}}}\end{align}} \right.\).

Find\({\bf{E}}\left( {{\bf{Y}}\left| {\bf{X}} \right.} \right)\)and\({\bf{Var}}\left( {{\bf{Y}}\left| {\bf{X}} \right.} \right)\).

Suppose thatXis a random variable for which the m.g.f. is as follows:\(\psi \left( t \right) = \frac{1}{5}{e^t} + \frac{2}{5}{e^{4t}} + \frac{2}{5}{e^{8t}}\)for鈭掆垶< t <鈭�.Find the probability distribution ofX. Hint:It is a simple discrete distribution.

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Suppose that a class contains 10 boys and 15 girls, and suppose that eight students are to be selected randomly from the class without replacement. Let X denote the number of boys selected, and let Y denote the number of girls selected. Find E(X 鈭� Y ).
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