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Chapter 4: Q 5E (page 264)

Suppose that a point\({{\bf{X}}_{\bf{1}}}\)is chosen from the uniform distribution on the interval\(\left( {{\bf{0,1}}} \right)\)and that after the value\({{\bf{X}}_{\bf{1}}}{\bf{ = }}{{\bf{x}}_{\bf{1}}}\)is observed, a point\({{\bf{X}}_{\bf{2}}}\)is chosen from a uniform distribution on the interval\(\left( {{{\bf{x}}_{\bf{1}}}{\bf{,1}}} \right)\). Suppose further that additional variables\({{\bf{X}}_{\bf{3}}}{\bf{,}}{{\bf{X}}_{\bf{4}}}{\bf{,}}...\)are generated in the same way. Generally,\({\bf{j = 1,2,}}...{\bf{,}}\)after the value\({{\bf{X}}_{\bf{j}}}{\bf{ = }}{{\bf{x}}_{\bf{j}}}\)has been observed,\({{\bf{X}}_{{\bf{j + 1}}}}\)is chosen from a uniform distribution on the interval\(\left( {{{\bf{x}}_{\bf{j}}}{\bf{,1}}} \right)\). Find the value of\({\bf{E}}\left( {{{\bf{X}}_{\bf{n}}}} \right)\).

Short Answer

Expert verified

\(E\left( {{X_n}} \right) = 1 - \frac{1}{{{2^n}}}\)

Step by step solution

01

Given information

\({X_1}\) is chosen from the uniform distribution on the interval is \(\left( {0,1} \right)\). After that, the value\({X_1} = {x_1}\) is observed, and a point \({X_2}\) is chosen from a uniform distribution with intervals.

02

Finding the expected value

For any given value \({x_{n - 1}}\) of \({X_{n - 1}}\), \(E\left( {{X_n}\left| {{x_{n - 1}}} \right.} \right)\) will be the midpoint of the interval\(\left( {{x_{n - 1}},1} \right)\)

Therefore,\(E\left( {{X_n}\left| {{x_{n - 1}}} \right.} \right) = \frac{{1 + {X_{n - 1}}}}{2}\)

It follows that,

\(\begin{align}E\left( {{X_n}} \right) &= E\left( {E\left( {{X_n}\left| {{X_{n - 1}}} \right.} \right)} \right)\\ &= \frac{1}{2} + \frac{1}{2}E\left( {{X_{n - 1}}} \right)\end{align}\)

Similarly,

\(\begin{align}E\left( {{X_{n - 1}}} \right) &= E\left( {E\left( {{X_{n - 1}}\left| {{X_{n - 2}}} \right.} \right)} \right)\\ &= \frac{1}{2} + \frac{1}{2}E\left( {{X_{n - 2}}} \right)\end{align}\)

Since, \(E\left( {{X_1}} \right) = \frac{1}{2}\)

So,

\(\begin{align}E\left( {{X_n}} \right) &= \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... + \frac{1}{{{2^n}}}\\ &= 1 - \frac{1}{{{2^n}}}\end{align}\)

Therefore, \(E\left( {{X_n}} \right) = 1 - \frac{1}{{{2^n}}}\)

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Most popular questions from this chapter

Suppose that an observed value of X is equally likely to come from a continuous distribution for which the pdf is for from one for which the pdf is g. Suppose that \(f\left( x \right) > 0\) for \(0 < x < 1\) and \(f\left( x \right) = 0\) otherwise, and suppose also that \(g\left( x \right) > 0\) for \(2 < x < 4\) and \(g\left( x \right) = 0\) otherwise. Determine:

  1. the mean and
  2. the median of the distribution of X.

Let\({\bf{\alpha > 0}}\). A decision-maker has a utility function for money of the form

\({\bf{U}}\left( {\bf{x}} \right){\bf{ = }}\left\{ {\begin{align}{}{{{\bf{x}}^{\bf{\alpha }}}}&{{\bf{if}}\,{\bf{x > 0,}}}\\{\bf{x}}&{{\bf{if}}\,{\bf{x}} \le {\bf{0}}{\bf{.}}}\end{align}} \right.\)

Suppose that this decision maker is trying to decide whether or not to buy a lottery ticket for \(1. The lottery ticket pays \)500 with a probability of 0.001, and it pays $0 with a probability of 0.999. What would the values of 伪 have to be for this decision-maker to prefer buying the ticket to not buying it?

Suppose that\({\bf{X}}\)and\({\bf{Y}}\)are random variables such that

\({\bf{Var}}\left( {\bf{X}} \right){\bf{ = 9}}\),\({\bf{Var}}\left( {\bf{Y}} \right){\bf{ = 4}}\),and\({\bf{\rho }}\left( {{\bf{X,Y}}} \right){\bf{ = - }}\frac{{\bf{1}}}{{\bf{6}}}\).Determine

(a)\({\bf{Var}}\left( {{\bf{X + Y}}} \right)\)and(b)\({\bf{Var}}\left( {{\bf{X - 3Y + 4}}} \right)\).

LetXbe a random variable for which \(E\left( X \right) = \mu \), and\(Var\left( X \right) = {\sigma ^2}\), and let c be an arbitrary constant. Show that \(E\left( {{{\left( {X - c} \right)}^2}} \right) = {\left( {\mu - c} \right)^2} + {\sigma ^2}\).

Let X be a random variable with c.d.f F. Suppose that\(a < b\)are numbers such that both a and b are medians of X.

  1. Prove that\(F\left( a \right) = \frac{1}{2}\).
  2. Prove that there exists a smallest \(c \le a\)and a largest\(d \ge b\)such that every number in the closed interval\(\left( {c,d} \right)\) is a median of X.
  3. If X has a discrete distribution, prove that \(F\left( d \right) > \frac{1}{2}\).
See all solutions

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