Chapter 10, Problem 67RE

(a)

To determine

The expected pay off of an infinite series.

Answer to Problem 67RE

The value of $\text{Expected payoff}=0\left(\frac{1}{2}\right)+1{\left(\frac{1}{2}\right)}^2+2{\left(\frac{1}{2}\right)}^3+3{\left(\frac{1}{2}\right)}^4+\mathrm{......}+n{\left(\frac{1}{2}\right)}^{n+1}+\mathrm{....}$

Explanation of Solution

Given information:

Toss a fair coin which the heads and tails are equally likely. When it comes up head on winning the dollar hence the game is over as soon as it comes up for the tails.

Formula used:

The Probability of occurring the tail in first chance, $T=\frac{1}{2}$

Calculation:

Let the event when head appear be H and when tail appears being T.

Probability of occurring the tail in first chance, $T=\frac{1}{2}$ probability of occurring the tail in second, $HT={\left(\frac{1}{2}\right)}^2$

Similarly, $HHT={\left(\frac{1}{2}\right)}^3$

Payoff when tail occur in first time $=0.\frac{1}{2}$

Payoff when tail occur in second time $=1.{\left(\frac{1}{2}\right)}^2$

Therefore,

  $\text{Expected payoff}=0\left(\frac{1}{2}\right)+1{\left(\frac{1}{2}\right)}^2+2{\left(\frac{1}{2}\right)}^3+3{\left(\frac{1}{2}\right)}^4+\mathrm{......}+n{\left(\frac{1}{2}\right)}^{n+1}+\mathrm{....}$

Conclusion:

The value of $\text{Expected payoff}=0\left(\frac{1}{2}\right)+1{\left(\frac{1}{2}\right)}^2+2{\left(\frac{1}{2}\right)}^3+3{\left(\frac{1}{2}\right)}^4+\mathrm{......}+n{\left(\frac{1}{2}\right)}^{n+1}+\mathrm{....}$

(b)

To determine

The value of the series $\frac{1}{1-x}$

Answer to Problem 67RE

The value is $\frac{1}{\left(1+x^2\right)}=1+2x+3x^2+\mathrm{....}+n{x}^{n-1}+\mathrm{...}$

Explanation of Solution

Given:

The series is

  $\frac{1}{1-x}=1+x+x^2+\mathrm{....}+x^n+\mathrm{...}$

Formula used:

Maclaurin series is used.

Calculation:

Maclaurin series generated by function $\frac{1}{1-x}$ is given by

  $\frac{1}{1-x}=1+x+x^2+\mathrm{....}+x^n+\mathrm{...}$

Differentiating the above expression

  $\begin{array}{l}\frac{1}{\left(1+x^2\right)}=\frac{d}{dx}\left(1+x+x^2+x^3+\mathrm{....}+x^n+\mathrm{...}\right)\\ \frac{1}{\left(1+x^2\right)}=1+2x+3x^2+\mathrm{....}+n{x}^{n-1}+\mathrm{...}\end{array}$

Therefore,

  $\frac{1}{\left(1+x^2\right)}=1+2x+3x^2+\mathrm{....}+n{x}^{n-1}+\mathrm{...}$

Conclusion:

The value is $\frac{1}{\left(1+x^2\right)}=1+2x+3x^2+\mathrm{....}+n{x}^{n-1}+\mathrm{...}$

(c)

To determine

The value of the series $x^2\frac{x^2}{{\left(1-x\right)}^2}$

Answer to Problem 67RE

The value is $x^2\frac{x^2}{{\left(1-x\right)}^2}=x^2+2x^3+3x^4+\mathrm{......}+n{x}^{n+1}+\mathrm{....}$

Explanation of Solution

Given information:

The series is

Formula used:

Multiplication is used.

Calculation:

Multiplying the above series by

  $\begin{array}{l}{x}^2\frac{x^2}{{\left(1-x\right)}^2}=x^2\left(1+2x+3x^2+\mathrm{......}+n{x}^{n-1}+\mathrm{.....}\right)\\ x^2\frac{x^2}{{\left(1-x\right)}^2}=x^2+2x^3+3x^4+\mathrm{......}+n{x}^{n+1}+\mathrm{....}\end{array}$

Therefore,

  $x^2\frac{x^2}{{\left(1-x\right)}^2}=x^2+2x^3+3x^4+\mathrm{......}+n{x}^{n+1}+\mathrm{....}$

Conclusion:

The value is $x^2\frac{x^2}{{\left(1-x\right)}^2}=x^2+2x^3+3x^4+\mathrm{......}+n{x}^{n+1}+\mathrm{....}$

(d)

To determine

The expected payoff of the game.

Answer to Problem 67RE

The expected payoff is $1

Explanation of Solution

Given information:

The series is

  $\frac{1}{1-x}=1+x+x^2+\mathrm{....}+x^n+\mathrm{...}$

Formula used:

Division is done.

Calculation:

If $x=\frac{1}{2}$ the formula in pan (c) matches the nonzero terms of the series in part (a)

Since,

  $\frac{{\left(\frac{1}{2}\right)}^2}{{\left[1-\frac{1}{2}\right]}^2}=1$

Therefore expected payoff is $1

Conclusion:

The expected payoff is $1