Q. 7.67 - A First Course in Probability

Question

Consider a gambler who, at each gamble, either wins or loses her bet with respective probabilities $p$ and $1-p$. A popular gambling system known as the Kelley strategy is to always bet the fraction $2p-1$of your current fortune when $p>12$. Compute the expected fortune after$n$ gambles of a gambler who starts with $x$units and employs the Kelley strategy.

Step-by-Step Solution

Answer

The starts with $x$units value are $E\left(X_n\right)={\left[2p^2+2(1-p{)}^2\right]}^n·x$ .

Step 1: Given Information

Compute the expected fortune after $x$ gambles of a gambler who starts with $x$ units and employs the Kelley strategy.

Step 2: Explanation

Define random variables ${\left(X_n\right)}_{n\ge 0}$that marks the amount of money that we have in time$n$. Suppose that we have $X_{n-1}$ money and we invest it.

With probability$p$, we get $(2p-1)X_{n-1}$ of additional money and we have $(2p-1) X_{n-1}+X_{n-1}$ of money in time $n$.

On the other hand, if we lose the bet, we will have $-(2p-1)X_{n-1}+X_{n-1}$

of money in time $n$.

Step 3: Explanation

Therefore, 

$E\left(X_n\mid X_{n-1}\right)=p\left(2p{X}_{n-1}\right)+(1-p)\left(2(1-p)X_{n-1}\right)$

Substitute, 

$=X_{n-1}\left[2p^2+2(1-p{)}^2\right]$

In order to write and calculate more clearly, define $\alpha :=2p^2+2(1-p{)}^2$. 

Step 4: Explanation

Applying the expectation to the previous equation, we have

$E\left(X_n\right)=\alpha E\left(X_{n-1}\right)$

Which yields,

$E\left(X_n\right)={\alpha }^{n}E\left(X_0\right)$

Substitute,

$={\left[2p^2+2(1-p{)}^2\right]}^n·x$

Since we are given that we start with$x$ units.

Step 5: Final answer

The start with $x$ units value found to be $E\left(X_n\right)={\left[2p^2+2(1-p{)}^2\right]}^n·x$.