$$ \left.\begin{array}{l} ext { The function } \ A(x)=P(1+b x)^{m}(1-x)^{n} \ ext { models the final value } A(x) ext { of a bankroll after } N=m+n ext { consecutive wagers on an event with given odds } b ext { to } 1 ext { against. } \ ext { In this model, the true odds are } m ext { to } n ext { in favor of the event. In other words, } m ext { represents the number of wins, } n ext { the number of losses, and } N=m+n ext {. } m ext {. } n ext {. } \ ext { is the total number of wagers. The value } P ext { represents the initial bankroll and } x ext { represents the fraction of the bankroll wagered at each of the } N \ ext { consecutive events. } \ ext { To learn more about the bankroll model and how to find a maximum value for this function with the Kelly Formula, check out the following two videos: } \ ext { Moneyline Odds and the Kelly Formula }[3][+] \ ext { Mathematics of Gambling and the Kelly Formula }[\mathbf{3 l}[~\end{array} ight] $$ In this example, we seek the value of $$ x $$ on the interval $$ [0,1] $$ that will maximize $$ A(x) $$ for given values of $$ P, b, m $$ and $$ n $$.

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$$ \left.\begin{array}{l}	ext { The function } \ A(x)=P(1+b x)^{m}(1-x)^{n} \ 	ext { models the final value } A(x) 	ext { of a bankroll after } N=m+n 	ext { consecutive wagers on an event with given odds } b 	ext { to } 1 	ext { against. } \ 	ext { In this model, the true odds are } m 	ext { to } n 	ext { in favor of the event. In other words, } m 	ext { represents the number of wins, } n 	ext { the number of losses, and } N=m+n 	ext {. } m 	ext {. } n 	ext {. } \ 	ext { is the total number of wagers. The value } P 	ext { represents the initial bankroll and } x 	ext { represents the fraction of the bankroll wagered at each of the } N \ 	ext { consecutive events. } \ 	ext { To learn more about the bankroll model and how to find a maximum value for this function with the Kelly Formula, check out the following two videos: } \ 	ext { Moneyline Odds and the Kelly Formula }[3][+] \ 	ext { Mathematics of Gambling and the Kelly Formula }[\mathbf{3 l}[~\end{array}ight] $$ In this example, we seek the value of $$ x $$ on the interval $$ [0,1] $$ that will maximize $$ A(x) $$ for given values of $$ P, b, m $$ and $$ n $$.

UpStudy AI · Accepted Answer

We begin with the function
$$
A(x)=P(1+bx)^m(1-x)^n,
$$
where $$ P, b, m, n $$ are given constants and $$ x $$ is in the interval $$[0,1]$$. To find the value of $$ x $$ that maximizes $$ A(x) $$, it is sufficient to maximize its natural logarithm because the logarithm is a monotonically increasing function. Define
$$
f(x)=\ln A(x)=\ln P+m\ln(1+bx)+n\ln(1-x).
$$
Since $$ \ln P $$ is constant, maximizing $$ f(x) $$ is equivalent to maximizing
$$
g(x)=m\ln(1+bx)+n\ln(1-x).
$$

**Step 1. Differentiate $$ g(x) $$ with respect to $$ x $$.**

We compute the derivative:
$$
g'(x)=\frac{m\cdot b}{1+bx}+\frac{n\cdot (-1)}{1-x}=\frac{mb}{1+bx}-\frac{n}{1-x}.
$$

**Step 2. Set the derivative equal to zero.**

To find the critical point, solve
$$
\frac{mb}{1+bx}-\frac{n}{1-x}=0.
$$
This can be rewritten as
$$
\frac{mb}{1+bx}=\frac{n}{1-x}.
$$

**Step 3. Solve for $$ x $$.**

Cross-multiply to eliminate the denominators:
$$
mb(1-x)=n(1+bx).
$$
Expand both sides:
$$
mb - mb\,x = n + nb\,x.
$$
Collect the terms involving $$ x $$ on one side:
$$
mb - n = mb\,x + nb\,x.
$$
Factor the $$ x $$ on the right-hand side:
$$
mb - n = x\,(mb+nb)=x\,b(m+n).
$$
Solve for $$ x $$:
$$
x=\frac{mb - n}{b(m+n)}.
$$

**Step 4. Consider domain restrictions.**

Since $$ x $$ must lie in the interval $$[0,1]$$, the value derived,
$$
x^*=\frac{mb - n}{b(m+n)},
$$
is the optimal fraction if it belongs to $$[0,1]$$. If it does not (for instance, if $$ mb < n $$ then $$ x^* $$ becomes negative), the maximum of $$ A(x) $$ will occur at one of the endpoints, typically $$ x=0 $$ when betting is not favorable.

Thus, the value of $$ x $$ that maximizes $$ A(x) $$ on $$[0,1]$$ is
$$
x^*=\frac{mb-n}{b(m+n)}.
$$
To maximize $$ A(x) = P(1+bx)^m(1-x)^n $$ on the interval $$[0,1]$$, set the derivative of $$ \ln A(x) $$ equal to zero and solve for $$ x $$. The optimal $$ x $$ is $$ x^* = \frac{mb - n}{b(m+n)}. $$ Ensure $$ x^* $$ is within $$[0,1]$$; otherwise, the maximum occurs at $$ x=0 $$.

In this example, we seek the value of on the interval that will maximize for given values of and .

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