Let $Y\sim \mathrm{Binom}(n,p)$.
$b>0$
$0<x\le1$
Let $R=E[(1+xb)^\frac{Y}{n}(1-x)^{1-\frac{Y}{n}}]$.
Is there a closed-form expression for $R$ as a function of $(x,b,n,p)$?
Is there a closed-form expression for $arg{_x}max(R)$ as a function of $(b,n,p)$?
This comes up in analysis of fractional betting.
The calculation of $R$ is quite routine - just a direct application of Binomial theorem:
$$ \begin{align} R &= E\left[(1 + xb)^{\frac {Y} {n}}(1 - x)^{1 - \frac {Y} {n}}\right] \\ &= \sum_{y=0}^n (1 + xb)^{\frac {y} {n}}(1 - x)^{1 - \frac {y} {n}} \binom {n} {y}p^y(1-p)^{n-y} \\ &= \sum_{y=0}^n \binom {n} {y} \left[(1 + xb)^{\frac {1} {n}}p\right]^y \left[(1 - x)^{\frac {1} {n}}(1-p)\right]^{n-y} \\ &= \left[(1 + xb)^{\frac {1} {n}}p + (1 - x)^{\frac {1} {n}}(1-p) \right]^n \end{align}$$
The remaining optimization problem is just a calculus problem. Note maximizer of $R$ is equivalent to the maximizer of $R^{\frac {1} {n}}$, and thus we just consider $$ \frac {\partial R^{\frac {1} {n}}} {\partial x} = \frac {bp} {n}(1 + xb)^{-(1 - \frac {1} {n})} - \frac {1 - p} {n}(1-x)^{-(1 - \frac {1} {n})} $$ Setting it equals to zero, we obtain $$ \begin{align} && bp(1 + xb)^{-(1 - \frac {1} {n})} &= (1 - p)(1-x)^{-(1 - \frac {1} {n})} \\ & \Rightarrow & (bp)^{\frac {n} {n-1}}(1 - x) &= (1-p)^{\frac {n} {n-1}}(1+xb) \\ & \Rightarrow & (b(1-p)^{\frac {n} {n-1}} + (bp)^{\frac {n} {n-1}})x &= (bp)^{\frac {n} {n-1}} - (1-p)^{\frac {n} {n-1}} \\ & \Rightarrow & x &= \frac {(bp)^{\frac {n} {n-1}} - (1-p)^{\frac {n} {n-1}} } {b(1-p)^{\frac {n} {n-1}} + (bp)^{\frac {n} {n-1}} } \\ \end{align}$$
The remaining parts will be checking when it falls into $[0, 1]$ and checking if this critical point is really a maximum.