Problem. Let $\rho\colon[-1,\infty)\to\mathbb{R}$ be a function such that $\int_{-1}^\infty\rho(x)\,dx=1.$ Let $G\colon[0,1]\to\mathbb{R}$ be a function that is defined with $G(f) := \int_{-1}^\infty \log(1+fx)\,\rho(x)\,dx.$ Show that $G$ is either monotonically increasing, or has one local maximum on the interval $(0,1)$ (depending on how “fat” the “tail” of the function $\rho$ is).
Real life motivation. Suppose that we start with capital $V_0$, and we can repeatedly invest it in some venture with a percentage return given by a fixed probability distribution. This distribution is continuous, but it never takes values below $-1$ (we can never lose more than our investment). Its density function is the function $\rho$ as shown above.
Now, let $X_1,X_2,\ldots$ be independent random variables having this distribution. Furthermore, assume that we always invest a fixed fraction $f$ of our current capital. After $n$ investments, then, our capital will be $V_n = V_0(1+fX_1)\cdots(1+fX_n).$
We wish to measure how fast our capital is growing, and to that end we define $G := \lim_{n\to\infty} \frac{1}{n}\log\frac{V_n}{V_0}.$ Clearly, the faster $V_n$ grows, the bigger $G$ is. And by the weak law of large numbers, we see that $G$ is indeed the integral shown above. Our goal is to say something useful about the fraction $f$ of our current capital that we invest each time.