Coldplay Canoodling - Extra Credit

Now, everyone at the concert spends at least some time canoodling. In particular, each member of a couple wants to spend some fraction of the time canoodling, where this fraction is randomly and uniformly selected between $0$ and $1$. This value is chosen independently for the two members of each couple, and the actual time spent canoodling is the product of these values. For example, if you want to canoodle during half the concert and your partner wants to canoodle during a third of the concert, you will actually canoodle during a sixth of the concert.

Meanwhile, the camera operators love to show canoodling couples. So instead of randomly picking couples to show on the jumbotron, they randomly pick from among the currently canoodling couples. (The time shown on the jumbotron is very short, so a couple’s probability of being selected is proportional to how much time they spend canoodling.)

Looking around the concert, you notice that the kinds of couples who most frequently appear on the jumbotron aren’t constantly canoodling, since there are very few such couples. Indeed, the couples who most frequently appear on the jumbotron spend a particular fraction $C$ of the concert canoodling. What is the value of $C$?

In this Extra Credit problem, we want to find the mode of the following distribution $$f_S(C) = \mathbb{P} \{ \text{shown canoodling}, \text{canoodling ratio} = C\} = k C \mathbb{P} \{ \text{canoodling ratio} = C \},$$ where the final equation comes from the assertin that a couple's probability of being selected is proportional to how much time they spend canoodling.

Let $X, Y \sim U(0,1)$ represent the canoodling propensity of the two members of the couple, then \begin{align*} \mathbb{P} \{ \text{canoodling ratio} = C \} &= \mathbb{P} \{ X \cdot Y = C \} \\ &= \frac{d}{dC} \mathbb{P} \{ X \cdot Y \leq C \} \\ &= \frac{d}{dC} \left( \int_0^1 \int_0^{\min \{ 1, C / x \}} \,dy \, dx \right) \\ &= \frac{d}{dC} \left( \int_0^1 \min \{ 1, \frac{C}{x} \} \,dx \right) \\ &= \frac{d}{dC} \left( \int_0^C 1\,dx + \int_C^1 \frac{C}{x} dx \right) \\ &= \frac{d}{dC} \left( C - C\ln C \right) \\ &= - \ln C.\end{align*} We already have $f_S(C) = - k C \ln C,$ and though it is irrelevant for answering the question, out of an abundance of rigor (??), let's find $k.$ From the law of total probability we get $$1 = \int_0^1 f_S(C) \,dC = k \int_0^1 (- C \ln C) \,dC = \left[ -\frac{C^2}{2} \ln C \right]_{C=0}^{C=1} + \int_0^1 \frac{C^2}{2} \,\frac{dC}{C} = \frac{k}{4},$$ so $k = 4.$

So the distribution is $f_S(C) = -4 C \ln C$ and we want to find the mode of this distribution. Since $$f^\prime_S(C) = -4 \ln C -4 = -4(1 + \ln C)$$ has only one zero in $[0,1]$ and $f_S(0) = f_S(1) = 0,$ we see that the most frequently appearing canoodling ratios on the Jumbotron is $$C^* = \frac{1}{e} \approx 0.367879441171\dots,$$ which gives $$f_S(C) \leq f_S(C^*) = \frac{4}{e}, \,\forall C \in [0,1].$$