A less than half-full tank

Frustrated with my old calculator, I toss it in the trash and buy a new one. But now I’m concerned this second calculator is also “tanked.” As before, every value of $A$ between $0$ and $1$ is equally likely, at first.

I ask my friend to generate one random number using this second calculator. My friend does so, and smirks. “I won’t tell you what the number is,” my friend says, “but it’s somewhere between $0$ and $0.5.$” On average, what can I expect the value of $A$ (for this second calculator) to be?

Similar to the classic problem, we will define terms with $A \sim U(0,1)$ at first and $X|A=a \sim U(0,a).$ In this case, we want to get the distribution of $A$ conditional on $0 \leq X \leq \frac{1}{2},$ that is, $$f_{A\mid 0 \leq X \leq 1/2} (a) = \lim_{da \downarrow 0} \frac{\frac{1}{da} \mathbb{P} \left\{ a -\frac{da}{2} \leq A \leq a + \frac{da}{2}, 0 \leq X \leq \frac{1}{2} \right\}}{ \mathbb{P} \{ 0 \leq X \leq \frac{1}{2} \} }.$$

Let's take the denominator first, like last time to see that \begin{align*}\mathbb{P} \{ 0 \leq X \leq \frac{1}{2} \} &= \int_0^1 \int_0^{1/2} \frac{1}{t} \chi_{[0,t]}(x) \,dx \,dt \\ &= \int_0^{1/2} \int_0^1 \frac{1}{t} \chi_{[0,t]}(x) \,dt \,dx \\ &= \int_0^{1/2} \left(\int_0^x 0 \,dt + \int_x^1 \frac{dt}{t} \right) \,dx \\ &= \int_0^{1/2} -\ln x \,dx \\ &= \frac{1}{2} - \frac{1}{2} \ln \frac{1}{2} = \frac{1}{2} \left( 1 + \ln 2 \right).\end{align*}

Let's assume that $a \gt \frac{1}{2},$ in which case, for small values of $da$ we have $\frac{1}{2} \leq a - \frac{da}{2},$ so we have $\chi_{[0,t]}(x) = 1$ for all $0 \leq x \leq 1/2$ and $t \in (a-da/2, a+da/2).$ Therefore, if $a \gt \frac{1}{2},$ then we have the numerator equal to \begin{align*}\frac{1}{da} \mathbb{P} \{ a - da/2 \leq A \leq a + da/2, 0 \leq X \leq 1/2 \} &= \frac{1}{da}\int_{a-da/2}^{a+da/2} \int_0^{1/2} \frac{1}{t} \chi_{[0,t]}(x) \,dx \, dt \\ &= \frac{1}{da} \int_{a-da/2}^{a+da/2} \frac{dt}{2t} = \frac{1}{2da} \left( \frac{a + \frac{da}{2}}{a - \frac{da}{2}} \right) \\&= \frac{1}{2} \left. \frac{d}{dt} \ln t \right|_{t=a} + O(da) = \frac{1}{2a} + O(da).\end{align*} On the other hand, if $a \lt \frac{1}{2},$ then if $da$ is small enough, then for every $t \in (a - da/2, a+da/2),$ we have $t \lt 1/2,$ so we have the denominator as \begin{align*}\frac{1}{da} \mathbb{P} \{ a - da/2 \leq A \leq a + da/2, 0 \leq X \leq 1/2 \} &= \frac{1}{da} \int_{a-da/2}^{a+da/2} \int_0^{1/2} \frac{1}{t} \chi_{[0,t]}(x) \,dx \,dt \\ &= \frac{1}{da} \int_{a-da/2}^{a+da/2} \left( \int_0^t \frac{1}{t}\, dx + \int_t^{1/2} 0 \,dx \right) \,dt \\ &= \frac{1}{da} \int_{a-da/2}^{a+da/2} 1 \,dt = 1\end{align*} Therefore, we see that as $da \downarrow 0,$ the numerator becomes the piecewise function $n(a) = \min \{ 1, \frac{1}{2a} \},$ so putting everything together, we see that $$f_{A\mid 0 \leq X \leq \frac{1}{2}} (a) = \frac{ n(a) }{ \frac{1}{2} (1 + \ln 2) } = \min \left\{ \frac{2}{1 + \ln 2}, \frac{1}{a(1 + \ln 2)} \right\}.$$

Therefore, if all we have is the knowledge that the draw from the tanked calculator is in the lower half of the interval, then the expected value of the tank parameter is \begin{align*}\mathbb{E} \left[ A \mid 0 \leq X \leq \frac{1}{2} \right] &= \int_0^{1/2} a \cdot \frac{2}{1 + \ln 2} \,da + \int_{1/2}^1 a \cdot \frac{1}{a(1 + \ln 2)} \,da \\ &= \left.\frac{2}{ 1 + \ln 2} \frac{a^2}{2} \right|_{a=0}^{a=1/2} + \frac{1}{1 + \ln 2} \cdot \frac{1}{2} \\ &= \frac{3}{4(1 + \ln 2)} \approx 0.442962081862\dots\end{align*}