A half-full tank

I think the random number generator on my calculator might be malfunctioning. Oh no!

Under normal conditions, it should generate random numbers between $0$ and $1.$ But my suspicion is that the calculator is “tanked,” meaning it only generates random numbers between $0$ and some value $0 \lt A \lt 1.$ Beyond that, I have no knowledge regarding the value of $A.$ At the moment, it’s equally likely to be any value from $0$ to $1$.

As an experiment, I ask the calculator to generate one random number. It produces a value of exactly $0.5.$ (While this is, admittedly, infinitely unlikely, let’s roll with it!) Based on this result, what can I expect the value of $A$ to be, on average?

Let's first define some terms. Let's assume that the tank paramater $A \sim U(0,1),$ and since the calculator is tanked that random draws are all independently and identically distributed as $X \sim U(0,A),$ that is, $$f_{X|A=a}(x) = \lim_{dx \downarrow 0} \frac{1}{dx} \mathbb{P} \{ x - \frac{dx}{2} \leq X \leq x + \frac{dx}{2} \mid A = a\} = \frac{1}{a} \chi_{[0,a]}(x).$$ Using Bayesian theorem, we see that \begin{align*}f_{A|X=x} (a) &= \lim_{da \downarrow 0} \frac{1}{da} \mathbb{P} \{ a - \frac{da}{2} \leq A \leq a + \frac{da}{2} \mid X = x \} \\ &= \lim_{da \downarrow 0} \frac{ \mathbb{P} \{ a - \frac{da}{2} \leq A \leq a + \frac{da}{2}, X = x \} }{ da \mathbb{P} \{ X = x \} } \\ &= \lim_{da \downarrow 0} \frac{ \frac{1}{da} \int_{a - da/2}^{a+da/2} \frac{1}{t} \chi_{[0,t]}(x) \,dt }{ \int_0^1 \frac{1}{t} \chi_{[0,t]}(x) \,dt }.\end{align*} In particular, the denominator can be calculated as $$\int_0^1 \frac{1}{t} \chi_{[0,t]}(x) \,dt = \int_0^x \frac{1}{t} \chi_{[0,t]}(x) \,dt + \int_x^1 \frac{1}{t} \chi_{[0,t]}(x) \,dt = \int_0^x 0 \,dt + \int_x^1 \frac{dt}{t} = -\ln x.$$ The numerator meanwhile is equivalent to $\frac{1}{a} \chi_{[0,a)} (x),$ since if $x \lt a,$ then there is some $\alpha \lt 2(a-x)$ for which for any $0 \lt da \lt \alpha,$ $\chi_{[0,t]}(x) = 1$ for all $t \in (a-\frac{da}{2}, a + \frac{da}{2}),$ so that $$\frac{1}{da} \int_{a - da/2}^{a+da/2} \frac{1}{t} \chi_{[0,t]}(x) \,dt = \frac{1}{da} \int_{a-da/2}^{a+da/2} \frac{dt}{t} = \frac{ \ln \left( \frac{a + \frac{da}{2}}{a - \frac{da}{2}} \right) }{da} = \left.\frac{d}{dt} \ln t \right|_{t=a} + O(da) = \frac{1}{a} + O(da).$$ Similarly, if $x \gt a,$ then there is some $\alpha^\prime \lt 2(x-a)$ for which for any $0 \lt da \lt \alpha^\prime,$ then $\chi_{[0,t]}(x) = 0$ for all $t \in (a - \frac{da}{2}, a + \frac{da}{2} ),$ so that $$\frac{1}{da} \int_{a-da/2}^{a+da/2} \frac{1}{t} \chi_{[0,t]}(x) \,dt = 0.$$ Putting this all together, we see that $$f_{A\mid X=x}(a) = \begin{cases} -\frac{1}{a \ln x}, &\text{if $x \lt a \leq 1$;}\\ 0, &\text{if $0 \leq a \lt x.$}\end{cases}$$

Therefore, in particular, since we observed that $X = \frac{1}{2},$ we have the expected value of the tank parameter $A$ as \begin{align*}E[A \mid X = \frac{1}{2}] &= \int_0^1 a f_{A \mid X=\frac{1}{2}} (a) \,da \\ &= \int_0^{1/2} a \cdot 0 \,da + \int_{1/2}^1 a \cdot \left(- \frac{1}{a \ln \frac{1}{2}}\right) \,da \\ &= \int_{1/2}^1 \frac{da}{ln 2} \\& = \frac{1}{2\ln 2} \approx 0.72134752044\dots\end{align*}