conditional expectation of uniform distribution given the realization of product of uniform distributions

Question

Assume that $A \sim U(0,1)$, and $B \sim U(0,b)$ with $b<1$, and A and B are independent. Can we calculate (closed form for) the expected value of $A$ given that we observe the realization of AB, call it z: $$E[A|AB=z] = ?$$

Answer 1

Yes we can.

\begin{eqnarray*} f_{A,AB}(a,z) &=&f_{A,B}(a,z/a)\frac1a \\ &=& f_{A}(a)f_{B}(z/a)\frac1a \\ &=& \frac1{ab}\;, \end{eqnarray*}

so

\begin{eqnarray*} E[A\mid AB=z] &=& \frac{\int_{z/b}^1\mathrm daf_{A,AB}(a,z)a}{\int_{z/b}^1\mathrm daf_{A,AB}(a,z)} \\ &=& \frac{\int_{z/b}^1\mathrm da\frac1b}{\int_{z/b}^1\mathrm da\frac1{ab}} \\ &=&\frac{1-\frac zb}{\log b-\log z}\;. \end{eqnarray*}

Answer 2

Let $c \in (0,1)$, $A \sim \mathcal U(0,1), B \sim \mathcal U(0,c)$ be independent. Then we can form a random vector $(A,B)$ with density function being the product of densities, that is $g_{(A,B)}(a,b) = \frac{1}{c}\chi_{(0,1)}(a) \chi_{(0,c)}(b)$.

Not, we'll try to transform $(A,B)$ to vector $(A,AB)$ by function $\phi : \mathbb R^2 \to \mathbb R^2$, $\phi(a,b)=(a,ab)$. Finding the inverse: $a=x, b=xy$, so $x=a, y=\frac{b}{a}$, by that $\phi^{-1}(x,y) = (x,\frac{y}{x})$. Note that $x \in (0,1), y \in (0,c)$, so there's no dividing by $0$. By transformation formula: $g_{(A,AB)}(x,y) = g_{(A,B)}(\phi^{-1}(x,y)) | \det D\phi^{-1}(x,y) | $

We get $g_{(A,B)}(\phi^{-1}(x,y)) = \frac{1}{c}\chi_{(0,1)}(x)\chi_{(0,c)}(\frac{y}{x}) = \frac{1}{c} \chi_{(0,1)}(x) \chi_{(0,xc)}(y)$

$\det D\phi^{-1}(x,y) = \frac{1}{x} $

So, $g_{(A,AB)}(x,y) = \frac{1}{cx}\chi_{(0,1)}(x) \chi_{(0,xc)}(y)$ is the density of our vector $(A,AB) =:(X,Y)$

Note that if vector $(X,Y)$ has density, then $$ \mathbb E[f(X,Y)|Y=y] = \frac{\int_{-\infty}^\infty g_{(X,Y)}(x,y)f(x,y)dx}{\int_{-\infty}^\infty g_{(X,Y)}(x,y)dx}$$

In our case we need $$ g_Y(y) = \int_{-\infty}^{\infty} \frac{1}{cx} \chi_{(0,1)}(x) \chi_{(0,cx)}(y) dx = \int_0^1 \frac{1}{xc}\chi_{(0,cx)}(y)dx = \int_{\frac{y}{c}}^1 \frac{1}{cx}dx = \frac{1}{c} \ln(1) - \ln(\frac{y}{c}) = \frac{\ln(c) - \ln(y)}{c}$$

Simirarly:

$$ \int_{-\infty}^\infty \frac{x}{cx} \chi_{(0,1)}(x) \chi_{(0,cx)}(y)dx = \frac{1}{c} \int_{\frac{y}{c}}^1 dx = \frac{1 -\frac{y}{c}}{c} $$

Putting it together $\mathbb E[X|Y] = \frac{c - Y}{c(\ln(c) - \ln(Y))}$, and getting back to $A,B$, we have:

$$ \mathbb E[A|AB] = \frac{c-AB}{c(\ln(c) - \ln(AB))} $$