To find conditional expectation of two uniformly distributed random variable.

Question

$X \sim U(0,1)$

$Y\sim U(-1,1) $

Transformation given

$U=X$

$W=X.Y$

To find $E(U|W)$

$f_{U,W}(u,w)=\dfrac{1}{2u} , 0<U<1, \ \ \ -u<W<u$

$E(U|W)=\int_{U}u f(U|W)du=\int_{w}^{1}u \frac{f(u,w)}{f_W(w)}du=\int_{w}^{1}u \frac{1}{2uf_W(w)}du$

Calculating $f_W(w)=\int_{w}^{1}\frac{1}{2u}du=\frac{\log(\frac{1}{w})}{2}$

$\int_{w}^{1}u \frac{1}{2u\bigg(\frac{\log(\frac{1}{w})}{2}\bigg)}du=\int_{w}^{1}\frac{1}{ log(\frac{1}{w})}du$

I am not sure if I am on right track can anyone help me ?

Answer 1

Lettuce cheque.

$\begin{align}f_{\small U,W}(u,w)&=\begin{Vmatrix}\dfrac{\partial\langle u,w/u\rangle}{\partial \langle u,w\rangle}\end{Vmatrix}f_{X,Y}(u,w/u) \\[1ex]&=\tfrac 1{2u}\mathbf 1_{0\leq u\leq 1, -u\leq w\leq u} \\[1ex]&=\tfrac 1{2u}\mathbf 1_{-1\leq w\leq 1,\lvert w\rvert \leq u\leq 1}\end{align}$

So, indeed, you look good except for the absence of absolute values:

$\begin{align}\mathsf E(U\mid W) &=\dfrac{\int_{\Bbb R}u\, f_{\small U,W}(u,W)~\mathrm d u}{\int_{\Bbb R} f_{\small U,W}(u,W)~\mathrm d u} \\[2ex]&=\dfrac{\int_{\lvert W\rvert}^1 \tfrac u{2u}~\mathrm d u}{\int_{\lvert W\rvert}^1 \tfrac 1{2u}~\mathrm d u} \\[2ex]&=\dfrac{\int_{\lvert W\rvert}^1 1~\mathrm d u}{\int_{\lvert W\rvert}^1 \tfrac 1{u}~\mathrm d u} \\[2ex]&=\dfrac{1-\lvert W\rvert}{-\ln\lvert W\rvert}\end{align}$