compute conditional expectation given density function

Question

Let $X$ be a random variable with density

$$ p(x) = \frac{3}{4}\mathbb{1}_{(0,1)}(x) + \frac{1}{4}\mathbb{1}_{(-1,0)}(x). $$ Define $Y := |X|$. How can I show that

$$ \mathbb{E}(X|Y) = \frac{Y}{2}. $$

Answer 1

Note that $$ X = X\cdot\mathsf 1_{\{X\geqslant0\}} + X\cdot\mathsf 1_{\{X<0\}}, $$ $X=Y$ on $\{X\geqslant0\}$, and $X=-Y$ on $\{X<0\}$. It follows that \begin{align} \mathbb E[X\mid Y] &= \mathbb E[X\cdot\mathsf 1_{\{X\geqslant0\}} + X\cdot\mathsf 1_{\{X<0\}}\mid Y]\\ &= \mathbb E[Y\cdot\mathsf 1_{\{X\geqslant0\}}\mid Y] +\mathbb E[-Y\cdot\mathsf 1_{\{X<0\}}\mid Y]\\ &= Y\cdot \mathbb E[\mathsf 1_{\{X\geqslant0\}}-1_{\{X<0\}}\mid Y]\\ &= Y\cdot(\mathbb E[\mathsf 1_{\{X\geqslant0\}}] - E[\mathsf 1_{\{X<0\}}])\\ &= Y\cdot(\mathbb P(X\geqslant0)-\mathbb P(X<0))\\ &= Y\cdot(3/4 - 1/4)\\ &= \frac Y2. \end{align}