Conditional expectation of indicator function

Question

Could someone confirm if the following is correct. If not why? \begin{equation} E[\mathbb{1_{X\leq x}}|Y]=P[X|Y]=\frac{P[X,Y]}{P[Y]} \end{equation} Thank you.

Answer 1

Assume that $Y:(\Omega,\mathcal F)\to(E,\mathcal E)$, that is, that the random variable $Y$ is $E$-valued (in most of the cases, $(E,\mathcal E)=(\mathbb R,\mathcal B(\mathbb R))$ but the natural setting of the question is more general). On the other hand, the random variable $X$ must be real valued, thus, $X:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$, and the measurable space $(\Omega,\mathcal F)$ is endowed with a probability measure $\Pr$.

By definition, for each real number $x$, $E(\mathbf 1_{X\leqslant x}\mid Y)=\Pr(X\leqslant x\mid Y)$ (the two notations are equivalent) is a real valued random variable $u(Y):\Omega\to\mathbb R$ where the measurable function $u:E\to\mathbb R$ must be such that, for every bounded measurable function $v:E\to\mathbb R$, one has $$E(v(Y):X\leqslant x)=E(v(Y)\mathbf 1_{X\leqslant x})=E(v(Y)u(Y)). $$ This defines the random variable $u(Y)$ uniquely, up to null events. One can show that $0\leqslant u(Y)\leqslant1$ almost surely. Furthermore, if $\Pr(Y=y)\ne0$ for some given $y$ in $E$ then $u(y)=\Pr(X\leqslant x\mid Y=y)$, that is, $u(y)=\Pr(X\leqslant x, Y=y)/\Pr(Y=y)$.

Thus, for each real number $x$, $E(\mathbf 1_{X\leqslant x}\mid Y)=\Pr(X\leqslant x\mid Y)$ is a real valued random variable, that is, a measurable function from $\Omega$ to $\mathbb R$ while, for each $y$ in $E$ such that $\Pr(Y=y)\ne0$, $\Pr(X\leqslant x\mid Y=y)$ is a real number. The only function $\mathbb R\to\mathbb R$ in the picture is $u$ when $Y$ is real valued, otherwise $u:E\to\mathbb R$. In particular, for $y$ in $E$, $\Pr(X\leqslant x\mid Y)(y)$ is undefined, while, for $\omega$ in $\Omega$, $\Pr(X\leqslant x\mid Y)(\omega)$ is simply $u(Y(\omega))$.

A simple case is when $(X,Y)$ has a joint density $f$ with respect to $\mathrm{Leb}\otimes\mu$, for some measure $\mu$ on $E$, and $f$ is, say, positive everywhere on $\mathbb R\times E$. Then, $\Pr(X\leqslant x\mid Y)=u(Y)$ almost surely, where the function $u$ is defined, for every $y$ in $E$, by $$u(y)=\frac1{f_Y(y)}\int_{-\infty}^xf(z,y)\mathrm dz,\qquad f_Y(y)=\int_{-\infty}^{+\infty}f(z,y)\mathrm dz.$$

Answer 2

This is true: $\operatorname{\large E}[{\large 1}_{(X\leq x)}\mid Y] = \operatorname{\large P}[X\leq x\mid Y]$

This is also correct: $\operatorname{\large P}[X\leq x\mid Y\leq y] = \dfrac{\operatorname{\large P}[X\leq x, Y\leq y]}{ \operatorname{\large P}[Y\leq y]}$

However: $ \operatorname{\large P}[X\leq x\mid Y]\neq \operatorname{\large P}[X\leq x\mid Y\leq y]$

Because the former is a random variable, while the later is a conditional probability density.

Thus: $\operatorname{\large E}[{\large 1}_{(X\leq x)}\mid Y](y) = \operatorname{\large P}[X\leq x\mid Y = y]$

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For more detail on the notation, see also: http://en.wikipedia.org/wiki/Conditioning_(probability)#Conditional_expectation