In our probability course we defined for a continuous random variable $X$ with $E[X^2]<+\infty$ and an event $F\subset\Omega$ with $P(F)>0$ that $E[X\mid F]:=\frac{1}{P(F)}E[X\mathbb{1}_{F}]$. I'm comfortable with the notion of conditional expectation and I understand that this definition is inspired by the case where $X$ is discrete, but we only defined expectation for either continuous, either discrete variables, and not for mixed variables. Is there a way to calculate $E[X\mid \{X>a\}]$ where $a\in\mathbb{R}$ and $f$ the everywhere positive density-function of $X$ without having to define expectation for non-continuous, non-discrete random variables via measure theory (its a quite elementary course)?

  • $\begingroup$ I would say it is impossible if you want to exclude measure theroy as a whole. You can use as little measure theory as possible, for example you can decompose $X$ into a continuous part $X^c$ and a discrete part $X^d$. $\endgroup$ – Cettt Jan 18 '17 at 15:21

Let the cdf of $X$ be denoted by


and assume that $f_X$, the derivative of $F_X$ exists.

Then the conditional cdf given that $X>a$ is

$$F_{X\mid X>a}(x)= \begin{cases} P(X<x\mid X>a),&\text{ if } x>a\\ 0,&\text{ otherwise} \end{cases}=$$

$$= \begin{cases} \frac{P(a<X<x)}{P(X>a)},&\text{ if } x>a\\ 0,&\text{ otherwise} \end{cases}= $$ $$= \begin{cases} \frac{F_X(x)-F_X(a)}{1-F_X(a)},&\text{ if } x>a\\ 0,&\text{ otherwise.} \end{cases} $$

Then the corresponding conditional density is

$$f_{X\mid X>a}(x)=\frac{d F_{X\mid X>a}}{dx}=$$ $$= \begin{cases} \frac{f_X(x)}{1-F_X(a)},&\text{ if } x>a\\ 0,&\text{ otherwise.} \end{cases} $$

Finally, the conditional expectation is

$$E[X\mid X>a]=\frac1{1-F_X(a)}\int_a^{\infty}xf_X(x)\ dx.$$


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