0
$\begingroup$

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)?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

Let the cdf of $X$ be denoted by

$$F_X(x)=P(X<x)$$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.