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)?
How to calculate $E[X\mid \{X>a\}]$ without measure theory?
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0I 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 $X^d$. – 2017-01-18
1 Answers
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Let the cdf of $X$ be denoted by
$$F_X(x)=P(X and assume that $f_X$, the derivative of $F_X$ exists. Then the conditional cdf given that $X>a$ is $$F_{X\mid X>a}(x)=
\begin{cases}
P(X $$=
\begin{cases}
\frac{P(a Then the corresponding conditional density is $$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.$$