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If a random variable $X$ has a density $f$, then the expected value can be simplified: $$\mathbb{E}[X]=a+∫_{a}^{b}(1-F(x))dx,$$

where $F$ is the cumulative distribution function, $F(x)=\Pr(X≤x)$.

My question is: Does this simplification should work for all probability distributions, even for those that are not absolutely continuous with respect to Lebesgue-measure on $[a,b]$?

If $X$ is any real-valued random variable with support $[a,b]$ and $F(x)=\Pr(x≤X)$, is it always true that $$ \mathbb{E}[X]=a+∫_{a}^{b}(1-F(x))dx$$

The answer to this question would be simple if one could generally extend integration-by-parts to Lebesgue-Stieltjes integration. However, this is not possible; see Wikipedia.

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It is better to write capital $X$ for the random variable and lower-case $x$ for the argument to the c.d.f. or the p.d.f., instead of using the same thing for both usages. Among other advantages, you avoid the awkwardness of such expressions as "$\mathbb{P}(x \le x\prime)$", instead writing $\Pr(X\le x)$. (And notice that the code for $\Pr$ is \Pr.) –  Michael Hardy Jun 12 '12 at 0:07
    
It does work for all Borel-measurable distributions that are supported on some subset of $[a,\infty)$, where $a$ is finite. I don't think you can do it if the support has no finite lower bound. –  Michael Hardy Jun 12 '12 at 0:10

2 Answers 2

up vote 1 down vote accepted

Let's use $X$ for the random variable, keeping $x$ for the variable of integration. In general, we have a probability measure $\mu$ on $I = [a,b]$ and

$$\eqalign{E[X] &= \int_I x\ d\mu(x) = a + \int_I (x-a) d\mu(x)\cr &= a + \int_I \int_a^x 1 \ dt \ d\mu(x) = a + \int_a^b \int_{[t,b]} 1 \ d\mu(x)\ dt \cr &= a + \int_a^b (1 - F(t)) \ dt\cr}$$

(note that $\int_{[t,b]} 1 \ d\mu(x) = 1 - F(t-)$, but that is $1 - F(t)$ (Lebesgue) almost everywhere)

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"The answer to this question would be simple if one could generally extend integration-by-parts to Lebesgue-Stieltjes integration."

This comment moves me to add another answer: Riemann--Stieltjes integration is enough for this. Look it up and/or make it an exercise to prove it. I.e. suppose you have a probability distribution on the set of Borel-measurable subsets of $\mathbb{R}$. Let $X$ be a random variable so distributed and let $F$ be the c.d.f., so $F(x)=\Pr(X\le x)$. Then $$ \mathbb{E}(X) = \int_{-\infty}^\infty x \, dF(x), $$ where that is a Riemann--Stieltjes integral. If the support of the distribution has a finite a lower bound, then applying integration by parts to the Riemann--Stieltjes integral does it.

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