# Notation for Expectation?

## Main Question or Discussion Point

How do I read, interpret the following definitions for the expectation of a random variable X?
Assume the integral is over the entire relevant space for X.

(1) E(X) = ∫ x dP
(2) E(X) = ∫ x dF(X)

If I asked you to calculate (1) or (2) for an arbitrary X, how does it look?
My only other understanding of E(X) is to do pdf times x, integrate, plug in bounds, but that's assuming X is nice enough to have such a pdf. I appreciate any replies!

## Answers and Replies

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Simon Bridge
Science Advisor
Homework Helper
http://en.wikipedia.org/wiki/Expected_value
... there are plenty of examples of expectation value calculations online - have you had a look?

Note: none of your notations look right.
The first looks looks like it's trying to be the Lebesgue Integral definition - which is more general, the second looks like it is trying to be the regular student definition while the second

i.e.

1. $$E[X]=\int_{\Omega}XdP$$

2. $$E[X]=\int_{-\infty}^\infty xp(x)dx$$

Okay I see. Would this be the other definition:

E(X) = ∫ X dF(X) = ∫ x f(x) dx, where dF(X) = (dF/dx)dx = f(x)dx. But this is assuming dF/dx exists, or perhaps I should say this is assuming dF/dx is useful. What if F is a singular continuous distribution?

mathman
Science Advisor
E(X) = ∫xdF(x) is the most general form. When dF/dx exists, you can use it.

Simon Bridge
Science Advisor
Homework Helper
What if F is a singular continuous distribution?
i.e. dF/dx does not exist?
Then you cannot use that definition.

Does "expectation value" make sense in the absence of a probability density function?

jbunniii
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Gold Member
(2) E(X) = ∫ x dF(X)
This is a Stieltjes integral. This integral is meaningful for any probability distribution, even when the cdf ##F## is not differentiable.

In the case where ##F## is differentiable and the density function ##p(x) = dF/dx## integrates back to ##F(x)##, it is equivalent to ##E(X) = \int x p(x) dx##.

In the case where ##F## is a "staircase" (the probability is all concentrated at discrete points), it is equivalent to ##E(X) = \sum_n a_n P(X = a_n)##, where ##a_n## are the x-coordinates of the jumps.

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How about for a d.f. F that is singularly continuous (neither absolutely continuous nor discrete)?

jbunniii
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How about for a d.f. F that is singularly continuous (neither absolutely continuous nor discrete)?
##E(X) = \int x dF(x)## is still meaningful in that case, but it may not be possible to write it in terms of a standard Riemann integral or a sum as in the two cases I noted above. Sometimes it can be written as a Riemann integral plus a sum, however (so called "mixed" continuous/discrete distribution). Do you have a specific distribution in mind?

Yes, the Cantor distribution! Funny that you ask, I'm actually trying to work out a few problems based on the Cantor distribution right now. Wikipedia gives the Cantor distribution's expectation as 1/2 based on a symmetry argument. I was wondering if there is any other way to calculate it. I'm curious to know, because I am also asked to find ∫(x^2)dF(X) for the Cantor distribution, and I'm pretty stuck.