How to Interpret P(dw) in Probability Measure Integrals?

[David1234]

What does it mean by
\int f(w) P (dw)

I don't really understand P (dw) here. Does it mean P (x: x \in B(x, \delta)) for infinitely small \delta?

For example, with P(x)=1/10 for x=1, 2, ..., 10. How can we interpret this in term of the above integral

Thanks...

 

[John Creighto]

I never took measure theory. Therefor my totally naive answer would be P(dw) is just a distribution function. Of course I'm probably completely wrong given I don't even know what a measure is.

 

[David1234]

P(dw) is like a distribution fuction... may be. I am confused about P(dw), is it probability of dw? Then what is dw? Following the above example, say, we have f(w)=1 for w=1 and 0 otherwise. What is the meaning of dw here and hence value of P(dw) at w=1? I guess the above integral would give value = 1/10.

 

[HallsofIvy]

I have never seen "P(dw)". I think you mean what I would call dP(w)= P'(w)dw- the derivative of the cumulative probability distribution and so the probability density function. In that case, \int F(w)dP= \int F(w)P'(w)dw is the expected value of F.

 

[David1234]

I guess if P(w) has a derivative we can write it that way. I got the expression from a textbook by Patrick Billingsley. Generally, when P(w) is not differentiable (as shown in the example), we can not write the expression in that form.

 

[statdad]

The notation
<br /> \int_\Omega X(\omega}) \, \mathcal{P}(dw)<br />

is used in probability to indicate the expectation of the random variable X
with respect tot the probability measure (distribution) \mathcal{P} over the probability space \Omega.

If \Lambda is any measurable set, then

<br /> \int_\Lambda X(\omega) \, \mathcal{P}(dw) = E[X \cdot 1_{\Lambda}]<br />

If the probability space is the real line with measure \mu, then

<br /> \int_\Lambda X(\omega) \, \mathcal{P}(dw) = \int_\Lambda f(x) \, \mu(dx)<br />

is the Lebesgue-Stieltjes integral of f with respect to the
probability measure \mu.

In more traditional form, if F is the distribution function of \mu, and \Lambda is an interval (a,b), then

<br /> \int_\Lambda X(\omega) \, \mathcal{P}(dw) = \int_\Lambda f(x) \, \mu(dx) = \int_{(a,b)} f(x) \, dF(x)<br />

If the probability measure doesn't have any atoms, the final integral is just a Lebesgue integral. If there are atoms, you need to take care to specify the interval according to whether the endpoints are or are not included - e.g.

<br /> \int_{a+0}^{b+0} f(x) \,dF(x), \quad \int_{a-0}^{b-0} f(x) \, dF(x)<br />

and so on.

Billingsley is one of the "classic" probability texts. Chang's "A Course in Probability Theory" is another - I studied from it many years ago, and have the second edition. His writing is a little terse, but there is a lot packed into his book.

 

[David1234]

Thanks a lot for the detail answer. I guess by Chang you mean Kai Lai Chung... :)

 

[statdad]

Yes, I did mean Kai Lai Chung - I would give a general description of my typing ability, but the description wouldn't be "safe for work".
Sorry for the confusion - glad the answer helped.

 

[Focus]

I have never seen "P(dw)". I think you mean what I would call dP(w)= P'(w)dw- the derivative of the cumulative probability distribution and so the probability density function. In that case, \int F(w)dP= \int F(w)P&#039;(w)dw is the expected value of F.

This isn't really correct. P may not be differentiable. When you say \int_B f(x) P(dx), you are referring to the Lebesgue integral of f with respect to P. It is the same as saying \int_B f dP. You are just telling where the arguments lie so there is no confusion. I have to disagree with statdad in that it is not the Stieltjes integral, it is just the plain old Lebesgue integral. For Stieltjes you want to take a distribution function F of P and then you work it out as \int_B f(x) dF(x)=\int_B f(x) P(dx).

Billingsley is a nice textbook and also I would recommend Ash, Real Analysis and Probability.

 

[David1234]

Thanks...

I will have a look at "Real Analysis and Probability" by Ash.