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I get the following question from Zastawniak's Probability Through Problems:

Assume that the distribution function of a random variable $X$ on a probability space $(\Omega,{\mathcal A},P)$ is defined as $$P^X(A)=\frac{1}{3}\delta_0(A)+\frac{2}{3}P_2(A) $$ for any Borel subset $A$ of ${\mathbb R}$, where $\delta_0$ is the Dirac measure and $P_2$ is an absolutely continuous probability measure with density $$ f(x)=\frac{1}{2}1_{[1,3]}. $$ Show that $P^X$ is a probability measure on $({\mathbb R},{\mathcal B})$.

This can be done by a couple line using the definition of probability measure. Here are my questions:

  • $P^X$ seems kind of combine a discrete and a continuous random variable. Can $X$ be written as a linear combination of two random variable $X_1$ and $X_2$ such that $P^{X_1}=\delta_0$, $P^{X_2}=P_2$?
  • Can any one come up with a textbook with detailed discussion of such random variables?
  • How can I calculate $$ \int_{\Omega}XdP? $$ One more step I can come up with is $$ \int_{\mathbb R}XdP^X $$ But I have no idea how to deal with $d(\frac{1}{3}\delta_0(A)+\frac{2}{3}P_2(A))$ theoretically.
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2 Answers 2

up vote 1 down vote accepted

I think you're looking for Lebesgue's Decomposition Theorem for measures. This decomposes a given measure into a a part absolutely continuous w.r.t. Lebesgue measure (your f(x) part), a discrete part (your dirac mass), and also a singular continuous part (your measure doesn't have one of these, but such things exist, i.e. based off of the Cantor function).

For integrating, you just split the measure into these two parts, do the regular integration on the second part, and do summation (i.e. discrete integration) for the first part.

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The theoretical tool for answering the third question is the following proposition which is an exercise in this note.

(Linearity in $\mu$ ) Let $(X,{\mathcal B},\mu)$ be a measure space, and let $f:X\to[0,\infty]$ be measurable.

Then

  1. $$\int_{X}fd(c\mu)=c\times\int_Xfdu \quad \text{for every} \quad c\in[0,+\infty].$$

  2. If $\mu_1,\mu_2,\cdots$ are a sequence of measures on $\mathcal{B}$ , then $$ \int_Xfd\sum_{n=1}^{\infty}\mu_n=\sum_{n=1}^{\infty}\int_Xfd\mu_n $$

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