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We know that if $P$ is a probability in the measurable space $(\mathbb{R}, \mathcal{B})$, then we can define a distribuition function $F$ as following: $$F(x):= P((-\infty,x])), \quad \forall x \in \mathbb{R}$$ By the Carathéodory's extension theorem we can prove that the converse is also valid, that is: if $F$ is a distibuition function, then there is a unique $P_{\small{F}}$ in $(\mathbb{R}, \mathcal{B})$ that $$F(x)= P_{\small{F}}((-\infty,x])),\quad \forall x \in \mathbb{R}$$ In other words, there is a one-to-one correspondence between the set of probabilities in $(\mathbb{R}, \mathcal{B})$ and the set of all distribuition functions.

$Remake:$:: If $X:(\Omega,\mathcal{F}, \mu) \to \mathbb{R}$ is a random variable, we can define a probability in $(\mathbb{R}, \mathcal{B})$ and a distribuition function, respectively, as following: $$P_{\small{X}}(B) := \mu(X \in B),\quad B \in \mathcal{B}$$ $$F_{\small{X}}(x) := P_{\small{X}}((-\infty, x]))= \mu(X \leq x), \quad x \in \mathbb{R}$$

A random variable is continue (absolutly?) if there is a density (non negative) function $f$ such that $$F_{\small{X}}(x) = \int_{-\infty}^{x}f(y)dy$$ I dont will write the Lebesgue integral definition, but it is supossed that the classical definition is well known. I would like to remember its notations
$$E(X) = \int_{\Omega}Xd\mu = \int_{\Omega}X^{+}d\mu - \int_{\Omega}X^{-}d\mu$$ Although I consider the one-to-one correspondence between $F_{X}$ and $P_X$, I can not find, besides the notations justificatives, good arguments for the following equivalences (for each iquality)

$$\int_{\Omega}Xd\mu = \int_{\mathbb{R}} xdF_{\small{X}}(x) = \int_{\mathbb{R}} xP_{\small{X}}(x) = \int_{\mathbb{R}} x f(x)dx$$ Some justificatives?

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Consider measurable spaces $(\Omega_1,\mathcal{A}_1)$ and $(\Omega_2,\mathcal{A}_2)$ and a measure $\mu$ on $(\Omega_1, \mathcal{A_1})$. Any measurable function $X:\Omega_1\to \Omega_2$ induces a measure $\mu_X$ on $\Omega_2$ through $\mu_X(A_2)=\mu( X^{-1}(A_2))$ for $A_2 \in \mathcal{A}_2$, the pushforward measure. If $g:\Omega_2 \to \mathbb{R}$ is an integrable function, then $\int_{\Omega_2} g ~\mathrm{d}(\mu_X) = \int_{\Omega_1} g \circ X ~\mathrm{d}\mu$. Take for $(\Omega_2,\mathcal{A}_2)$ the real numbers with the borel sigma algebra and for $g$ the identity. Then you have

$$\int_{\Omega_1}X~\mathrm{d}\mu = \int_ {\Omega_1} id(X) \mathrm{d}\mu = \int_{\mathbb{R}} id ~\mathrm{d}\mu_X.$$

This gives you the equivalence between the first and the third expression in your last serie of equations: you can calculate the expectation through integration with respect to the pushforward measure. The second and third expression just differ by notation: $dF_X$, stresses that this is the Lebegue-Stieltje integral. The last expression is a property of absolutly continuous measures: if $\mu=f dx$ is a absolutly continuous measure with density $f$, then $\int_{\mathbb{R}} g d\mu = \int_{\mathbb{R}} g \cdot f d\lambda$, for all integrable $g$, where $\lambda$ is the onedimensional Lebegue measure.

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  • $\begingroup$ The equivalence between the first and the third expression is ok. Now, about the justification $(\mu = fdx)$ of the last expression: can I afirm that it is consequence of the radon nikodym theorem? $\endgroup$ – orrillo Jul 10 '17 at 23:28
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    $\begingroup$ No, Radon-Nikodym states that, if a (sigma-fin.) measure $\mu$ is absolutely continuous with respect to a (sigma-fin.) measure $\lambda$, then there is a measurable, positive function $f$ with $\mu = f d\lambda$. If you are given a measure $\mu = f d\lambda$, e.g. $\mu(A)=\int_A f d\lambda$ for all measurable $A$, then you can prove that $\int g d\mu = \int g \cdot f d\lambda$ by algebraic induction (you have to show first, that the identity holds for indicators $\int 1_A fd\mu = \mu(A) = \int_A f d\lambda = \int 1_A \cdot f d\lambda$, then for lin. combinations and limits). $\endgroup$ – crankk Jul 11 '17 at 6:12
  • $\begingroup$ Sorry, I am some confused. First, remember that for any borelian $B$, $P_{\small X}(B) = \int_B f(x) dx $. More precisaly, $P_{\small X}(B) = \int_B f(x) d\lambda(x)$, where $\lambda$ is the lebesgue measure. Radon Nikodyn says that if $P_{\small X} << \lambda$ - that is $\lambda(B) = 0$ implies $P_{\small X}(B) = \int_B f(x) d\lambda(x) = 0$ - then there is a unique $f$ that $d P_{\small X}/d\lambda = f$, in other words, $d P_{\small X} = f d\lambda$. With this consequence, I can say that $\int_{\mathbb{R}} xdP_{\small{X}}(x) = \int_{\mathbb{R}} x f(x)dx$. Is my reasoning correct? $\endgroup$ – orrillo Jul 13 '17 at 4:50
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    $\begingroup$ And the way you show it is by algebraic induction: first you show that it is true for indicator functions. Than assume that it is true for two functions $g_1$ and $g_2$, show that it is true for a linear combination of the two. Then assume it is true for a (monotonic,dominated) converging sequence $f_n$, show that it is true for the limit. $\endgroup$ – crankk Jul 13 '17 at 6:02
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    $\begingroup$ No. $P_X$ is absolutely continuous. Hence Radon-Nikodym tells you that there is a positive measurable $f$ with $P_X=f d\lambda$. Now IF you have any measure of the form $\mu= f d\lambda$, than there is a theorem how to calculate integrals: If $g$ is integrable (w.r.t. $\lambda$ or $\mu$), than $$\int g d\mu = \int g \cdot f d\lambda$$. This you can prove by algebraic induction. If your measure is not absolutely continuous than "there is no f" and of course no such identity. $\endgroup$ – crankk Jul 14 '17 at 9:33

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