Vrftsjtryk

I'm trying to prove rigorously that $int_{Omega}X;dP=int_{-infty}^{infty}xf(x);dx$. Where $f$ is the pdf of the random variable $X$.

I can't find a proof on the wikipedia article, or if it's there then it's disguised enough that I can't recognize it. Basically what I've got is a sort of semi-rigorous (and probably incorrect) proof of the equality, but maybe somebody could help me flesh out the details.

Proof:

Beginning from the definition that $E[X]:=int_{Omega}X;dP$.

Given a random variable $X:Omegarightarrowmathbb{R}$, we have the induced measure on $(mathbb{R}, mathscr{B}(mathbb{R}))$ given by $P({Xin A})$. Then by the Radon-Nikodym theorem there exists a measurable function $f:mathbb{R}rightarrow [0,infty)$ such that $$P({Xin A}) = int_Afdmu,$$

where $dmu$ is the Lebesgue measure. From here it gets a bit hand-wavy. Basically since this measure was induced by the random variable $X$, then on $mathbb{R}$ this random variable is simply given by the identity function $g(x)=x$. And thus by definition we write the expected value of $g$ with respect to our Radon-Nikodym produced measure in the form $$int_{-infty}^{infty}x;dBig(int_AfdmuBig).$$

Now by the Fundamental Theorem of Calculus, this becomes $$int_{-infty}^{infty}xf(x);dx.$$

What does everyone think about this?

I'll try to provide a little more general result. Let $(Omega, mathcal{E}, P)$ be a probability space and let $Xcolon Omegalongrightarrow mathbb{R}$ be a random variable, i.e for each $Iin mathcal{B}$, $X^{-1}(I)inmathcal{E}$, where $mathcal{B}$ is the usual Borel $sigma-$algebra on $mathbb{R}$. Let us write $mu:=mu_{X}$ for the probability distribution of $X$, i.e for the measure defined on $mathcal{B}$ by $mu(I):=P(X^{-1}(I))$ for each $Iinmathcal{B}$. Then the following holds.

Theorem (Abstract-Concrete Formula): Let $phicolonmathbb{R}longrightarrow mathbb{R}$ be a borelian function, i.e $phi^{-1}(I)inmathcal{B}$ for every $Iinmathcal{B}$, and write $phi(X)$ for the composition $phicirc X$. Suppose at least one between the integrals $$int_{Omega} phi(X) dPquadtext{and}quad int_{mathbb{R}}phi (x) dmu $$
exists (resp. exists and it is finite). Then also the other one exists (resp. exists and it is finite) and it holds that $$int_{Omega} phi(X) dP=int_{mathbb{R}}phi (x) dmu .$$
In particular, $phi(X)$ is summable with respect to $P$ if and only if $phi$ is summable with respect to $mu$.

(When I say that a Lebesgue integral for a measurable function exists, I allow that it is not finite). The proof of this fact is quite straightforward but it requires some measure theory results such as the approximation theorem with simple functions and the Lebesgue's monotone convergence theorem. Indeed, suppose first that $phi$ is a (finitely) simple and positive function. Then also $phi(X)$ is simple (and positive) and (therefore) both mentioned integrals always exist. Writing $phi=sum_{i=1}^{n} c_{i}1_{E_{i}}$, where $n=vert phi(mathbb{R})vert$, $phi (mathbb{R})={c_{1},cdots,c_{n}}$ and $E_{i}:=phi^{-1}({c_{i}})$, we get $$int_{mathbb{R}}phi (x) dmu=sum_{i=1}^{n}c_{i}mu (E_{i})=sum_{i} c_{i}P(X^{-1}(E_{i}))=sum_{i} c_{i}P(X^{-1}(phi^{-1}({c_{i}})))=int_{Omega} phi(X) dP.$$ Assume now $phi$ is a non-negative borelian function. Then there exists a non-decreasing sequence $(phi_{n})_{nin mathbb{N}}$ of simple, positive functions such that $limlimits_{nto infty}phi_{n}(x)=phi (x)$ for every $xinmathbb{R}$. By monotone convergence theorem we get immediately that $$int_{mathbb{R}}phi (x) dmu=int_{mathbb{R}}(lim_{ntoinfty}phi_{n} (x)) dmu =lim_{ntoinfty} int_{mathbb{R}}phi_{n} (x) dmu=lim_{ntoinfty} int_{Omega}phi_{n} (X) dP=int_{Omega} phi(X) dP.$$ Finally, suppose only $phicolon mathbb{R}longrightarrowmathbb{R}$ is a borelian function, with no further restrictions. Then one can write $phi=phi^{+}-phi^{-}$ (here for notation). Suppose $int_{mathbb{R}}phi (x) dmu$ exists: then at least one between $int_{mathbb{R}}phi^{+} (x) dmu$ and $int_{mathbb{R}}phi^{-}(x) dmu$ (say, the first one) must be finite and hence also $int_{Omega}(phi(X))^{+} dP$ is finite, i.e $int_{Omega}phi(X) dP$ exists. It is now clear that we can conclude with our thesis.

Corollary: In the previous situation, if $E[X]<+infty$, then $$E[X]:=int_{Omega} X dP=int_{mathbb{R}} x dmu.$$ In particular, if $mu$ is absolutely continuous with respect to Lebesgue's Measure (on $mathbb{R}$) and has density $f$, then we get $E[X]=int_{mathbb{R}} xf(x) dx$ (by (a straightforward consequence of) Radon-Nikodym theorem).

If $Xgeqslant0$ almost surely, Fubini yields
$$
E[X]=int_Omega Xmathrm dP=int_Omegaint_0^inftymathbf 1_{tleqslant X}mathrm dtmathrm dP=int_0^inftyint_Omegamathbf 1_{tleqslant X}mathrm dPmathrm dt=int_0^infty P[Xgeqslant t]mathrm dt.
$$
Now, for each $t$,
$$
P[Xgeqslant t]=int_0^infty mathbf 1_{xgeqslant t}f(x)mathrm dx,
$$
hence Fubini again yields
$$
E[X]=int_0^inftyint_0^infty mathbf 1_{xgeqslant t}f(x)mathrm dxmathrm dt=int_0^inftyint_0^infty mathbf 1_{xgeqslant t}mathrm dtf(x)mathrm dx=int_0^infty xf(x)mathrm dx.
$$
If $X$ is real valued with $P[Xgt0]cdot P[Xlt0]ne0$, use the identity $E[X]=aE[Y]-bE[Z]$ where $a=P[Xgt0]$, $b=P[Xlt0]$, $Y$ has the distribution of $X$ conditioned on $Xgt0$ and $Z$ has the distribution of $-X$ conditioned on $Xlt0$, that is, $a=1-F(0)$, $b=F(0)$,
$$
f_Y(y)=frac{f(y)}{a}mathbf 1_{ygt0},qquad f_Z(z)=frac{f(-z)}{b}mathbf 1_{zgt0}.
$$
This yields
$$
E[X]=aint_0^infty yfrac{f(y)}{a}mathrm dy-bint_0^infty zfrac{f(-z)}{b}mathrm dz=int_{-infty}^{+infty}xf(x)mathrm dx.
$$

For non-negative $X$, argue as in @Did's proof the well-known result $E[X]=int_0^infty P[Xge t],dt$. For general integrable $X$ we have by definition $E[X]=E[X^+]-E[X^-]$. Since $X^+$ is non-negative, its expectation equals
$$
begin{align}
int_0^infty P[X^+ge t],dt&=int_0^infty P[Xge t],dt\
&=int_{t=0}^inftyint_{x=0}^infty I[xge t]f(x),dx,dtstackrel{rm Fubini}=int_{x=0}^inftyunderbrace{int_{t=0}^infty I[tle x]dt}_x, f(x)dx
end{align}
$$
and similarly the expectation of $X^-$ equals
$$
begin{align}
int_0^infty P[X^-ge t],dt&=int_0^infty P[Xle -t],dt\
&=int_{t=0}^inftyint_{x=-infty}^0 I[xle -t]f(x),dx,dt
stackrel{rm Fubini}=int_{x=-infty}^0underbrace{int_{t=0}^infty I[tle -x]dt}_{-x}, f(x)dx.
end{align}
$$
Adding these together gives $E[X]=int_{-infty}^0 xf(x)dx +int_0^infty xf(x)dx$.