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Fubini-Tonelli

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Theorem
MeasureTheory

Measure Theory

Let (X,M,μ),(Y,N,ν)(X,\mathscr{M},\mu),(Y,\mathscr{N},\nu ) be σ-finite measure spaces and let f:X×YCf:X\times Y\to\mathbb{C} be P\mathscr{P}-measurable. 1. Fubini: If f:X×Y[0,+]f:X\times Y\to[0,+\infty], set φ:X[0,+]\varphi:X\to[0,+\infty] and ψ:Y[0,+]\psi:Y\to[0,+\infty] as φ(x)=Yfxdν\varphi(x)=\int\limits _{Y}f_{x} \, d\nu and ψ(y)=Xfydμ\psi(y)=\int\limits _{X}f^{y} \, d\mu. Then, φ\varphi is M\mathscr{M}-measurable and ψ\psi is N\mathscr{N}-measurable and we have Xφdμ=Yψdν\int\limits _{X}\varphi \, d\mu=\int\limits _{Y}\psi \, d\nu That is XYf(x,y)dν(y)dμ(x)=YXf(x,y)dμ(x)dν(y)=X×Yfd(μ×ν)\int\limits_{X} \int\limits_{Y}f(x,y) \, d\nu(y) \, d\mu(x) = \int\limits_{Y} \int\limits_{X}f(x,y) \, d\mu(x) \, d\nu(y) =\int\limits _{X\times Y}f \, d(\mu\times \nu) 2. Tonelli: Let φ(x)=Yfxdν\varphi^{*}(x)=\int\limits _{Y}|f_{x}| \, d\nu. If Xφdμ<\int\limits _{X}\varphi^{*} \, d\mu<\infty then X×Yfd(μ×ν)<\int\limits _{X\times Y}|f| \, d(\mu \times \nu)<\infty similarly for ψ(y)\psi^{*}(y). 3. If X×Yfd(μ×ν)<\int\limits _{X\times Y}|f| \, d(\mu \times \nu)<\infty then fxL1(ν),fyL1(μ)f_{x}\in L^{1}(\nu),f^{y}\in L^{1}(\mu) and XYf(x,y)dν(y)dμ(x)=YXf(x,y)dμ(x)dν(y)\int\limits _{X}\int\limits _{Y}f(x,y) \, d\nu(y) \, d\mu(x)=\int\limits _{Y}\int\limits _{X}f(x,y) \, d\mu(x) \, d\nu(y)

Probability

For (E,E,P1),(F,F,P2)(E,\mathcal{E},\mathbb{P}_{1}),(F,\mathcal{F},\mathbb{P}_{2}) probability spaces and product measure P\mathbb{P} on (E×F,EF,P)(E\times F,\mathcal{E}\otimes \mathcal{F},\mathbb{P}). If f:E×FRf: E\times F\to \mathbb{R} is EF\mathcal{E}\otimes \mathcal{F}-measurable and if either f0f\ge 0 or fL1f\in L^{1} then E×FfdP=EFfdP2dP1=FEfdP1dP2\int\limits _{E\times F}f \, d\mathbb{P}=\int\limits _{E}\int\limits _{F}f \, d\mathbb{P_{2}} \, d\mathbb{P}_{1}=\int\limits _{F}\int\limits _{E}f \, d\mathbb{P}_{1} \, d\mathbb{P}_{2} with yEfy(x)dP1(x)y\mapsto \int\limits _{E}f_{y}(x) \, d\mathbb{P}_{1}(x) measurable.

Using this theorem we then covered convolutions and how allows us to find the distribution of sums of independent rvs.

Suppose X,YX,Y are independent rvs with distributions μ,ν\mu,\nu respectively. Then the distribution of X+YX+Y is given by μν\mu*\nu, where (μν)(H)=Rμ(Hy)ν(dy).(\mu*\nu)(H)=\int\limits _{\mathbb{R}}\mu(H-y) \, \nu(dy) .Equivalently, assume X,YX,Y have densities f,gf,g, then X+YX+Y has density fgf*g.

SDEs

Let (X,M,μ),(Y,N,ν)(X,\mathscr{M},\mu),(Y,\mathscr{N},\nu ) be σ-finite measure spaces and let f:(X×Y,P,μ×ν)(R,B(R))f:(X\times Y,\mathscr{P}, \mu \times \nu)\to(\mathbb{R},\mathcal{B}(\mathbb{R}))be P\mathscr{P}-measurable. Assume X(Yf(x,y)dν(y))dμ(x)<\int\limits _{X}\left( \int\limits _{Y}|f(x,y)| \, d\nu(y) \right) \, d\mu(x)<\infty (i.e. fL2(X×Y,P,μ×ν)f\in L^{2}(X\times Y,\mathscr{P}, \mu \times \nu)) then, XYf(x,y)dν(y)dμ(x)=YXf(x,y)dμ(x)dν(y)=X×Yfd(μ×ν)\int\limits_{X} \int\limits_{Y}f(x,y) \, d\nu(y) \, d\mu(x) = \int\limits_{Y} \int\limits_{X}f(x,y) \, d\mu(x) \, d\nu(y) =\int\limits _{X\times Y}f \, d(\mu\times \nu)

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