A3. Let (Ω, F, P) be a probability space and let H ∈ F with P(H) > 0. For any arbitrary
A ∈ F, let
PH(A) =
P(A ∩ H)
P(H)
Show that (Ω.F, PH) is a probability space.
We need to show that $PH$ is a probability on the sigma-algebra F. Indeed:
PH(A)=P(A∩H)P(H)⩾0PH(Ω)=P(Ω∩H)P(H)=P(H)P(H)=1Ai∩Aj=∅:PH(∑Ai)=P(∑Ai∩H)P(H)=P(∑(Ai∩H))P(H)==∑P(Ai∩H)P(H)=∑P(Ai∩H)P(H)=∑PH(Ai)PH\left( A \right) =\frac{P\left( A\cap H \right)}{P\left( H \right)}\geqslant 0\\PH\left( \varOmega \right) =\frac{P\left( \varOmega \cap H \right)}{P\left( H \right)}=\frac{P\left( H \right)}{P\left( H \right)}=1\\A_i\cap A_j=\emptyset :PH\left( \sum{A_i} \right) =\frac{P\left( \sum{A_i}\cap H \right)}{P\left( H \right)}=\frac{P\left( \sum{\left( A_i\cap H \right)} \right)}{P\left( H \right)}=\\=\frac{\sum{P\left( A_i\cap H \right)}}{P\left( H \right)}=\sum{\frac{P\left( A_i\cap H \right)}{P\left( H \right)}}=\sum{PH\left( A_i \right)}PH(A)=P(H)P(A∩H)⩾0PH(Ω)=P(H)P(Ω∩H)=P(H)P(H)=1Ai∩Aj=∅:PH(∑Ai)=P(H)P(∑Ai∩H)=P(H)P(∑(Ai∩H))==P(H)∑P(Ai∩H)=∑P(H)P(Ai∩H)=∑PH(Ai)
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