Question #314463

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.

1
Expert's answer
2022-03-20T06:40:23-0400

We need to show that $PH$ is a probability on the sigma-algebra F. Indeed:

PH(A)=P(AH)P(H)0PH(Ω)=P(ΩH)P(H)=P(H)P(H)=1AiAj=:PH(Ai)=P(AiH)P(H)=P((AiH))P(H)==P(AiH)P(H)=P(AiH)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)}


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