Answer to Question #314464 in Statistics and Probability for Mari bless

Question #314464

Let F and G be two sigma-fields on Ω. Prove that F ∩ G is also a sigma-field on Ω.


Show by example that F ∪ G may fail to be sigma-field if Ω = {1, 2, 3}.



1
Expert's answer
2022-03-21T00:25:55-0400

"\\emptyset \\in F,\\emptyset \\in G\\Rightarrow \\emptyset \\in F\\cap G\\\\A\\in F\\cap G\\Rightarrow A\\in F,A\\in G\\Rightarrow \\bar{A}\\in F,\\bar{A}\\in G\\Rightarrow \\bar{A}\\in F\\cap G\\\\A_i\\in F\\cap G\\Rightarrow A_i\\in F,A_i\\in G\\Rightarrow \\bigcup{A_i}\\in F,\\bigcup{A_i}\\in G\\Rightarrow \\bigcup{A_i}\\in F\\cap G"

Thus "F\\cap G" is a sigma-algebra.

Let

"F=\\left\\{ \\emptyset ,\\left\\{ 1 \\right\\} ,\\left\\{ 2,3 \\right\\} ,\\left\\{ 1,2,3 \\right\\} \\right\\} \\\\G=\\left\\{ \\emptyset ,\\left\\{ 2 \\right\\} ,\\left\\{ 1,3 \\right\\} ,\\left\\{ 1,2,3 \\right\\} \\right\\} \\\\F\\cup G=\\left\\{ \\emptyset ,\\left\\{ 1 \\right\\} ,\\left\\{ 2 \\right\\} ,\\left\\{ 1,3 \\right\\} ,\\left\\{ 2,3 \\right\\} ,\\left\\{ 1,2,3 \\right\\} \\right\\}"

We see that "\\left\\{ 1 \\right\\} \\in F\\cup G,\\left\\{ 2 \\right\\} \\in F\\cup G,\\left\\{ 1 \\right\\} \\cup \\left\\{ 2 \\right\\} =\\left\\{ 1,2 \\right\\} \\notin F\\cup G"

Thus "F\\cup G" is not a sigma-algebra.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS