Homework Help is Here – Start Your Trial Now!
Math
Probability
Theorem 2.10 Suppose X, Y, Xn are integrable. (i) If X = a a.s., then E(X N) = a, a.s. (ii) E(aX+bY|N) = aE(X\N) + bE(Y|N), a.s. (iii) If X < Y a.s., then E(X|N)< E(Y|N), a.s. (iv) |E(X\N)| < E(\X||N), a.s.

Theorem 2.10 Suppose X, Y, Xn are integrable. (i) If X = a a.s., then E(X N) = a, a.s. (ii) E(aX+bY|N) = aE(X\N) + bE(Y|N), a.s. (iii) If X < Y a.s., then E(X|N)< E(Y|N), a.s. (iv) |E(X\N)| < E(\X||N), a.s.

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
See similar textbooks
icon
Related questions
Question

Please prove i - iv

Theorem 2.10 Suppose X, Y,Xn are integrable. (i) If X = a a.s., then E(X\N) = a, a.s. — а, (ii) E(aX +bY|N) = aE(X|N) + bE(Y|N), a.s. (iii) If X < Y a.s., then E(X|N) < E(Y|N), a.s. (iv) |E(X|N)| < E(\X||N), a.s.
Transcribed Image Text:Theorem 2.10 Suppose X, Y,Xn are integrable. (i) If X = a a.s., then E(X\N) = a, a.s. — а, (ii) E(aX +bY|N) = aE(X|N) + bE(Y|N), a.s. (iii) If X < Y a.s., then E(X|N) < E(Y|N), a.s. (iv) |E(X|N)| < E(\X||N), a.s.
Expert Solution
steps

Step by step

Solved in 5 steps with 4 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Similar questions
Recommended textbooks for you
A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
Probability
ISBN:
9780134753119
Author:
Sheldon Ross
Publisher:
PEARSON
A First Course in Probability
A First Course in Probability
Probability
ISBN:
9780321794772
Author:
Sheldon Ross
Publisher:
PEARSON
SEE MORE TEXTBOOKS