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Chapter 3: Q. 3.30 (page 111)

Show that for any events Eand F,

P(E∣E∪F)≥P(E∣F)

Hint: Compute P(E∣E∪F)by conditioning on whether F occurs.

Short Answer

Expert verified

The result is that P(E∣E∪F)is weighted average betweenP(E∣F)and1

Step by step solution

01

Prove

Demonstrate as all eventualities E,F

P(E∣E∪F)=P[E∣F∩(E∪F)]P(F∣E∪F)+PE∣Fc∩(E∪F)PFc∣E∪F

If it's not clear, evaluate that equations via enlarging the correct hand side by a probability density statement.

F∩(E∪F)=F

Fc∩(E∪F)=E∩Fc

⇒PE∣Fc∩(E∪F)=PE∣EFc=1

02

Weighted Average

That's also sufficient to ascertain the disparity, as

P(E∣E∪F)=P(E∣F)P(F∣E∪F)+1⋅PFc∣E∪F

P(F∣E∪F)+PFc∣E∪F=1

P(E∣F)≤1

thus localid="1649659759663" P(E∣E∪F)is that the weighted combination of such constants localid="1649659766251" P(E∣F)and localid="1649659771667" 1, so just between that two integers,

localid="1649659823011" P(E∣F)≤P(E∣E∪F)≤1

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