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Let A1,A2,…,Anbe arbitrary events, and define

Ck={at least kof the Aioccur}. Show that

∑k=1nPCk=∑k=1nPAk

Hint: Let Xdenote the number of the Aithat occur. Show

that both sides of the preceding equation are equal to E[X].

Short Answer

Expert verified

The arbitrary events is showed as∑k=1nPCk=E(X)=∑k=1nPAk.

Step by step solution

01

Given Information

PCk=P(X≥k)as arbitrary events inA1,A2,…,An.

02

Explanation

We have that,PCk=P(X≥k)

Hence,∑k=1nPCk=∑k=1nP(X≥k)=∑k=0∞P(X>k)=E(X)

where the last equality is the famous expression of the mean of non-negative discrete random variable. On the other hand, define random variables Iito be indicators whether event Aihas occurred or not.

03

Explanation

We have that, X=∑k=1nIk

Because of the linearity of the mean, we have that,

E(X)=∑k=1nEIk=∑k=1nPIk=1=∑k=1nPAk

so we have showed that,∑k=1nPCk=E(X)=∑k=1nPAk.

04

Final answer

The arbitrary events is showed as∑k=1nPCk=E(X)=∑k=1nPAk.

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