Chapter 7: Q. 7.5 (page 359)

Let $A_{1}, A_{2}, \dots, A_{n}$ be arbitrary events, and define
$C k =$ {at least $k$ of the $A i$ occur}. Show that
$\sum_{k = 1}^{n} P (C_{k}) = \sum_{k = 1}^{n} P (A_{k})$
Hint: Let $X$ denote the number of the $A i$ that occur. Show
that both sides of the preceding equation are equal to $E [X]$ .

Short Answer

Expert verified

The arbitrary events is showed as $\sum_{k = 1}^{n} P (C_{k}) = E (X) = \sum_{k = 1}^{n} P (A_{k})$ .

Step by step solution

Given Information

$P (C_{k}) = P (X \geq k)$ as arbitrary events in $A_{1}, A_{2}, \dots, A_{n}$ .

Explanation

We have that, $P (C_{k}) = P (X \geq k)$

Hence, $\sum_{k = 1}^{n} P (C_{k}) = \sum_{k = 1}^{n} P (X \geq k) = \sum_{k = 0}^{\infty} P (X > k) = E (X)$

where the last equality is the famous expression of the mean of non-negative discrete random variable. On the other hand, define random variables $I_{i}$ to be indicators whether event $A_{i}$ has occurred or not.

Explanation

We have that, $X = \sum_{k = 1}^{n} I_{k}$

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

$E (X) = \sum_{k = 1}^{n} E (I_{k}) = \sum_{k = 1}^{n} P (I_{k} = 1) = \sum_{k = 1}^{n} P (A_{k})$

so we have showed that, $\sum_{k = 1}^{n} P (C_{k}) = E (X) = \sum_{k = 1}^{n} P (A_{k})$ .

Final answer

The arbitrary events is showed as $\sum_{k = 1}^{n} P (C_{k}) = E (X) = \sum_{k = 1}^{n} P (A_{k})$ .

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Let A1,A2,…,Anbe arbitrary events, and defineCk={at least kof the Aioccur}. Show that∑k=1nPCk=∑k=1nPAkHint: Let Xdenote the number of the Aithat occur. Showthat both sides of the preceding equation are equal to E[X].

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Explanation

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