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I'm currently studying probability theory with the textbook Probability and Random Processes for Electrical and Computer Engineers (John Gubner) and had a question regarding multiple random variables.

More specifically, this comes from Example 2.10 of section 2.3 Multiple Random Variables - Independence on page 72.

A webpage server can handle $r$ requests per day. Find the probability that the server gets more than $r$ requests at least once in $n$ days. Assume that the number of requests on day $i$ is $X_i \sim \text{Poisson}(\lambda)$ and that $X_1, \dots , X_n$ are independent.

Solution:

$$ \begin{align} P\left(\bigcup_{i = 1}^n \{ X_i \gt r \}\right) & = 1 - P\left( \bigcap_{i = 1}^n \{X_i \le r \} \right) \\\ & = 1 - \prod_{i = 1}^nP\big(X_i \le r \big) \\\ & = 1 - \prod_{i = 1}^n \left( \sum_{i = 1}^r \frac{\lambda^k e^{-\lambda}}{k!} \right) \\\ & = 1 - \left( \sum_{i = 1}^r \frac{\lambda^k e^{-\lambda}}{k!} \right)^n \end{align} $$

I understand the solution, but what's confusing me is how the first line was derived. That is:

$$ P \left( \bigcup_{i = 1}^n \{X_i \gt r \} \right) = 1 - P\left( \bigcap_{i = 1}^n \{ X_i \le r \} \right)$$

I've tried working it out with a simple example of two random variables, but that's also not really making a lot of sense.

How was this equality derived?

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1 Answer 1

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$P(E)=1-P(E^{c})$. The complement of the set $\bigcup_{i=1}^{n} \{X_i >r\}$ is exactly the set $\bigcap_{i=1}^{n} \{X_i \leq r\}$ .

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  • $\begingroup$ Thanks for the quick answer. Is that complement that result of DeMorgan's Law? I'm just having trouble intuitively grasping how that result occurs. For the first example you gave it makes sense that $P(E) = 1 - P(E^c)$, but it doesn't come to me as easily for the case with sets. $\endgroup$
    – Sean
    Commented Sep 15, 2019 at 5:53
  • $\begingroup$ Yes, I am using De Morgan's Laws. See, for example en.wikipedia.org/wiki/De_Morgan%27s_laws $\endgroup$ Commented Sep 15, 2019 at 6:01
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    $\begingroup$ Basically $(\bigcup_i S_i)^\complement=\bigcap_i (S_i^\complement)$ as per DeMorgan's Law. $$\begin{align}x\in(\bigcup_i S_i)^\complement&\iff \neg\exists i~.x\in S_i\\&\iff \forall i~.x\notin S_i\\&\iff x\in\bigcap_i S_i^\complement\end{align}$$ And, of course, the complement of the event of $\{X_i>r\}$ is that $\{X_i\leq r\}$ $\endgroup$ Commented Sep 15, 2019 at 6:10

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