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i am trying to show the equality but i dont get any further with the exercise. Can someone give me a hint? Here is the task:

Let $A_1,A_2...,A_n$ be events ($A_k \in \mathcal{F}$)

a) Express the following events with $A_1,....,A_n$

i): $M_k$ = "at least $k$ of the events occur"

ii): $G_k$ = "exactly $k$ of the events occur"

and

b) Let $Z = \sum_{k=1}^{n} \mathbb{I}_{A_k}$ the number of events that have occured. Show that:

$$ \mathbb{E}[Z] = \sum_{k=1}^{n} \mathbb{P}[A_k] = \sum_{k=1}^{n} k\mathbb{P}[G_k] = \sum_{k=1}^{n} \mathbb{P}[M_k] $$

My approach is:

a_i) $M_k = \bigcup_{i=1}^n \bigcap_{j=1}^{k} A_{ij}$

a_ii) $G_k = M_k \, \cap \, M_{k+1}^{c} = \big( \bigcup_{i=1}^n \bigcap_{j=1}^{k} A_{ij} \big) \, \cap \, \big( \bigcap_{i=1}^{n} \bigcup_{j=1}^{k+1} A_{ij}^{c} \big) $

b) $$ \mathbb{E}[Z] = \mathbb{E}[\sum_{k=1}^{n} \mathbb{I}_{A_k}] = \sum_{k=1}^{n} \mathbb{E}[\mathbb{I}_{A_k}] = \sum_{k=1}^{n} \mathbb{P}[A_k] $$

now i dont now how to argue. It would be nice, if someone could help me.

Thanks in advance!

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  • $\begingroup$ I think you have the right idea for $M_k$ but you definitely need to get your indices right. $\endgroup$ – Fimpellizieri Dec 5 '19 at 15:57
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To continue of what you've written argue in this way: the expected value of number of occurred events is the number of events multiplied by its probability, i.e.

$$\mathbb{E}Z = \sum_{k=1}^n k \mathbb{P}(G_k).$$

Then, since each next $M_k$ contains previous ones hence you get $$ \sum_{k=1}^n k \mathbb{P}(G_k) = \sum_{k=1}^n \mathbb{P}(M_k), $$ as desired.

This is exactly the idea behind the following argument.

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  • $\begingroup$ thank you pointguard0 ! $\endgroup$ – Kaya Dec 5 '19 at 16:19

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