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"


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!

  • $\begingroup$ 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

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.