# Show the equality of E[Z] = $\sum_{k=1}^{n} P[M_K]$ (see the task)

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.

• I think you have the right idea for $M_k$ but you definitely need to get your indices right. – Fimpellizieri Dec 5 '19 at 15:57

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