I've got a question concerning the expected value of a sum which starts at a certain value given through a random variable.
More precisely:
$$G(n):=P[X \geq K]$$ where $X \in Bin(n,p)$ and $K$ is a random variable with expected value $\mu$ and Support $\{1,...,m\}$ where $m \leq n$. Now I am interested in $G(n)$.
I tried to do the following calculation:
$$G(n)=P[X \geq K]= \sum_{k=1}^m P[X \geq k] P[K=k] = E[P[X \geq K]] = E[ \sum_{k=K}^n {{n} \choose {k}} p^k (1-p)^{n-k} ]$$
Now, I'm not sure whether my calculation from above makes any sense.. And if, I'm not sure whether now I can conclude that this is equal to $$\sum_{k=\mu}^n {{n} \choose {k}} p^k (1-p)^{n-k} = P[X \geq \mu]$$
I would really appreciate if you could take a look at my calculations and tell me whether they are wrong (and maybe why) or why they are correct. I have no idea how to justify the last step I made, when I replace $E[\sum_{k=K} ...]$ by $\sum_{k=\mu} ...$.
Thank you very much for your help..
The first part is basically correct, which leads to $$ G(n)=E\left[ \sum_{k=K}^n a_n(k) \right],\qquad a_n(k)={n\choose k} p^k (1-p)^{n-k}. $$ Now, rewrite the random variable in the RHS as $$ \sum_{k=0}^n a_n(k)\mathbf 1_{K\leqslant k}, $$ and deduce that $$ G(n)=\sum_{k=0}^n a_n(k)P(K\leqslant k). $$ There is no general expression of the RHS as a function of $\mu$ only. An equivalent formulation is $G(n)=E[b_n(K)]$ where $$ b_n(i)=\sum_{k=i}^na_n(k). $$