I'm aware that this is in regards to the same question set as Find the Mean for Non-Negative Integer-Valued Random Variable and Expected value equals sum of probabilities
my issue is understanding and unfortantly either of those two help my understanding.
Let T be a non-negative integer-valued random variable with $E(T) < \infty$. prove that
$$E(T) = \sum_{k=1}^{\infty} P(T \geq k)$$
so my understanding of this question is shocking infact it wasn't untill i had read the above that i relised $P(T \geq k) = \sum_{i=k}^{\infty}P(T=i)$ and why this made sense. (completely slipped my mind)
i've tried to follow one of the aboves answers and i need to make sure my understanding of whats going on is concrete and correct.
from my understanding the key to this is remembering that infinite double series can be written as a single series which is going to be the expectation. in all honesty its the reworking of the lower and upper limits which has me concerned.
we start with $$E(T) = \sum_{k=1}^{\infty} P(T \geq k) = \sum_{k=1}^{\infty}\sum_{i=k}^{\infty}P(T=i)= \sum_{k=0}^{\infty}\sum_{i=k}^{\infty}P(T=i)$$
this is just straight substitution of the above identity and shifting the lower index by 1 which if i remember rightly we can do because this is a convergent series. from reading the above the next step i believe is.
$$\sum_{k=0}^{\infty}\sum_{i=k}^{\infty}P(T=i) = \sum_{i=0}^{\infty}\sum_{k=0}^{i}P(T=i)$$ which looks like a rearrangement of the series. i will tend to infinity anyway so this looks like we're just taking particular values at a time in order to end up with i inside the summation. but this is the step in which im unsure about and so any explination would be great. from here it looks like
$$\sum_{i=0}^{\infty}\sum_{k=0}^{i}P(T=i) = \sum_{i=0}^{\infty} iP(T=i) = E[T]$$
which answers the set question but not my question as to why this is allowed.
again any and all help would be great. thanks in advance.
Best Answer
In the step $$ \sum_{k=1}^{\infty}\sum_{i=k}^{\infty}P(T=i)= \sum_{k=0}^{\infty}\sum_{i=k}^{\infty}P(T=i) $$ the shifting of the lower index by $1$ is not allowed, and ends up incorrectly adding an additional $1$ to the sum, and you won't end up with $E(T)$. If you don't shift the lower index, you can continue to rearrange the order of summation as before, and your final sum is $$ \sum_{i=1}^{\infty}\sum_{k=1}^{i}P(T=i) = \sum_{i=1}^{\infty} iP(T=i) $$ At this point you can shift the lower index to $i=0$ because you're adding zero to the sum. The result will then be $$\sum_{i=1}^{\infty} iP(T=i)=\sum_{i=0}^{\infty} iP(T=i)=E(T).$$