This is a question from my lecture notes.
""" Persons arrive at a copy machine according to a Poisson process with rate λ=1 per minute. The number of copies made is uniformly distributed between 1 and 10. Each copy requires 3 seconds. Find the average waiting time in the queue and the average system waiting time. """
I know how to do the problem, but I am having trouble understanding why the following calculation of an expectation is correct. $$ E[X^2] = \sum_{i = 1}^{10}(3i)^2 * \space Pr(X = 3i) $$
I don't understand why there is a $(3i)^2$ and why it's $Pr(X =3i)$ term in this expected value. I would set this up:
$$ 3E[X^2] = \sum_{i = 1}^{10}(i)^2 * \space Pr(X = i) $$ Can anyone please provide some intuition? I've looked online and through my textbooks and I can't find anything that helps me with this intuition. I know it has something to do with the 3 seconds in the problem statement, but I can't figure out how to make sense of this and therefore generalize this problem.
Thanks in advance!
I am guessing that you are trying to compute variance of $X$.
The problem is we did not define $X$.
In your lecture note, $X$ seems defined to be waiting time. Waiting time comes in multiple of 3 in this setting.
We have $Var[X]=E[X^2]-E[X]^2$
On the other hand, it seems that you are trying to wokr with the number of copies.To avoid confusion, let's define it to be $Y$ instead. In particular, we have the relationship $$X=3Y$$ and if we square them and take expectation, we have
$$E[X^2]=3^2E[Y^2].$$
Again, we can compute $Var[Y]=E[Y^2]-E[Y]^2$
A point to consider is suppose you really prefer to work with $Y$ (the number of copies), can you compute $Var[X]$?
$$Var[X]=E[(3Y)^2]-E[3Y]^2=9(E[Y^2]-E[Y]^2)=3^2Var[Y]$$