You have a set of consecutive n natural numbers {1,2,3...,n}. m different numbers are drawn from these n numbers. Calculate the variance of the sum of these m numbers?.
My try:
Since each draw of m numbers are different and as likely as any other draw, would it not be true that it is a uniform distribution with range of X from the minimum of (1,2,3,...,m) with X being $\frac{m(m+1)}{2}$ to a maximum of (n,n-1,n-2,..n-(m-1)) with X being $\frac{m(2n-m+1)}{2}$ This mamimum and maximum form the range of the uniform distribution and that I can find the Expected value and Variance with these formulas $\frac{(a+b)}{2}$ and $\frac{(b-a)^2}{12}$ where a is the minimum and b is the maximum thus found.
Let me know if this is the right approach
I figured a solution for this after some research which is quite complicated and would love for someone to simplify in the lines that I have presented.
$$Var(X) = \sum_{i=0}^{m(n-m)+1}\dfrac{ a_i q^{i}}{{n\choose m}}$$
$a_iq^i = {n\choose m}_q$ the gaussian binomial coefficient and $a_i$ is the coefficient representing the frequency of the sum starting from $\frac{m(m+1)}{2}$ to $\frac{m(2n-m+1)}{2}$
We define $q^i = \dfrac{\left(\frac{m(m-n)}{2} + i\right)^2}{4}$