Variance of values that $x_1$ gets in the solutions of $x_1+x_2+...+x_k = n$

Question

I know that the there are ${n-1}\choose{k-1}$ solutions for the equation $x_1+x_2+...+x_k = n$ which all $x_i\ge 1$, is there any way to calculate the variance of the values that $x_1$ gets$?

Answer 1

Take $X_1,\dots,X_k$ to be a random vector uniformly distributed on the solutions to $x_1+\dots+x_k=n$ among positive integers. The probability that $X_1=x$ for $x=1,\dots,n-(k-1)$ is the number of solutions to $X_2+\dots+X_k=n-x$ divided by the total number of solutions. So this is $\frac{{n-x-1 \choose k-2}}{{n-1 \choose k-1}}$. So noting that the mean is just $n/k$ by symmetry, the variance is

$$\sum_{x=1}^{n-(k-1)} \frac{x^2 {n-x-1 \choose k-2}}{{n-1 \choose k-1}} - (n/k)^2.$$

I'll leave the simplification of that sum to you.