Limit of probability in stars and bars

Question

Let $n$ and $r$ be integers. Compute the number of solutions to the equation $$x_1 + x_2 + \cdots + x_r = n,$$ where $x_i \geq 0$ are integers. Next, assuming the uniform distribution on the space of solutions, find $P(x_1 = a)$ and its limit as $r\to\infty, n\to\infty, n/r\to \rho > 0$.

The first part of the question is just stars and bars. One can easily get ${}n + r - 1 \choose r - 1$ .

Now I'm not so sure how to find $P(x_1 = a)$ or the limit. This is from a Probability and Measure Theory book, so I am trying to prove everything formally (particularly the limit portion). I tried to come up with something to condition on, and I got nowhere. I'm pretty sure it'll be a function of $n$ and $r$ because of the limit question that follows it. Does anyone have any ideas?

Answer 1

The number of cases where $X_1=a$ is the same as the number from your original calculation but replacing $n$ by $n-a$ and replacing $r$ by $r-1$, so ${n-a + r - 2 \choose r - 2}$

and with the uniform distribution this makes $\mathbb P(X_1=a) = \dfrac{{n-a + r - 2 \choose r - 2}}{{n + r - 1 \choose r - 1}}$ which you could rewrite

I will leave finding the formal limit to you, but it would not surprise me if you ended up with a Poisson distribution with parameter $\rho$, i.e. $\lim \mathbb P(X_1=a) = \frac{\rho^a }{a!}e^{-\rho}$

Answer 2

If $X_1=a$, then we will be left with $r-1$ non-negative integers that sum to $n-a$. Hence the cardinality of the set $(X_1=a)$ is $$ \binom{n-a+r-2}{r-2} $$ by stars and bars. Since we have a uniform distribution on the space of solutions it follows that $$ P(X_1=a)=\binom{n-a+r-2}{r-2}\bigg/\binom{n + r - 1}{ r - 1}. $$