Let $n \in N$ and let $X^{(n)}=(X_1^{(n)},X_2^{(n)},X_3^{(n)})$ be multinomial distributed with parameters $3n;\frac{1}{3},\frac{1}{3},\frac{1}{3}$. Compute the approximation of $\mathbf{P}(X_1^{(n)}=X_2^{(n)}=X_3^{(n)})=n$ using the Sterling-Formula.
I believe I have understood what the multinomial distribtion is doing and also how the binomial distribution is just a special case of it. The multinomial coefficient is just the number of the permutations possible for a given combination we are looking for. We use this only when we want a fixed number of occurrences for the the elements but don't care about their position. The second part is just each class's probability to the power of how many occurrences. For the very easy base problem I know how to use it.
But in this problem, where does this "to the power of n" come from and how is it linked to the given parameter $3n$. For now I don't want to look at the approximation as it is just replacing $n!$ with the Sterling term. I can show you what I have got so far, but I have to be honest, all I did was plug in parameters, and I'm pretty certain it is not correct, because I'm basically ignoring that power to the n thing
$$ \mathbf{P}((X_1^{(n)}=X_2^{(n)}=X_3^{(n)})=n)= {3n\choose n,n,n} \cdot \frac{1}{3}^n \cdot \frac{1}{3}^n \cdot \frac{1}{3}^n = {\frac{(3n)!}{3n!}} \cdot \frac{1}{9}^n$$
Any help is much appreciated!!
Best Answer
$X^{(n)}$ is not a power of $n$ but a label, which with the rest of the definition here in effect says that you throw $3n$ balls independently into $3$ labelled urns (each ball going into each urn with equal probability) and count the number of balls in each urn. The question is then asking for the probability that there are exactly $n$ balls in each urn.
You may have a typo in your final expression, which I would have thought could be : $$\mathbf{P}((X_1^{(n)}=X_2^{(n)}=X_3^{(n)})=n)= {3n\choose n,n,n} \cdot \frac{1}{3}^n \cdot \frac{1}{3}^n \cdot \frac{1}{3}^n = {\frac{(3n)!}{(n!)^3}} \cdot \frac{1}{27^n}$$
To answer the question, you now need to replace the factorials using the Stirling approximation