# Multinomial Distribution: Maximum Possibility

I am reading Albert N Shiryaev's Probability. There is one question from Chapter I §2.

Problem 2: Show that for the multinomial distribution $\{P(A_{n1},..., A_{nr})\}$ the maximum probability is attained at a point $(k_1, ..., k_r)$ that satisfies the inequalities $np_i-1< k_i{\le}(n+r-1)p_i, i=1,...,r$.

The probability of Multinomial distribution can be found from wiki

May I have any hint on how to prove that? If possible, can we have the expression for this maximum possibility? I am thinking of proving it using Lagrange multiplier. However, the Lagrange multiplier may only useful for continuous case?

Thanks,

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Try the case $r=2$ (Binomial) first. Look at the ratio of consecutive terms. Some manipulation of binomial coefficients is needed. For the future, use $p_1$, $p_2$, not $p$ and $1-p$, also $k_1$, $k_2$, not $k$ and $n-k$. – André Nicolas Jul 21 '11 at 6:04
Thanks for the hint. Previous I over think on the problem. I was trying to analyse the part $p_i^k/k!$. Now I try to think it as a whole. I can prove binomial case. For multinomial, I reach the following step and am stuck again. May I have more hints? $\frac{k_i+1}{p_i}\ge\frac{k_j}{p_j},\forall i,j \in \{1,2, .., r\} \Rightarrow np_i-1<k_i\le (n+r-1)p_i$. – ygao Jul 25 '11 at 7:47
Maybe assume that we have maximum where some $k_i$ is "too small" (other case, "too big"). Then since the sum of the $k_i$ is fixed, as is the sum of the probabilities, some $k_j$ is "too big". See what happens when we increase $k_i$ by $1$, decrease $k_j$ by $1$, leaving all the rest alone. A ratio calculation like in the $n=2$ case should work. – André Nicolas Jul 25 '11 at 11:58