1. ## Combinatorial inequality

I have to prove that if $\displaystyle n_{1}+n_{2}=n$ and$\displaystyle \frac{n_{1}}{n}=p$, then prove $\displaystyle \binom rk (p-\frac{k}{n})^k (q-\frac{r-k}{n})^{r-k}<\frac{\binom{r}{k}\binom{n-r}{n_{1}-k}}{\binom {n}{n_{1}}}<\binom rk p^k q^{r-k}(1-\frac{r}{n})^{-r}$
I have tried and simplified it to show that this reduces to proving $\displaystyle (n_{1}-k)^k (n-n_{1}-r+k)^{r-k}<\frac{\binom{r}{k}\binom{n-r}{n_{1}-k}}{\binom {n}{n_{1}}}<{n_{1}}^k(n-n_{1})^{r-k}(n-r)^{-r}$

But I cant see how to proceed further to prove this.

2. ## Re: Combinatorial inequality

Originally Posted by earthboy
I have to prove that if $\displaystyle n_{1}+n_{2}=n$ and$\displaystyle \frac{n_{1}}{n}=p$, then prove $\displaystyle \binom rk (p-\frac{k}{n})^k (q-\frac{r-k}{n})^{r-k}<\frac{\binom{r}{k}\binom{n-r}{n_{1}-k}}{\binom {n}{n_{1}}}<\binom rk p^k q^{r-k}(1-\frac{r}{n})^{-r}$
I have tried and simplified it to show that this reduces to proving $\displaystyle (n_{1}-k)^k (n-n_{1}-r+k)^{r-k}<\frac{\binom{r}{k}\binom{n-r}{n_{1}-k}}{\binom {n}{n_{1}}}<{n_{1}}^k(n-n_{1})^{r-k}(n-r)^{-r}$

But I cant see how to proceed further to prove this.
Do we know that q = 1 - p ? And so q = n2/n ? where n2 = n sub 2

Also, in your second inequality rCk should have been divided out from the middle expression as well as the left and right.

3. ## Re: Combinatorial inequality

Originally Posted by Debsta
Do we know that q = 1 - p ? And so q = n2/n ? where n2 = n sub 2

Also, in your second inequality rCk should have been divided out from the middle expression as well as the left and right.
Yes, we know that $\displaystyle q=1-p$ and so $\displaystyle q=n_{2}/n$.
Thanks for the correction,$\displaystyle \binom rk$ should have been divided out in the middle.
so we have to prove : $\displaystyle (n_{1}-k)^k (n-n_{1}-r+k)^{r-k}<\frac{\binom{n-r}{n_{1}-k}}{\binom {n}{n_{1}}}<{n_{1}}^k(n-n_{1})^{r-k}(n-r)^{-r}$

4. ## Re: Combinatorial inequality

Originally Posted by earthboy
Yes, we know that $\displaystyle q=1-p$ and so $\displaystyle q=n_{2}/n$.
Thanks for the correction,$\displaystyle \binom rk$ should have been divided out in the middle.
so we have to prove : $\displaystyle (n_{1}-k)^k (n-n_{1}-r+k)^{r-k}<\frac{\binom{n-r}{n_{1}-k}}{\binom {n}{n_{1}}}<{n_{1}}^k(n-n_{1})^{r-k}(n-r)^{-r}$
Hello,
You have to multiply the middle term $q^k$ by $n^k$ and $n^{r-k}$

5. ## Re: Combinatorial inequality

Originally Posted by Vinod
Hello,
You have to multiply the middle term $q^k$ by $n^k$ and $n^{r-k}$
Hello,you have to multiply last term by $n^r$

6. ## Re: Combinatorial inequality

Originally Posted by Vinod
Hello,you have to multiply last term by $n^r$
can you clarify a bit. I tried doing the manipulations but I am getting stuck in the middle, for both inequalities.

7. ## Re: Combinatorial inequality

Originally Posted by earthboy
can you clarify a bit. I tried doing the manipulations but I am getting stuck in the middle, for both inequalities.
Hello,

This is a limit theorem for the hypergeometric distribution. If n is large and $\frac{n_1}{n}=p$, then the probability $q_k$ is close to, more precisely,

$\binom{r}{k}(p-\frac{k}{n})(q-\frac{r-k}{n})^{r-k}\leq q_k\leq \binom{r}{k} p^k q^{r-k}(1-\frac{r}{n})^{-r}$ where

$q_k=\frac{\binom{r}{k}(\binom{n-r}{r-k}}{\binom{n}{r}}$

$q_k$ is most probably close to $\binom{r}{k}p^k q^{r-k}$