Probable direction of deviations from the expected value in binomial and hypergeometric cases

Question

Suppose I have an urn with N marbles, with frequencies p and q for red and black marbles, and with p > 0,5. I take a sample of r marbles.

It sounds intuitive to say that deviations from the mean should be expected to occur in the direction of the more frequent marble (say for example, if I have 90 red and 10 black, then with r = 10, 10 red should be more likely than 8 red, 2 black). That is, that $Prob(\hat{p} > p) > Prob(\hat{p} < p)$. Or equivalently, $Prob(\hat{p} \ge p) > Prob(\hat{p} \le p)$.

But that is not always the case. Take p = 0,51 and r = 2, for example. What's going on here is that the expected value is not a possible sample value, but it is close to it, so $\hat{p}=0,5$ will count as a case where $\hat{p} < p$. So let's get rid of this sort of cases by taking as an additional premise that $p .r \in \mathbb{Z}$.

¿Will the result hold in this case?

What would be needed to answer positively is that

• For sampling with replacement: $$\sum_{k=p.r}^{r} \binom{r}{k} p^k (1-p)^{r-k} > \sum_{k=0}^{p.r} \binom{r}{k} p^k (1-p)^{r-k}$$

• For sampling without replacement: $$\frac{\binom{p.N}{p.r} \binom{N - p.r}{r - p.r}}{N \choose r} > \frac{\binom{(1-p).N}{(1-p).r} \binom{N - (1-p).r}{r - (1-p).r}}{N \choose r}$$ (could also use a cumulative hypergeometric distribution but this seems easier)

Answer 1

Here's a positive answer for the binomial distribution in the case $p=1-\frac1{n}$. (I suppose you could check some other cases like $p=1-\frac2n$ in a similar way.)

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We need to prove $$\binom{n}{n}p^n > \sum_{k=0}^{n-2}\binom{n}k p^k(1-p)^{n-k},$$ $$\left(1-\frac1n\right)^n > \sum_{k=0}^{n-2}\binom{n}k \left(1-\frac1n\right)^k\left(\frac1n\right)^{n-k},$$ $$\left(n-1\right)^n > \sum_{k=0}^{n-2}\binom{n}k \left(n-1\right)^k.$$ The last right-hand side is actually, per Wolfram Alpha $$ \sum_{k=0}^{n-2} (n-1)^k \binom{n}{k} = n^n-n (n-1)^{n-1}-(n-1)^n$$ so we need $$n^n<2(n-1)^n+n(n-1)^{n-1}$$ $$\left(1+\frac1{n-1}\right)^n<2+\frac{n}{n-1} = 3 + \frac1{n-1}$$ or, with $m=n-1$, $$\left(1+\frac1m\right)^m < \frac{3+\frac1m}{1+\frac1m} = \frac{3m+1}{m+1} = 3-\frac2{m+1}=3-\frac2n$$ This is true: the left-hand side is bounded above by $e$, and the right-hand side surpasses $e$ by $m=7$, and it can also be checked for $m\in\{1,\dots,6\}$.

Answer 2

It's not true for sampling without replacement. Consider e.g. $N=9$, $p = 2/3$, $r = 6$. Note that the possible numbers of red balls in the sample are $3,4,5,6$, with $4$ corresponding to $\widehat{p} = p$. Since $P(3) = P(5) + 2 P(6)$, we have $P(\widehat{p} > p) = P(5) + P(6) < P(3)$.