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Cartucho
Cartucho
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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 9990 red and 110 black, then with r = 10, 10 red should be more likely than 98 red, 12 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)

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 99 red and 1 black, then with r = 10, 10 red should be more likely than 9 red, 1 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)

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)

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Cartucho
Cartucho
  • 41
  • 4

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

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 99 red and 1 black, then with r = 10, 10 red should be more likely than 9 red, 1 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)