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My notes say the following:

Theorem

If $X \sim HGeom(w, b, n)$ and $N = w + b \to \infty$ such that $p = \dfrac{w}{w + b}$ remains fixed, then the PMF of $X$ converges to the $Bin(n, p)$ PMF.

$HGeom(\cdot)$ is the hypergeometric distribution.

What I'm confused about here is how $p = \dfrac{w}{w + b}$ remains fixed if $N = w + b \to \infty$?

I would greatly appreciate it if people could please take the time to clarify this.

Community
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The Pointer
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2 Answers2

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Let's just solve $$p = \frac{w}{w+b}$$ for $b:$ $$b = w\cdot \frac{1-p}{p}$$ so $p$ being constant while $w+b \to \infty$ means we want them to be related as $b = w\cdot \frac{1-p}{p}$

If that seems circular to you, we can remove $p$ entirely by noting that $\frac{1-p}{p}$ is simply a non-negative real number, so we simply want $b$ and $w$ to vary directly (i.e. one is a positive real multiple of the other) as both values go to infinity.

Brian Moehring
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Let me try to give you the statistical/probabilistic interpretation of this fact.

Notation:

$w$ is the number of success states in the population

$N$ is the population size and $N = w + b$

$n$ is the number of draws (i.e. quantity drawn in each trial)

$p = \frac{w}{w + b} = \frac{w}{N}$

$N = w + b \to \infty$

Background:

Suppose there is a population of size $N$ with $w$ units labelled as "success" and $N-w = b$ labelled as "failure". A sample of size $n$ is drawn without replacement. The random variable $X$ is defined as the number of "successes" in the sample. Then $X \sim HGeom(w, b, n)$

Now consider the following situation:

We still have the population of size $N$ with $w$ units labelled as "success" and $N-w = b$ labelled as "failure", but now we take a sample of size $n$ drawn with replacement. Then, with each draw, the units remaining to be drawn look the same: still $w$ "successes" and $N-w = b$ "failures". Thus, the probability of drawing a "success" on each single draw is $p = \frac{w}{N}$ and this doesn't change.

When we were drawing without replacement, the proportions of successes would change, depending on the result of previous draws. For example, if we were to obtain a "success" on the first draw, then the proportion of "successes" for the second draw would be $\frac{w-1}{N-1}$, whereas if we were to obtain a "failure" on the first draw the proportion of successes for the second draw would be $\frac{w}{N-1}$

Define the random variable $Y$ as the number of "successes" in the sample, when we are drawing with replacement. Then $Y$ is a binomial random variable:

$Y \sim Bin(n,p) $

Now think about the following proposition given in your book:

If the population size $N \rightarrow \infty$ in such a way that the proportion of successes $\frac{w}{N} \rightarrow p$ and $n$ is held constant, then the hypergeometric probability mass function approaches the binomial probability mass function.

Can you see the connection now? If not, let me know. I'll try to explain further.

PTDS
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