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Say, we know that the probability of an object having some property equals exactly $P$. We are given a sample (of size $N$) of these objects - in fact, that is a Binomial distribution with probability of success $P$ and size $N$. Say, in this sample we have $K$ objects with this property.

Now we take a subsample (with size $n$) of this sample and we want to estimate $P$ - overall proportion of the objects with this property. I want to find the expectation and variance of $p = \frac{k'}{n}$, where $k'$ is amount of objects with this property among $n$ objects.

Actually, I was able to prove that $\mathbb{E}p = P$, but I completely struggle to find $Var(p)$, can somebody help me?

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  • $\begingroup$ Is your subsample with our without replacement? Why does the sampling procedure in the first paragraph play any role in the question, which seems to assume only that the population from which your sample of size $n$ is taken has $K$ successes and $N-K$ non-successes? Do you intend the expectation and variance of $p$ to be taken over the distribution of both samples? $\endgroup$
    – whuber ♦
    Commented Nov 2, 2019 at 19:27
  • $\begingroup$ @whuber As for second question, I actually made a mistake in description (edited now) - we want to estimate $P$. And yes, expectation and variance should be taken over both distributions here. Subsample is without replacement. $\endgroup$
    – Elijah
    Commented Nov 2, 2019 at 19:44
  • $\begingroup$ In what way do you perceive the ultimate $n$ observations as not being fully equivalent to an iid sample of the original binomial distribution? $\endgroup$
    – whuber ♦
    Commented Nov 2, 2019 at 19:47
  • $\begingroup$ The probability of success in the subsample is a random variable $\frac{K}{N}$, not $P$. Although $\mathbb{E}\frac{K}{N} = P$ $\endgroup$
    – Elijah
    Commented Nov 2, 2019 at 20:02

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This subsample is identical to taking an iid sample of $n$ observations from a Bernoulli$(P)$ distribution, whence $k^\prime$ has a Binomial$(n,P)$ distribution.

Let's prove this rigorously. To do so, let the sequence $X_1, X_2, \ldots, X_N$ of iid random variables, with common (discrete) distribution $F,$ model the original sample of $N$ values. Let $A = (a_j),\,j=1,2,\ldots,n$ be any $n$-element sequence of the indexes $1,\ldots, N.$ Then a fortiori the sequence $X_{a_1}, X_{a_2}, \ldots, X_{a_n}$ is a sequence of iid random variables with common distribution $F,$ whence its sum

$$S_A(X) = \sum_{j=1}^n X_{a_j}$$

is distributed according to $F^{*n} = F * F * \cdots * F$ (the sum of $n$ iid values from $F$).

For any $k$ and $A$ denote the chance that $S_A(X)=k$ by $f(k).$ The point, as emphasized by this notation, is that $f(k)$ does not depend on $A.$

Now let $\mathbb{P}$ be any probability distribution on $\mathfrak{S}(N, n),$ the collection of length-$n$ permutations of the set $\{1,2,\ldots, N\}$. Select $A$ (an ordered subsample) according to $\mathbb{P}.$ For any possible value $k$ of the sum, the event that the sum equals $k$ decomposes into the disjoint union over all $A\in \mathfrak{S}(N, n),$ so its chance is the sum of the chances associated with each $A:$

$${\Pr}_{X,A}(S_A(X) = k) = \sum_{A\in \mathfrak{S}(N, n)} \mathbb{P}(A) {\Pr}_X(S_A(X)=k\mid A) = \sum_{A\in \mathfrak{S}(N, n)} \mathbb{P}(A) f(k) = f(k)$$

because (also axiomatically) $\sum_{A\in \mathfrak{S}(N, n)} \mathbb{P}(A) = 1.$

This has demonstrated that you don't even have to subsample randomly: you can subsample using any probability distribution you want over the set of possible subsamples, even to the point of not selecting the subsample randomly at all. The distribution of the sum is still $F^{*n}.$

When the $X_i$ take values in $\{0,1\}$ to indicate the absence or presence of a property, respectively, then $F$ is the Bernoulli$(P)$ distribution and therefore $F^{*n}$ is the Binomial$(n,P)$ distribution, QED.

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