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Let $V=\{v_1,\ldots,v_n\}$ be a set of $n$ elements and $N\sim Binomial(k, p)$ with $k<n$.

We denote by $S_{N}:=\sum_{v\in V_N}v$ the r.v. of the sum of uniform samples $V_{N}\subset V$ of size $N$. To clarify, $V_N$ takes elements from $V$ with probability $p$ without replacement.

I am trying to calculate $\text{Var}\left[S_N\right]$

I've been trying to relate this variance to $\text{Var}\left[\bar{v}\right]$, as the variance of the average of samples with fixed size $k$ is the variance of the average of the population divided by k:

$$\text{Var}\left[S_k\right] = \frac{\text{Var}\left[\bar{v}\right]}{k}$$

I've tried with the Law of Total Variance, but it doesn't seem to close.

In addition, from Wald's equation, I've got that

$$\text{E}\left[S_N\right] = np\bar{v}$$

but $\text{E}\left[S_N^2\right]$ doesn't seem to close either.

This question is related to this other question but I'm interested in the variance instead of the expectation. In my question, however, $v_i$ are not random variables and there is no order in the sums.

synack
  • 974

1 Answers1

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I think I got it. If we model the problem thinking of the sum as the product of the values with $n$ i.i.d. $B_i\sim Bernoulli(p)$, we have:

$$\text{Var}[S_N] = \text{Var}\left[\sum_{i=1}^{n}v_iB_i\right]= \sum_{i=1}^n v_i^2\text{Var}[B_i]= p(1-p)\sum_{i=1}^n v_i^2$$

synack
  • 974