Product of Two Draws: With vs Without Replacement

Expectation · Medium · Free problem

From the numbers

, 2, \dots, 10$ you draw two numbers and multiply them. Compare two sampling schemes: (a) draw the two numbers WITHOUT replacement (two distinct numbers), versus (b) draw WITH replacement (the two draws are independent and could be equal). Which scheme gives the larger expected product? Answer intuitively and explain why, ideally without heavy computation.

Hints

  1. Express $E[XY]$ using the sum and sum of squares of
    $ through
    0$.
  2. With replacement the draws are independent: $E[XY] = \mu^2$. Without replacement they are negatively correlated.
  3. Without replacement, $E[XY] = (S^2 - Q)/(n(n-1))$, which is less than $\mu^2$ because it removes the squared (diagonal) terms.

Worked Solution

How to Think About It: Write the expected product in terms of the sum and the sum of squares. Let the draws be $X$ and $Y$. With replacement, $X$ and $Y$ are independent with the same mean $\mu$, so $E[XY] = \mu^2$. Without replacement, the two draws are negatively correlated (picking a high number first leaves a slightly lower-mean pool), which drags the expected product DOWN.

The Clean Argument: Let $S = 1 + \cdots + 10 = 55$ and $Q = 1^2 + \cdots + 10^2 = 385$, with $\mu = 5.5$.

  • With replacement: $E[XY] = E[X]E[Y] = \mu^2 = 5.5^2 = 30.25$.
  • Without replacement: by symmetry $E[XY] = \dfrac{(\sum_i i)^2 - \sum_i i^2}{10 \cdot 9} = \dfrac{S^2 - Q}{90} = \dfrac{55^2 - 385}{90} = \dfrac{3025 - 385}{90} = \dfrac{2640}{90} \approx 29.33.$

Why: With replacement is larger. Intuitively, drawing without replacement forbids the diagonal pairs $(i,i)$ -- in particular it removes the large product

0 \times 10 = 100$ and other big squares -- and introduces negative dependence, both of which lower the expected product.

Answer: WITH replacement gives the larger expected product ($30.25$ versus $\approx 29.33$). The intuition: without-replacement draws are negatively correlated and exclude the high squared terms, so the product's expectation is smaller.

Intuition

The key conceptual point is the sign of the dependence: sampling without replacement induces negative correlation between draws, which lowers the expectation of their product relative to the independent case. You can see it without arithmetic -- with replacement allows the big square terms like

0^2$, while without replacement forbids drawing the same number twice, stripping out exactly those large diagonal contributions. This negative-dependence-from-finite-sampling effect matters in finite-population statistics, sampling-based estimators, and any setting where draws compete for a fixed pool.

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