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Suppose $X$ is a continuous random variable for the weight of silver pennies. We then measure 338 pennies (in grams), leaving us with 338 observations.

The observed mean weight is 15.722 grams and the observed variance is 1.999 squared grams.

If we now take random and disjoint subsets of five observations from the 338 observed weights without replacement, what would happen to the mean and variance?

This is exercise 5.2 from Principles of Statistics by M.G. Bulmer. The answers are:

  1. New mean = $5 \times 15.722$
  2. New variance = $5 \times 1.999$

I understand why the new mean is five times the old one, for $\sum_{i=1}^{n}x_i$ does not change and there is one observation for every five observations in the original calculation. Thus, $\sum x_i$ is now divided by $\frac{338}{5}$, which is equivalent to $5 \bar{x}$.

However, why is the new variance five times the old one?

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Writing out the equations:

$$E(X_j) = \mu$$ $$V(X_j) = \sigma^2$$

$X_j$ are independent, identically distributed random variables.

Let $x_j$ be a sample from the Random Variable $X_j$. Let $y_i = \sum_j^n x_j$. The distribution on $x_j$ induces a distribution on $y_i$ such that we can treat $Y_i$ as a random variable.

$$E(Y_i) = E(\sum_j^n X_j) = \sum_j^n E(X_j) = \sum_j^n \mu = n\mu$$ $$V(Y_i) = V(\sum_j^n X_j) = \sum_j^n V(X_j) + \sum_{j \neq k} Cov(X_j, X_k)$$

Since $X_j$ and $X_k$ are independent, then $Cov(X_j, X_k) = 0$

$$V(Y_i) = \sum_j^n V(X_j) = \sum_j^n \sigma^2 = n \sigma^2$$

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  • $\begingroup$ Why is $V(Y) = V(\sum X_i)$ ? $\endgroup$
    – Arturo Sbr
    Feb 19 at 21:11
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    $\begingroup$ @Arturo It is simply substituting $Y=\sum X_i$ into the variance function. $\endgroup$
    – B.Liu
    Feb 19 at 22:37
  • $\begingroup$ @B.Liu This is exactly the part that confuses me. Is $Y_i = \sum X_i$? I am confused by this because $y_i$ is the disjoint sum of five different $x_i$'s. For instance, $y_1 = x_1 + x_2 + x_3 + x_4 + x_5$. $\endgroup$
    – Arturo Sbr
    Feb 19 at 23:11
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    $\begingroup$ @ArturoSbr If we substitute $n=5$ into the equations in the answer above, we have $Y=\sum_{i=1}^{5} X_n = X_1 + X_2 + X_3 + X_4 + X_5$, which is more or less the random variable version of what you defined in the question for the observations. Note the lack of subscript for $Y$ (as we don't really need them for the purpose of calculating the mean and variance). $\endgroup$
    – B.Liu
    Feb 20 at 2:10
  • $\begingroup$ @ArturoSbr There is a typo above - $\sum_{i=1}^5 X_n$ should be $\sum_{i=1}^5 X_i$. $\endgroup$
    – B.Liu
    Feb 20 at 3:27

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