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So the description I read is:

It is important to note that the formulas for calculating the variance and the standard deviation differ depending on whether you are working with a distribution of scores taken from a sample or from a population. The reason these two formulas are different is quite complex and requires more space than allowed in a short book like this. I provide an overly brief explanation here and then encourage you to find a more thorough explanation in a traditional statistics textbook. Briefly, when we do not know the population mean, we must use the sample mean as an estimate. But the sample mean will probably differ from the population mean. Whenever we use a number other than the actual mean to calculate the variance, we will end up with a larger variance, and therefore a larger standard deviation, than if we had used the actual mean. This will be true regardless of whether the number we use in our formula is smaller or larger than our actual mean. Because the sample mean usually differs from the population mean, the variance and standard deviation that we calculate using the sample mean will probably be smaller than it would have been had we used the population mean. Therefore, when we use the sample mean to generate an estimate of the population variance or standard deviation, we will actually underestimate the size of the true variance in the population because if we had used the population mean in place of the sample mean, we would have created a larger sum of squared deviations, and a larger variance and standard deviation. To adjust for this underestimation, we use n - 1 in the denominator of our sample formulas. Smaller denominators produce larger overall variance and standard deviation statistics, which will be more accurate estimates of the population parameters.

I understood none of this as it seems contradictory. It says that when using a number other than the actual mean (to me, e.g. the sample mean) will be different then the population mean causing a larger variance. In the middle it says the variance and sample mean will be smaller if the population mean had been used... Can you please explain which it is exactly and why so?

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  • $\begingroup$ Just so you know, regarding the formatting, it looks like you used the code format instead of the blockquote, which is why it looked funny. Is this copied verbatim? In particular, there were no paragraph breaks in all of that? :) $\endgroup$ – cardinal Apr 10 '12 at 0:19
  • $\begingroup$ @cardinal yup, I used code because when I tried to use blockquote I somehow got: > text text text text > text text text text text ` > text text text text >` although I regret the formatting above as well, somehow the >'s were getting pushed into the previous line. Yup! copied so I cannot talk about the formatting. Thanks, the new one looks great :D $\endgroup$ – Eiyrioü von Kauyf Apr 10 '12 at 0:24
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    $\begingroup$ Related: stats.stackexchange.com/questions/11421/… $\endgroup$ – Henry Apr 10 '12 at 0:46
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This seems just like sloppy choice of words. The sentence

Whenever we use a number other than the actual mean to calculate the variance, we will end up with a larger variance

seems to mean the term 'actual mean' to refer to the sample mean. I think this is where your confusion comes from. This is evidenced further from the sentence

the variance and standard deviation that we calculate using the sample mean will probably be smaller than it would have been had we used the population mean.

He apparently uses this intuitive motivation to show us why the sample variance is biased down, when he says

Therefore, when we use the sample mean to generate an estimate of the population variance or standard deviation, we will actually underestimate the size of the true variance

To see why this is true, define a function

$$ f(c) = \frac{1}{n} \sum_{i=1}^{n} (x_{i} - c)^2 $$

Using basic calculus, the minimizer of $f$ satisfies $f'(c) = 0$, which is equivalent to

$$ \sum_{i=1}^{n} (x_i - c) = 0 $$

therefore the minimizer is

$$ c^{\star} = \frac{1}{n} \sum_{i=1}^{n} x_{i}, $$

the arithmetic mean of the $x_{i}$'s. I'll leave it to you to check that this is a minimum and not a maximum.

In the case of data, $X_{1}, ..., X_{n}$ with sample mean $\overline{X}$, $f(\overline{X})$ is exactly the sample variance. Given what we've said above

$$ f(\overline{X}) \leq f(z) $$

for any other $z$, which includes the case where $z = \mu$, the population mean. This is why the sample variance is always less than the mean squared difference from the population mean (except in the case where $\overline{X} = \mu$, which occurs with probability zero when the $X_i$'s come from a continuous distribution).

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  • $\begingroup$ what is the minimizer ? I assume it's the value that makes the things in the sigma smaller. Why is it (c* I assume?) equal to the 1/n ($\Sigma$ $x_i$) above? $\endgroup$ – Eiyrioü von Kauyf Apr 10 '12 at 1:01
  • $\begingroup$ The minimizer is the value that $c^{\star}$ such that $f(c^{\star}) < f(z)$ for any other $z$. Since this is a monotonically increasing function as you move away from $c^{\star}$, there is only one minimum. I've shown $\frac{1}{n} \sum x_{i}$ is the minimizer by using the basic calculus fact that the value that satisfies $f'(z) = 0$ is a local extreme of the function $f(z)$ (here $f'(z)$ denotes the derivative of $f(z)$ with respect to $z$). $\endgroup$ – Macro Apr 10 '12 at 1:06
  • $\begingroup$ with regard to your first statement that I believe is where the confusion starts. By actual mean I believed him to be referring to the population mean $\endgroup$ – Eiyrioü von Kauyf Apr 10 '12 at 1:14
  • $\begingroup$ Based on the rest of the passage, I think the author was clearly referring to the sample mean $\endgroup$ – Macro Apr 10 '12 at 2:04

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