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Background

I have two estimates of variance and their associated standard errors calculated from sample sizes of $n=500$ and $n=10,000$ the results are $\hat{\sigma^2} (sd_{\hat{\sigma^2}})$: $$\hat{\sigma^2}_{n=500}=69 (6.4)$$

$$\hat{\sigma^2}_{n=10,000}=72 (1.5)$$

Question

If I say that variance increased by 3, what is the standard deviation around this estimate?

Notes

  1. $SD$ of var calculated using $SD_{\hat{\sigma^2}}=\sqrt{s^4(2/(n-1) + k/n)}$

  2. I suspect that the fact that I am estimating the sd of a variance is not relevant to the calculation, but may help in the interpretation of what I am doing.

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Could you clarify what you mean by "increased by 3?" Some issues: (1) "increased by 3" could mean "+3" or "*3"; (2) are you referring to the underlying variance $\sigma^2$ or its estimate $\hat{\sigma}^2$; (3) what could this increase possibly mean in either case? Specifically, if $\sigma^2$ changes, how do you suppose the data would change (and via them, the estimated variance)? If the estimate of $\sigma^2$ changes, how would that occur without changing the data altogether? – whuber♦ Mar 1 '11 at 16:52
@whuber: I think he simply meant that his estimates of the variances went from $69$ for the initial sample to $72$ for the large sample and $72-69=3$. – Henry Mar 1 '11 at 17:03
@Henry that is correct – David Mar 1 '11 at 17:44
@Henry Thanks; this should have been obvious but I missed it. – whuber♦ Mar 1 '11 at 17:49
@whuber if I understand your question: the estimate of $\sigma^2$ should change because the number of samples increases (thus the data are actually different), and I am assuming that the reason that the variance would increase is that there is a greater chance of sampling values in the tails of the distribution. – David Mar 1 '11 at 17:54
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1 Answer

up vote 5 down vote accepted

The standard deviation of the difference between two independent random variables is the square root of the sum of the squares of their individual standard deviations (easier to express as variances) so in this case $$\sqrt{6.4^2 + 1.5^2} \approx 6.6$$

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Would it not be necessary to weight the SDs given the disparity in group sizes? – pmgjones Mar 1 '11 at 17:26
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@propofol: I think you would weight them if you concluded that the two samples came from the same distribution and wanted to derive a new estimate of the standard deviation for the combined sample; more importantly, you would also need to take account of the difference in the estimates. But I was assuming that at this stage David was interested in the question of whether the estimates from the two samples were significantly different - on these numbers there is insufficient evidence to conclude there has been a change. – Henry Mar 1 '11 at 17:34
thanks for your answer; I am familiar with this for the case of adding two IRV's but did not realize that it was the case for calculating the difference as well. I guess it makes sense that it does. – David Mar 1 '11 at 17:52
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@David The difference of two RVs is the sum of one and the negative of the other. The variance of the negative of an RV equals the variance of the RV itself. – whuber♦ Mar 1 '11 at 18:01
@whuber uggh. I should have recognized that. Thanks for the help. – David Mar 2 '11 at 6:32

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