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We take two samples from $Z$ of size $n_1$ and $n_2$ and take the difference of the mean of these samples.

Both should have the same expected value, so the mean is zero.

But what is the variance of this difference?

hardmath
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snoram
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    What is $Z$? Is there independence within the samples? Across the samples? And what have you tried? – Stefan Hansen Feb 16 '14 at 11:53
  • The samples do not intersect. – snoram Feb 16 '14 at 11:59
  • I thought the variance would be simply the sum of the two sample variances: var(z)/n1^0.5+var(z)/n2^0.5. But it doesn't feel intuitive that the variance of the difference is bigger than of the individual sample. – snoram Feb 16 '14 at 12:01
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    Let me try again: What is $Z$? Is there independence within the samples? Across the samples? And how do you take a difference between two samples? Do you mean the difference of the two averages? Please be MUCH more clear. – Stefan Hansen Feb 16 '14 at 12:03
  • Sorry about the messy explanation. I mean the difference of the two sample means. I think independence in every way should ab assumed: within the samples and also across the samples. – snoram Feb 16 '14 at 12:07
  • For the third time, what is $Z$? – Stefan Hansen Feb 16 '14 at 12:43
  • It is not specified it just says that we have two samples from the same population. I think you can think of any population. – snoram Feb 16 '14 at 12:52
  • So your text just says "...two samples from Z" without specifying what Z is? In any case, without specifying from what distribution you sample from, you can't really say anything about the mean, variance or distribution of the difference of averages. – Stefan Hansen Feb 16 '14 at 13:00
  • Exactly, it just says "two sample from a population" with size n1 and n2 and means... I just called it Z. But doesn´t CLT say that the mean from any distribution is normally distributed provided the sample size is big enough? – snoram Feb 16 '14 at 13:04

2 Answers2

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Your answer in the comments is correct. The variance of the difference is the sum of the individual variances. Why shouldn't it be larger than either one? If you add two normal variables with unit variance, you get a variable with variance greater than one.

Ross Millikan
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Assume without loss of generality that the underlying distribution is centered and has variance $\sigma^2$, then one asks for the variance of $$ U=\sum_{k=1}^{n_1}\frac{Z_k}{n_1}+\sum_{k=1}^{n_2}\frac{-Z'_k}{n_2}, $$ where the random variables $Z_k$ and $Z'_k$ are independent and centered with variance $\sigma^2$. Thus $U$ is the sum of $n_1+n_2$ independent centered random variables, with variance either $\sigma^2/n_1^2$ for the $n_1$ first ones, or $\sigma^2/n_2^2$ for the $n_2$ last ones. Hence, $$ \mathrm{var}(U)=n_1\frac{\sigma^2}{n_1^2}+n_2\frac{\sigma^2}{n_2^2}=\left(\frac1{n_1}+\frac1{n_2}\right)\,\sigma^2. $$

Did
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