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I've encountered a problem that requires calculating the distribution of $$ X_1 - \overline{X} $$ where $X_i, i = 1, \cdots, n $ are samples from a normal population $N(\alpha,\sigma^2)$, and $\overline{X}$ being the sample mean.

I thought that $X_1 - \overline{X}$ should follow $N(0, \frac{n+1}{n}\sigma^2)$, but then I realised that $X_1$ and $\overline{X}$ are not independent, so the above induction is wrong.

But I don't know how to proceed from here. Any hints would be welcome!

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You can write $$X_1-\bar{X}=X_1-\frac{X_1+\dots+X_n}{n}=\left(1-\frac{1}{n}\right)X_1-\frac{X_2+\dots+X_n}{n}=\frac{n-1}{n}X_1-\frac{n-1}{n}\frac{X_2+\dots+X_n}{n-1}.$$

The first term, $\frac{n-1}{n}X_1$, is a constant times $X_1$, so it is a normal with mean $\frac{n-1}{n}\alpha$ and variance $\left(\frac{n-1}{n}\right)^2\sigma^2$.

The second term, $\frac{n-1}{n}\frac{X_2+\dots+X_n}{n-1}$, is a multiple of the sample mean of $X_2,\dots,X_n$. The sample mean of a sample of size $n-1$ has mean $\alpha$ and variance $\frac{\sigma^2}{n-1}$, so this random variable $\frac{n-1}{n}\frac{X_2+\dots+X_n}{n-1}$ has mean $\frac{n-1}{n}\alpha$ and variance $$\left(\frac{n-1}{n}\right)^2\frac{\sigma^2}{n-1}=\frac{n-1}{n^2}\sigma^2.$$

Now, the two terms, $\frac{n-1}{n}X_1$ and $\frac{n-1}{n}\frac{X_2+\dots+X_n}{n-1}$, are independent (the second does not contain $X_1$ in the sum), so their sum is also a normal random variable, with mean $$\frac{n-1}{n}\alpha-\frac{n-1}{n}\alpha=0$$ and variance $$\left(\frac{n-1}{n}\right)^2\sigma^2+\frac{n-1}{n^2}\sigma^2=\frac{n^2-n}{n^2}\sigma^2=\frac{n-1}{n}\sigma^2.$$

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  • $\begingroup$ Thanks! But I guess the variance is $\frac{n}{n-1}$, is that right? $\endgroup$ – R. Feng Mar 19 '17 at 9:22
  • $\begingroup$ The variance of what? $\endgroup$ – Anna SdTC Mar 19 '17 at 9:25
  • $\begingroup$ the final line of your solution $\endgroup$ – R. Feng Mar 19 '17 at 9:26
  • $\begingroup$ The variance of the sum of independent random variables is the sum of variances, and the two variances are what I computed in the previous two paragraphs. Is there a calculation mistake? $\endgroup$ – Anna SdTC Mar 19 '17 at 9:30
  • $\begingroup$ I think so, since $(n-1)^2+n-1=n^2-n$. $\endgroup$ – R. Feng Mar 19 '17 at 9:35

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