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Let $X_1, X_2, ... , X_n$ i.i.d random variables with probability density function $N(\theta, \theta^2)$. Show that

$$T(X) = \frac{X_{(1)}-X_{(n)}}{X_{(2)}-X_{(n)}}$$

is ancillary to $ \theta$.

My attempt: Since $\frac{X_i}{\theta} \sim N(1,1)$, which does not depends on theta, we have that if $$Y_i=\frac{X_i}{\theta}$$ the $ Y_i $ are iid $N(1,1)$ and, for example, $$Y_{(n)} = 1/\theta X_{(N)}$$ The same occurs to the other order statistics. Then, $$\frac{X_{(1)}-X_{(n)}}{X_{(2)}-X_{(n)}} =\frac{X_{(1)}/\theta-X_{(n)}/\theta}{X_{(2)}/\theta-X_{(n)}/\theta}$$ Does not depends on $\theta$.

Is that correct?

Thanks!

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Let's make sure that, if we define $Y_i = X_i/\theta$, then it follows that $Y_{(i)} = X_{(i)}/\theta$.

Working the simplest case, the density of $X_{(n)}$ is $$f_{X_{(n)}}(x_{(n)}) = nf_X(x_{(n)})[F_X(x_{(n)})]^{n-1} \tag{1}$$

Defining $Z\equiv X_{(n)}/\theta$, by applying the change of variable formula we obtain

$$f_Z(z) = \theta nf_X(\theta z)[F_X(\theta z)]^{n-1} \tag{2}$$

Let's now consider the densities related to $Y = X/\theta$. We have

$$f_Y(y) = \theta f_X(\theta y) \tag {3}$$

which indeed is a $N(1,1)$. Also,

$$F_Y(y) = F_X(\theta y) \tag{4}$$

(carefully note that the $\theta$ "outside" the density in $(3)$ has "disappeared" in $(4)$).

The density of the maximum-order statistic of $Y$, $Y_{(n)}$, is

$$f_{Y_{(n)}}(y_{(n)}) = nf_Y(y_{(n)})[F_Y(y_{(n)})]^{n-1}$$ and using $(3)$ and $(4)$ we get

$$f_{Y_{(n)}}(y_{(n)}) = \theta nf_X(\theta y_{(n)})[F_X(\theta y_{(n)})]^{n-1} \tag{5}$$

Comparing $(2)$ and $(5)$, we see that they are identical, since they have exactly the same functional form, and the variables in both cases enter in exactly the same way (and they also have the same support). Now, the fact that $Z \equiv X_{(n)}/\theta$ and $Y_{(n)}$ have the same distribution, does not, in principle and in general, guarantee that they are the same random variables, only that they are identically distributed (and analogous conclusions hold for the other order statistics involved). But this is all we need in order to prove what we need to prove (since the appearance of $\theta$ is misleading -it cancels out everywhere).

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  • $\begingroup$ I really appreciatte your work to improve my answer and correct my mistakes! Thank you very much $\endgroup$ – Giiovanna Aug 17 '14 at 16:02
  • $\begingroup$ I appreciate the kind words, Giiovanna. $\endgroup$ – Alecos Papadopoulos Aug 17 '14 at 16:08

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