Show that for a random sample of size $n$ from the distribution $f(x)=e^{-(x-\theta)} , x>\theta$ , $2n[X_{(1)}-\theta] \sim \chi^2_{2}$ distribution and $2\sum_{i=2}^{n}[X_{(i)}-X_{(1)}]$ also has the $\chi^2_{2n-2}$ distribution and is independent of the first statistic. Here, $X_{(i)}$ is defined as the $i$ th order statistic.

My approach:

I did the following series of transformations: $(X_1,X_2,..,X_n) \rightarrow (Y_1,Y_2,...,Y_n) \rightarrow (Y_{(1)},Y_{(2)},...,Y_{(n)}) \rightarrow (U_1,U_2,...U_n)$

where $Y_i=X_i-\theta$ , $U_1=2nY_{(1)}$ and $U_{i}=2(Y_{(i)}-Y_{(1)}) \ \text{for i =2,3,...n}$

SO, first the joint pdf of $X_1,X_2,...X_n$ is given by

$f(x_1,x_2,...x_n)=e^{-\sum_{i=1}^{n}(x_i-\theta)} I_{x_i > \theta}$

Again, you can see $f(y_1,y_2,..,y_n)=e^{-\sum y_i} I_{y_i>0}$ Now, the joint pdf of order statistics $f_{1,2,...n}(y_1,..y_n)=n!e^{-\sum y_i} I_{y_1<y_2<...<y_n}$ Now transforming to $U$, the jacobian of transformation comes to be $\frac{1}{n2^n}$ Thus, $f(u_1,u_2,..u_n)=\frac{(n-1)!}{2^n}e^{\frac{-\sum u_i}{2}}$ From here I can deduce $u_1 \sim \chi^2_{2}$ But I cannot deduce anything from the remaining. Help!

  • $\begingroup$ You know the PDF of $(U_2,\ldots,U_n)$ and you are after the PDF of $V=U_2+\cdots+U_n$, right? There are entirely standard techniques to do that, did you try one? $\endgroup$ – Did May 3 '18 at 6:25
  • $\begingroup$ But that is not working..Did I make any mistakes in the above calculations? Because one $(n-1)!$ is coming in and I don't know how to handle that $\endgroup$ – Legend Killer May 3 '18 at 6:31
  • $\begingroup$ Sorry but what is not working? Once again, if your computations are correct, you know that the PDF of $(U_1,U_2,\ldots,U_n)$ is $$f(u_1,u_2,\ldots,u_n)=2^{-n}(n-1)!\exp(-\tfrac12(u_1+u_2+\cdots+u_n))\mathbf 1_{u_1>0}\mathbf 1_{0<u_2<u_3<\cdots<u_n}$$ and you are after the PDF of $$V=U_2+\cdots+U_n$$ Please describe precisely that which is stopping you. $\endgroup$ – Did May 3 '18 at 6:57
  • $\begingroup$ Your joint pdf of $(X_1,\cdots,X_n)$ is not correct. $\endgroup$ – StubbornAtom May 3 '18 at 7:06
  • $\begingroup$ I am stuck because $\sum_{i=2}^{n} u_i$ is supposed to follow chi-squared $2n-2$ which is not coming by my calculations $\endgroup$ – Legend Killer May 3 '18 at 7:23

If you transform $(X_1,\cdots,X_n)\to(Y_1,\cdots,Y_n)$ such that $Y_i=X_i-\theta$,

then the $Y_i$'s would be i.i.d $\text{Exp}$ with mean $1$.

You have $Y_{(i)}=X_{(i)}-\theta$ for all $i$.

So, $\displaystyle\sum_{i=2}^n(X_{(i)}-X_{(1)})=\sum_{i=2}^n\left[(X_{(i)}-\theta)-(X_{(1)}-\theta)\right]=\sum_{i=2}^n\left(Y_{(i)}-Y_{(1)}\right)$

Note that $Y_i\stackrel{\text{i.i.d}}{\sim}\text{Exp}(1)$ for all $i\,\implies2Y_i\stackrel{\text{i.i.d}}{\sim}\chi^2_2$ for all $i$$\implies2nY_{(1)}\sim\chi^2_2$.

Also, $\displaystyle\sum_{i=1}^nY_i=\sum_{i=1}^nY_{(i)}=\sum_{i=1}^n(X_{(i)}-Y_{(1)})+nY_{(1)}=\sum_{i=2}^n(Y_{(i)}-Y_{(1)})+nY_{(1)}$

Finally note that $\displaystyle\sum_{i=1}^n2Y_i=\sum_{i=2}^n2(Y_{(i)}-Y_{(1)})+2nY_{(1)}$

The sum $\displaystyle\sum_{i=1}^n2Y_i\sim\chi^2_{2n}$, while $2nY_{(1)}\sim\chi^2_2$, and

$\sum_{i=2}^n2(Y_{(i)}-Y_{(1)})$ is distributed independently of $2nY_{(1)}$.

This implies that $\displaystyle\sum_{i=2}^n2(Y_{(i)}-Y_{(1)})\sim\chi^2_{2n-2}$, by the reproductive property of $\chi^2$ distribution.

For a complete proof, you have to fill in the details.

  • $\begingroup$ Yes, I know this..Can you just tell me where am I going wrong in my approach? $\endgroup$ – Legend Killer May 3 '18 at 7:47
  • $\begingroup$ @LegendKiller I would ask you to use the transformation $U_1=Y_{(1)}$ and $U_i=Y_{(i)}-Y_{(1)}$ for $i=2,3,\cdots,n$. $\endgroup$ – StubbornAtom May 3 '18 at 7:54
  • $\begingroup$ In that case , I am getting $f(u_1,u_2,..,u_n)=n!e^{-nu_1-\sum_{2}^{n}u_i}$ , how can I proceed from here? $\endgroup$ – Legend Killer May 3 '18 at 8:20
  • $\begingroup$ @LegendKiller That means $U_1$ is independent of $(U_2,\cdots,U_n)$ with the pdf of $U_1$ being $f_{U_1}(t)=ne^{-nt}\mathbf1_{t>0}$. Using a single transformation it is seen that $2nU_1=2nY_{(1)}=2n(X_{(1)}-\theta)\sim\chi^2_2$. We also have $\displaystyle\sum_{i=2}^n2Y_i=\sum_{i=2}^n2U_i+2nU_1$. Due to reasons mentioned earlier, $$\sum_{i=2}^n2U_i=2\sum_{i=2}^n(Y_{(i)}-Y_{(1)})=2\sum_{i=2}^n(X_{(i)}-X_{(1)})\sim\chi^2_{2n-2}$$ $\endgroup$ – StubbornAtom May 3 '18 at 10:08
  • $\begingroup$ But you worked this step beforehand, I am telling that from the joint distribution of $(u_2,...u_n)$ , how can you show that $2\sum_{i=2}^{n} U_i$ follows chi square $2n-2$? $\endgroup$ – Legend Killer May 3 '18 at 10:14

I think a more easy to follow (and simpler) proof would be to use a different change of variables.

We have the joint density of the order statistics $(U_1=X_{(1)},\cdots,U_n=X_{(n)})$

$$f_{\mathbf U}(u_1,\cdots,u_n)=n!\exp\left[-\sum_{i=1}^nu_i+n\theta\right]\mathbf1_{\theta<u_1<u_2<\cdots<u_n}$$

Now transform $(U_1,\cdots,U_n)\to(Y_1,\cdots,Y_n)$ such that $Y_i=(n-i+1)(U_i-U_{i-1})$ for all $i=1,2\cdots,n$ and taking $U_0=\theta$.

It follows that $\sum_{i=1}^nu_i=\sum_{i=1}^ny_i+n\theta$. The jacobian determinant comes out as $n!$.

So you get the joint density of $(Y_1,\cdots,Y_n)$

$$f_{\mathbf Y}(y_1,\cdots,y_n)=\exp\left[-\sum_{i=1}^ny_i\right]\mathbf1_{y_1,\cdots,y_n>0}$$

Not surprisingly, the spacings of successive order statistics from an exponential sample come out as independent . In fact, the $Y_i$'s are i.i.d exponential with mean $1$ for all $i=1,2,\cdots,n$.

This implies $2Y_i\stackrel{\text{i.i.d}}{\sim}\chi^2_2$ for all $i=1,2,\cdots,n$

So we have two independent variables $2Y_1$ and $\sum_{i=2}^n2Y_i$. Both have the chi-square distribution --- the former with $2$ degrees of freedom and the latter with $2n-2$ degrees of freedom.

It is now a matter of time to see that $2Y_1=2n(X_{(1)}-\theta)$ and $2\sum_{i=2}^nY_i=2\sum_{i=2}^n(X_{(i)}-X_{(1)})$.


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