0
$\begingroup$

I am given a set of iid random variables $\{X_i\}\sim f(x)$ with cdf $F(X)$ I am asked to find the distribution of the following $D_i=X_{(i)}-X_{(i-1)}$ where $X_{(i)}$ is the $i$th order statistic and $i\ge2$. My approach is

$$P(D_i=z)=P(X_{(i)}-X_{(i-1)}=z)=P(X_{(i)}=X_{(i-1)}+z)=\int P(X_{(i)}=y+z|X_{(i-1)}=y)P(X_{(i-1)}=y)dy$$

I know $$P(X_{(i-1)}=y)=\frac{n!}{(i-2)!(n-i+1)!}f(y)F(y)^{i-2}(1-F(y))^{n-i+1}$$

I was trying to find this probability of $P(X_{(i)}=y+z|X_{(i-1)}=y)$ through counting. Let $X_{(i-1)}=y$ then my probability is (where I am using $X$ as a the distribution representative)

$$P(X_i=y+z, X_j=y \hspace{3mm}\text{for some} \hspace{3mm}i,j\in\{1...n\},X<y+z \hspace{2mm}\text{for} \hspace{2mm} i-2 \text{obs}, X>y+z \hspace{2mm}\text{for} \hspace{2mm} (n-i) \text{obs})$$

Since there are $\frac{n!}{(i-2)!(n-i)!}$ combinations that will result in the above scenario and each has a probability of $$f(y)f(y+z)F(y+z)^{i-2}(1-F(y+z))^{n-i}$$ then $$P(D_i=z)=\int\frac{n!}{(i-2)!(n-i)!}f(y)f(y+z)F(y+z)^{i-2}(1-F(y+z)^{n-i}\frac{n!}{(i-2)!(n-i+1)!}f(y)F(y)^{i-2}(1-F(y))^{n-i+1}dy$$

I am having trouble reducing this large integral. So I am guessing if my approach is even correct or if there is a simpler approach?

I am trying to find the joint distribution of $P(D_i=d_i,...,D_n=d_n)$ which basically reduces to $P(X_{(n)}-X_{(1)}=\sum_{i=2}^{n}d_i)$ that is why I need the above distribution.

$\endgroup$
0
$\begingroup$

Yes, indeed, you are close.


The basis for evaluating the probability density function of the event $\{X_{(i)}=y\}$ is that it's outcomes consist of some arrangement of $(i-1)$ samples that are below $y$, one sample that is equal to $y$, and $(n-1)$ samples that are above $y$.  Since we may safely ignore ties as being so to near impossible, there are $n!/(i-1)!(n-i)!$ ways to arrange outcomes that satisfy this.$$f_{\small X_{(i)}}(y)=\dfrac{n! F_{\small X}(y)^{i-1} f_{\small X}(y)(1-F_{\small X}(y))^{n-1}}{(i-1)!1!(n-i)!}$$


Likewise, for $i\in\{2..n\}$, the event of $\{X_{(i-1)}=x,X_{(i)}=x+z\}$ is the event that $i-2$ samples are below $x$, one sample exactly $x$, one sample exactly $x+z$, and the remaining $n-i$ samples are above $x+z$.   Ignoring the zero density posibilities of ties, there are $n!/(i-2)!(n-i)!$ ways to arrange samples that satisfy this criteria.

So then considering that the event of $\{D_i=z\}$ is $\bigcup_x\{X_{(i-1)}=x,X_{(i)}=x+z\}$, we integrate the density over all real values for $x$.

$$\begin{align}f_{\small D_i}(z)&=\int_\Bbb R f_{\small X_{(i-1)},X_{(i)}}(x,x+z)~\mathrm d x\\&=\dfrac{n!}{(i-2)!1!1!(n-i)!}\int_{\Bbb R} F_{\small X}(x)^{i-2}\, f_{\small X}(x)\, f_{\small X}(x+z)\, (1-F_{\small X}(x+z))^{n-i}\,\mathrm d x\end{align}$$

$\endgroup$
2
  • $\begingroup$ It appears that I am calculating $P(X_{(i)}=y+z|X_{(i-1)}=y)$ the same way you are calculating $P((X_{(i)}=y+z, X_{(i-1)}=y)$ On this ocassion are they the same? $\endgroup$ Jul 31 '20 at 14:43
  • $\begingroup$ I believe I made the mistake by calculating $P(X_{(i)}=y+z|X_{(i-1)=y})=P(X_{(i)}=y+z, X_{(i-1)}=y)$. Where the former is not the event of interest since it restricts the sample space to chains that have $X_{(i-1)}=y$ where I want the event that both are true amongst all the possible chain arrangements. Thank you Graham $\endgroup$ Jul 31 '20 at 14:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.