# Order statistics and a convolution

1. Dec 3, 2007

### strangequark

1. The problem statement, all variables and given/known data

Hi, I'm having some problem with one of my final exam study questions, and I'm hoping someone can help me out a little.

here is the problem:

Let $$Y_{1},Y_{2},...,Y_{n}$$ denote random samples of numbers from a uniform distribution on the interval [0,1]. Denote the largest and smallest numbers as $$Y_{n}'$$ and $$Y_{1}'$$.

a. Find the pdf and cdf of $$Y_{1}'$$ and$$Y_{n}'$$.
b. Find $$E(Y_{n}'-Y_{1}')$$
c. Show that $$\lim_{n\to\infty}E(Y_{n}'-Y_{1}')=1$$
d. Find the pdf for $$Y_{2}$$ and find the pdf for $$Y=Y_{1}+Y_{2}$$

2. Relevant equations

the ith order statistic:

$$f_{Y_i}=\frac{n!}{(i-1)!(n-i)!}(F_{Y}(y))^{i-1}(1-F_{Y})^{n-i}f_{Y}(y)$$

convolution:
if $$X=Y_{1}+Y_{2}$$

$$f_{X}(x)=\int^{x}_{0}f_{Y_{1}}(y)f_{Y_{2}}(x-y)dy$$

3. The attempt at a solution

i have:
a.

ok, so im pretty sure that my pdf's will be:
$$f_{Y_{1}'}=\frac{n!}{(n-1)!}(1-y)^{n-1}=n(1-y)^{n-1}$$
and
$$f_{Y_{n}'}=\frac{n!}{(n-1)!}(y)^{n-1}=ny^{n-1}$$

b. cumulative functions easy from above.

c.
I get
$$E(Y_{n}')=\frac{n}{n+1}$$
and
$$E(Y_{1}')=\frac{1}{n+1}$$
so:
$$E(Y_{n}'-Y_{1}')=\frac{n-1}{n+1}$$

d. here's where I'm confused... for $$f_{Y_{1}}$$ and $$f_{Y_{2}}$$, I have:

$$f_{Y_{1}}=n(1-y)^{n-1}$$

Is my $$f_{Y_{2}}$$ correct?

$$f_{Y_{2}}=n(n-1)y(1-y)^{n-2}$$

if it is, how the heck do I calculate this integral??

$$\int^{x}_{0}n(n)(n-1)(1-y)^{n-1}(x-y)(1-(x-y))^{n-2}dy$$

I'm not seeing my mistake. The answer should be:

$$f_{X}(x)=1-|1-x|$$ for $$0\leqx\leq2$$

help???

Last edited: Dec 3, 2007
2. Dec 3, 2007

### EnumaElish

$f_{Y_{2}}(z)$ is n!/((n-k)!(k-1)!)F(z)k-1[(1-F(z)]n-kf(z) = n(n-1)y(1-y)n-2, so it is correct.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?