Order statistics and a convolution

[strangequark]

Homework Statement

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}' andY_{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}

Homework 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

The Attempt at a Solution

i have:
a.

ok, so I am 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?

 

[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.