(adsbygoogle = window.adsbygoogle || []).push({}); 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 [tex]Y_{1},Y_{2},...,Y_{n} [/tex] denote random samples of numbers from a uniform distribution on the interval [0,1]. Denote the largest and smallest numbers as [tex]Y_{n}'[/tex] and [tex]Y_{1}'[/tex].

a. Find the pdf and cdf of [tex]Y_{1}'[/tex] and[tex]Y_{n}'[/tex].

b. Find [tex]E(Y_{n}'-Y_{1}')[/tex]

c. Show that [tex] \lim_{n\to\infty}E(Y_{n}'-Y_{1}')=1[/tex]

d. Find the pdf for [tex]Y_{2}[/tex] and find the pdf for [tex]Y=Y_{1}+Y_{2}[/tex]

2. Relevant equations

the ith order statistic:

[tex]f_{Y_i}=\frac{n!}{(i-1)!(n-i)!}(F_{Y}(y))^{i-1}(1-F_{Y})^{n-i}f_{Y}(y)[/tex]

convolution:

if [tex]X=Y_{1}+Y_{2}[/tex]

[tex]f_{X}(x)=\int^{x}_{0}f_{Y_{1}}(y)f_{Y_{2}}(x-y)dy[/tex]

3. The attempt at a solution

i have:

a.

ok, so im pretty sure that my pdf's will be:

[tex]f_{Y_{1}'}=\frac{n!}{(n-1)!}(1-y)^{n-1}=n(1-y)^{n-1}[/tex]

and

[tex]f_{Y_{n}'}=\frac{n!}{(n-1)!}(y)^{n-1}=ny^{n-1}[/tex]

b. cumulative functions easy from above.

c.

I get

[tex]E(Y_{n}')=\frac{n}{n+1}[/tex]

and

[tex]E(Y_{1}')=\frac{1}{n+1}[/tex]

so:

[tex]E(Y_{n}'-Y_{1}')=\frac{n-1}{n+1}[/tex]

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

[tex]f_{Y_{1}}=n(1-y)^{n-1}[/tex]

Is my [tex]f_{Y_{2}}[/tex] correct?

[tex]f_{Y_{2}}=n(n-1)y(1-y)^{n-2}[/tex]

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

[tex]\int^{x}_{0}n(n)(n-1)(1-y)^{n-1}(x-y)(1-(x-y))^{n-2}dy[/tex]

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

[tex]f_{X}(x)=1-|1-x|[/tex] for [tex]0\leqx\leq2[/tex]

help???

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# Homework Help: Order statistics and a convolution

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