How to find the expected value of the first order statistic

Question

There are $Y_1, Y_2, \dots ,Y_n$ which are identically and independently distributed with pdf $4[(1-y)^3]$ for $0<y<1.$ We were asked to find the pdf of the first order statistic of which I got $f(Y(1))=n[4(1-y)^3]\cdot[(1+(1-y)^4)]^{(n-1)}.$ I got this by using the formula of the first order statistic.

I am confused on how to find the expected value of this since $n$ is undefined. I thought it would be the integral from $0$ to $1$ of $f(Y(1))* (Y(1)),$ but what is $Y(1)?$

Also I am confused on how to find the $P(Y(1)<0.1),$ which I am thinking is the integral from $0$ to $0.1,$ but how do we take this integral if $n$ is unknown? Since $n$ is unknown do we have to do integration by parts then?

Answer 1

You are told that the probability distribution function is $4(1- y)^3$. And, in your original statement, you are not asked about "Y(1)", you given $Y_1$, $Y_2$, …, $Y_n$, random variables. You ask about "the expected value of this" but what is "this"? You have n different random variables and can have any number of functions of them you could find the expected value of any of them. What is the exact statement of the problem?

Answer 2

The cdf of $Y_1$ is $F_Y(y)=1-(1-y)^4$. Therefore, the cdf of $Y_{(1)}$ is given by

$$ F_{Y_{(1)}}(y)=1-\mathsf{P}(Y_1>y,\ldots,Y_n>y)=1-(1-y)^{4n}, $$

and the corresponding pdf is

$$ f_{Y_{(1)}}(y)=\frac{d}{dy}F_{Y_{(1)}}(y)=4n(1-y)^{4n-1}. $$

The expectation of $Y_{(1)}$ is

$$ \mathsf{E}Y_{(1)}=\int_{0}^1yf_{Y_{(1)}}(y)dy=\int_{0}^14ny(1-y)^{4n-1}dy=\frac{1}{4n+1}. $$

Alternatively, you may compute it as follows:

$$ \mathsf{E}Y_{(1)}=\int_{0}^1(1-F_{Y_{(1)}}(y))dy=\int_{0}^1(1-y)^{4n}dy=\frac{1}{4n+1}. $$