- #1
bonfire09
- 249
- 0
Homework Statement
If ##X_1,..,X_n## is a random sample and has a density ##f_X(x)=6x(1-x)## for ##0<x<1##. Let ##X_{(1)}=\min(X_1,..,X_n)##. Find the ##E(X_{(1)})## (Expected Value)?
Homework Equations
There is an uploaded picture of the pdf I used.
The Attempt at a Solution
"I've been stuck on this for hours. If I try integrating it the long way in order to find the expected value I run into problems. " So I tried this way but it seems to be not correct to me. Any help would be great thanks.
So I need to find the ##E(X_{(1)})## (Expected Value) where the pdf of ##X_{(1)}## is
##f_{X_{(1)}}=n*[6x(1-x)*[1-3x^2+2x^3]^{n-1}]## where ##0<x<1##. I know that this looks like some form of a beta distribution.
What I did to find the parameters ##\alpha, \beta## I integrated the pdf of ##X_{(1)}## until I got something that looks like a beta distribution.
We see that $$\int_{0}^{1} 6nx(1-x)*[1-3x^2+2x^3]^{n-1} dx$$ $$=n\int_{0}^{1}6x(1-x)*[1-3x^2+2x^3]^{n-1} dx $$.
Let $u= 3x^2-2x^3$ so it follows that $u'=6x-6x^2$. Plugging this back in we get
$$=n\int_{0}^{1} [1-u]^{n-1} du =n\dfrac{\gamma{(1)}\gamma{(n)}}{\gamma{(n+1)}} \int_{0}^{1} \dfrac{\gamma{(n+1)}}{\gamma{(1)}\gamma{(n)}} [1-u]^{n-1} du.$$ Thus we see that ##X_{(1)}=\mathrm{Beta}(\alpha=1,\beta=n)##. So would this mean from this that ##E(X_{(1)})=\dfrac{1}{n+1}##?.
[/B]