Beta distribution

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]

Attachments

• 15.3 KB Views: 341

Related Calculus and Beyond Homework Help News on Phys.org
Ray Vickson
Homework Helper
Dearly Missed

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]
Nope. ##\int_0^1 (1-u)^{n-1} \, du = \frac{1}{n}## (very elementary calculus!), so your integral above = 1. That is as it should be, since you just evaluated ##\int_0^1 f_{X_{(1)}}(x) \, dx##. What you need to evaluate is another integral:
$$E X_{(1)} = \int_0^1 x f_{X_{(1)}} (x) \, dx = \int_0^1 6n x^2(1-x) (1 - 3x^2 + 2x^3)^{n-1} \, dx$$
Getting an answer for general ##n## does not look simple, but for specific (small-to-moderate) ##n## it can be done. Here are some examples:
$$\begin{array}{cc} n & E X_{(1)} \\ 2 & 13/35\\ 3 & 43/140\\ 4 & 191/715\\ \vdots & \vdots \end{array}$$

While all these integrals are elementary, they are messy enough that I used the computer algebra system Maple to do the work. It can do the integral for n = 100 in seconds, but manual computation would take many hours.

If you do not have access to Maple or Mathematica, you can use the free on-line program Wolfram Alpha (which is a kind of Mathematica lite), but there may be limits on the size of n it can handle.

Last edited:
Right I did try to evaluate ##\int_{0}^{1} xf_{x_{(1)}}## but I found it difficult to try to do. I tried to evaluate it on matlab/mathematica and even then it couldn't evaluate it. Not very sure how to go about evaluating the integral.Maybe I'm not seeing it but there could be some way to integrate this. I noticed that there is convolution going on but unfortunately the extra x messes things up.

Wolfram Alpha did give an answer but in a very messy form that I could not really use.

Last edited:
Ray Vickson
Homework Helper
Dearly Missed
Right I did try to evaluate ##\int_{0}^{1} xf_{x_{(1)}}## but I found it difficult to try to do. I tried to evaluate it on matlab/mathematica and even then it couldn't evaluate it. Not very sure how to go about evaluating the integral.Maybe I'm not seeing it but there could be some way to integrate this. I noticed that there is convolution going on but unfortunately the extra x messes things up.
You can write
$$1-3x^2+2x^3 = 2(1-x)^2 \left(x+\frac{1}{2} \right)$$
to get the density function ##f_n(x)## of ##X_{(1)}## as
$$f_n(x) = 6n 2^{n-1} x (1-x)^{2n-1} \left( x + \frac{1}{2} \right)^{n-1}$$
One way to deal with ##\int x f_n(x) \, dx## is to expand ##(x + 1/2)^{n-1}## in a binomial expansion, then integrate term-by-term; each such term will be beta-related. When I do it all in Maple, it manages to arrive at a final expression for ##EX_{(1)}## involving a hypergeometric function.

I'm assuming that I should be able to do this by hand then? Or will I have to use maple or some other software to this for me through iteration? Well thanks for you help. I'll try this method out tomorrow and see where I get.