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        riddles >> putnam exam (pure math) >> Statistics problem
        
(Message started by: BenVitale on Feb 1^st, 2008, 12:40pm)

Title: Statistics problem
Post by BenVitale on Feb 1^st, 2008, 12:40pm

Let X1,X2,.....,Xn denote a random sample from the uniform distribution on the interval (K,K+1).

Let r1=Xbar -1/2, and r2=Xn -n/(n+1)

Show that r1 and r2 are unbaised estimators of K.

Title: Re: Statistics problem
Post by Eigenray on Feb 1^st, 2008, 8:37pm

Do you mean r2 = max{X_i} - n/(n+1) ?

Title: Re: Statistics problem
Post by BenVitale on Feb 2^nd, 2008, 1:53am

I Assume that Xbar = 1/n *SUM X_j
and X_n = max{X_j}.

In both cases all we need is to show that the expectation of r1, r2 equals K

Title: Re: Statistics problem
Post by Icarus on Feb 2^nd, 2008, 10:06am

on 02/02/08 at 01:53:33, BenVitale wrote:

I Assume that ...
and X_n = max{X_j}.

Since "X_n" also represents one of the X_j, this is not a good choice of notation, unless you mean that X_1 <= X_2 <= ... <= X_n.

Title: Re: Statistics problem
Post by pex on Feb 2^nd, 2008, 10:30am

What's this doing in Putnam?

r₁ is very easy. E[X_i] = K + 1/2 for all i, so
E[X_mean] = (1/N) * N * (K + 1/2) = K + 1/2
and hence E[r₁] = K.

r₂ is standard too, but it requires a bit more work. The cumulative distribution function of each of the X_i is
F(x) = Pr(X_i<x) = {0 if x<K, x - K if K<x<K+1, 1 if K+1<x}.

The cumulative distribution function of the maximum is easy to find: we have
Pr(X_max<x) = Pr(all X_i<x) = F(x)ⁿ.

This means that the density function is n * F(x)^n-1 * F'(x). We'll only need it on the interval (K, K+1), where it equals
n * (x - K)^n-1 * 1 = n(x-K)^n-1.
Then, we can find the expectation of X_max as
integral(x from K to K+1) x*n(x-K)^n-1 dx
= n*integral(t from 0 to 1) (t+K)t^n-1 dt
= n*integral(t from 0 to 1) tⁿ dt + n*K*integral(t from 0 to 1) t^n-1dt
= n*1/(n+1) + n*K*(1/n)
= n/(n+1) + K.
Thus, E[r₂] = K as well.

Title: Re: Statistics problem
Post by BenVitale on Feb 3^rd, 2008, 9:34pm

Do you know which variable (r1 or r2) has the smaller variance. I calculated Var(r1)=1/12 however I'm pretty sure this is incorrect since this number is independent of n.

Title: Re: Statistics problem
Post by Eigenray on Feb 3^rd, 2008, 9:59pm

For independent variables Var(X+Y) = Var(X) + Var(Y), and Var(c X) = c² Var(X). So Var(r1) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif Var(X_i/n) = n*Var(X_i)/n² = 1/(12n).

On the other hand,

Var(r2) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif₀¹ (x-n/(n+1))² nx^n-1dx = n/[(n+1)²(n+2)],

which is less than 1/(12n) for n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 8.

Title: Re: Statistics problem
Post by BenVitale on Feb 5^th, 2008, 10:40am

it's interesting that r2 has a smaller variance than r1 for large n. Usualy things involving Xbar are better estimators.

Let E(x/y) denote the expected value of x given y.

How can we prove that E(xy)=E(yE(x/y))?

Title: Re: Statistics problem
Post by pex on Feb 5^th, 2008, 10:56am

on 02/05/08 at 10:40:50, BenVitale wrote:

Let E(x/y) denote the expected value of x given y.

How can we prove that E(xy)=E(yE(x/y))?

By conditioning, also known as the "Law of Iterated Expectations": generally, E(A) = E( E(A|B) ).

E(XY) = E( E(XY|Y) ) = E( Y E(X|Y) ), where the first equality uses the LIE and the second follows from the fact that given Y, the expected value of XY is just E(X|Y) * Y.