Author |
Topic: Statistics problem (Read 3187 times) |
|
Benny
Uberpuzzler
Gender: 
Posts: 1024
|
 |
Statistics problem
« on: Feb 1st, 2008, 12:40pm » |
 Quote  Modify
|
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.
|
|
 IP Logged |
If we want to understand our world — or how to change it — we must first understand the rational choices that shape it.
|
|
|
Eigenray
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 1948
|
|
Re: Statistics problem
« Reply #1 on: Feb 1st, 2008, 8:37pm » |
Quote Modify
|
Do you mean r2 = max{Xi} - n/(n+1) ?
|
|
IP Logged |
|
|
|
Benny
Uberpuzzler
Gender:
Posts: 1024
|
|
Re: Statistics problem
« Reply #2 on: Feb 2nd, 2008, 1:53am » |
Quote Modify
|
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
|
|
IP Logged |
If we want to understand our world — or how to change it — we must first understand the rational choices that shape it.
|
|
|
Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: Statistics problem
« Reply #3 on: Feb 2nd, 2008, 10:06am » |
Quote Modify
|
on Feb 2nd, 2008, 1:53am, 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.
|
|
IP Logged |
"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
|
|
|
pex
Uberpuzzler
Gender:
Posts: 880
|
|
Re: Statistics problem
« Reply #4 on: Feb 2nd, 2008, 10:30am » |
Quote Modify
|
What's this doing in Putnam?
r1 is very easy. E[Xi] = K + 1/2 for all i, so
E[Xmean] = (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 Xi is
F(x) = Pr(Xi<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(Xmax<x) = Pr(all Xi<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 Xmax as
integral(x from K to K+1) x*n(x-K)n-1 dx
= n*integral(t from 0 to 1) (t+K)tn-1 dt
= n*integral(t from 0 to 1) tn dt + n*K*integral(t from 0 to 1) tn-1dt
= n*1/(n+1) + n*K*(1/n)
= n/(n+1) + K.
Thus, E[r2] = K as well.
|
| « Last Edit: Feb 2nd, 2008, 10:30am by pex » |
IP Logged |
|
|
|
Benny
Uberpuzzler
Gender:
Posts: 1024
|
|
Re: Statistics problem
« Reply #5 on: Feb 3rd, 2008, 9:34pm » |
Quote Modify
|
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.
|
|
IP Logged |
If we want to understand our world — or how to change it — we must first understand the rational choices that shape it.
|
|
|
Eigenray
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 1948
|
|
Re: Statistics problem
« Reply #6 on: Feb 3rd, 2008, 9:59pm » |
Quote Modify
|
For independent variables Var(X+Y) = Var(X) + Var(Y), and Var(c X) = c2 Var(X). So Var(r1) =  Var(Xi/n) = n*Var(Xi)/n2 = 1/(12n).
On the other hand,
Var(r2) =  01 (x-n/(n+1))2 nxn-1dx = n/[(n+1)2(n+2)],
which is less than 1/(12n) for n  8.
|
|
IP Logged |
|
|
|
Benny
Uberpuzzler
Gender:
Posts: 1024
|
|
Re: Statistics problem
« Reply #7 on: Feb 5th, 2008, 10:40am » |
Quote Modify
|
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))?
|
|
IP Logged |
If we want to understand our world — or how to change it — we must first understand the rational choices that shape it.
|
|
|
pex
Uberpuzzler
Gender:
Posts: 880
|
|
Re: Statistics problem
« Reply #8 on: Feb 5th, 2008, 10:56am » |
Quote Modify
|
on Feb 5th, 2008, 10:40am, 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.
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print