• Support PF! Buy your school textbooks, materials and every day products Here!

Statistics- unbiased estimator #2

  • Thread starter Roni1985
  • Start date
  • #1
201
0

Homework Statement


There is some question, I solved it but am not sure I got the right answer.

Let Y1, Y2..... Yn be a random sample of size n from the pdf fY(y;[tex]\Theta[/tex]= [tex]\frac{1}{\Theta}[/tex]*e-y/[tex]\Theta[/tex] , y>0
Let [tex]\Theta[/tex]_hat=n*Ymin is tex]\Theta[/tex]_hat for [tex]\Theta[/tex] ?

Homework Equations



Ymin=n*(1-FY(y))n-1fY(y)

Also, it looks like an exponential distribution.


The Attempt at a Solution



What I need to find is E[n*Ymin]

I get [(-n)/(n+1)]/(1/ [tex]\Theta[/tex] )

Is this the correct answer ?
 

Answers and Replies

  • #2
85
0
where did you get the eqn for Ymin it makes no sense.
 
  • #3
85
0
i take that back give me a sec
 
  • #4
85
0
okay here it is. the PDF of Ymin is f(y)=(-n/theta)*exp(-yn/theta). To get this we need to use the following (questionable?) sequence of logic: the probability that ymin is less than y is 1 minus the probability that ymin >= x. for this inequality to hold we need Y(i)>=x for all i. i.e. Y(1)>=x AND Y(2)>=x ... Y(n)>=x. this will give us the CDF for Ymin from which we derive the PDF of Ymin given above. Now check that E(nYmin)= theta. (I actually get
-theta but there may be some minor issue with my calculation see what you get, if
E(nYmin) =theta (not negative theta) then it is unbiased.
 
  • #5
201
0
okay here it is. the PDF of Ymin is f(y)=(-n/theta)*exp(-yn/theta). To get this we need to use the following (questionable?) sequence of logic: the probability that ymin is less than y is 1 minus the probability that ymin >= x. for this inequality to hold we need Y(i)>=x for all i. i.e. Y(1)>=x AND Y(2)>=x ... Y(n)>=x. this will give us the CDF for Ymin from which we derive the PDF of Ymin given above. Now check that E(nYmin)= theta. (I actually get
-theta but there may be some minor issue with my calculation see what you get, if
E(nYmin) =theta (not negative theta) then it is unbiased.
Hello,

I don't really understand what you are trying to do here.
you found the pdf for Ymin ? we are given the pdf
Also, I got the CDF for Ymin, but am not sure I did it the right way.
And , I don't really understand what you got for the CDF..

Thanks.
 
  • #6
statdad
Homework Helper
1,495
35
You are given the distribution for the individual values - it is the distribution from which you take the sample. The minimum is a statistic, and you need its distribution.

The derivation outlined above is correct. If [tex] Y_{\min} [/tex] is the minimum of a sample, then to get its distribution (assuming the sample comes from a continuous distribution, as yours does). I'll use [tex] G, g [/tex] for the CDF and PDF of [tex] Y_{\min}[/tex], and [tex] F, f [/tex] for the CDF and PDF of the population.

[tex]
P(Y_{\min} \ge t) = P(X_1 \ge t \text{ and } X_2 \ge t \dots \text{ and } X_n \ge t) = (1-F(t))^n
[/tex]

because of independence, so the CDF of [tex] Y_{\min} [/tex] is

[tex]
G(t) \equiv P(Y_{\min} \le t) = 1 - (1 - F(t))^n
[/tex]

and the density [tex] Y_{\min}[/tex]is

[tex]
g(t) = G'(t) = n(1-F(t))^{n-1} f(t)
[/tex]

In your problem the original data come from an exponential distribution. Use the CDF for that in place of F, the PDF in place of f, to get the density of [tex] Y_{min} [/tex] in this particular case. The expected value of the minimum is

[tex]
\int t g(t) \, dt
[/tex]

and this will be a function of [tex] \theta [/tex]. You should be able to finish the problem from there.
 

Related Threads on Statistics- unbiased estimator #2

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
8K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
2
Views
4K
Replies
0
Views
2K
Replies
0
Views
875
  • Last Post
Replies
2
Views
915
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
904
  • Last Post
Replies
5
Views
2K
Top