Statistics- unbiased estimator #2

[Roni1985]

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 f_Y(y;\Theta= \frac{1}{\Theta}*e^-y/\Theta , y>0
Let \Theta_hat=n*Y_min is tex]\Theta[/tex]_hat for \Theta ?

Homework Equations

Y_min=n*(1-F_Y(y))^n-1f_Y(y)

Also, it looks like an exponential distribution.

The Attempt at a Solution

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

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

Is this the correct answer ?

 

[rsa58]

where did you get the eqn for Ymin it makes no sense.

 

[rsa58]

i take that back give me a sec

 

[rsa58]

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.

 

[Roni1985]

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.

 

[statdad]

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 Y_{\min} 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 G, g for the CDF and PDF of Y_{\min}, and F, f for the CDF and PDF of the population.

<br /> 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 <br />

because of independence, so the CDF of Y_{\min} is

<br /> G(t) \equiv P(Y_{\min} \le t) = 1 - (1 - F(t))^n<br />

and the density Y_{\min}is

<br /> g(t) = G&#039;(t) = n(1-F(t))^{n-1} f(t)<br />

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 Y_{min} in this particular case. The expected value of the minimum is

<br /> \int t g(t) \, dt<br />

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