# Statistics- unbiased estimator #2

• Roni1985
In summary, the conversation discusses a problem involving finding the expected value of a minimum sample from a random distribution, specifically from an exponential distribution. The conversation includes a derivation of the PDF and CDF for the minimum sample and discusses the correct answer for the expected value.

## 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;$$\Theta$$= $$\frac{1}{\Theta}$$*e-y/$$\Theta$$ , y>0
Let $$\Theta$$_hat=n*Ymin is tex]\Theta[/tex]_hat for $$\Theta$$ ?

## 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/ $$\Theta$$ )

Is this the correct answer ?

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

i take that back give me a sec

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.

rsa58 said:
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.

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.

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

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

$$G(t) \equiv P(Y_{\min} \le t) = 1 - (1 - F(t))^n$$

and the density $$Y_{\min}$$is

$$g(t) = G'(t) = n(1-F(t))^{n-1} f(t)$$

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

$$\int t g(t) \, dt$$

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

## 1. What is an unbiased estimator?

An unbiased estimator is a statistic that accurately estimates the population parameter it is intended to measure. This means that, on average, the estimator produces estimates that are equal to the true value of the parameter. In other words, there is no systematic over- or underestimation.

## 2. How is an unbiased estimator different from a biased estimator?

A biased estimator is a statistic that consistently over- or underestimates the true value of the population parameter it is intended to measure. This means that, on average, the estimator produces estimates that are different from the true value of the parameter. In other words, there is a systematic error in the estimation process.

## 3. Why is unbiasedness important in statistics?

Unbiasedness is important because it ensures that the estimates obtained from a sample accurately reflect the true value of the population parameter. This allows for reliable and valid conclusions to be drawn from the data.

## 4. Can an estimator be both biased and consistent?

Yes, an estimator can be both biased and consistent. Consistency means that as the sample size increases, the estimator produces estimates that are closer to the true value of the parameter. This does not guarantee unbiasedness, as the estimates may still be consistently over- or underestimating the true value.

## 5. How do you determine if an estimator is unbiased?

To determine if an estimator is unbiased, you can compare the expected value of the estimator (calculated using mathematical formulas or simulations) to the true value of the parameter. If the expected value is equal to the true value, then the estimator is unbiased. Additionally, unbiasedness can also be assessed using statistical tests or by comparing the estimator to other known unbiased estimators.