1. ## Unbiased Estimators

Let Y_1, Y_2,...,Y_n be a random sample of size n from the pdf f_Y(y;theta)=(1/theta)*e^(-y/theta), y>0:

a) Let theta hat=n*Y_min. Is theta hat unbiased for theta?

For this question, I'm not sure how to generate the pdf for Y_min. Once I get that, I assume I multiply by n, multiply by the other pdf, and solve the integral.

b) Is theta hat=(1/n)summation(from i=1 to n)Y_i unbiased for theta?

2. Originally Posted by eigenvector11
Let Y_1, Y_2,...,Y_n be a random sample of size n from the pdf f_Y(y;theta)=(1/theta)*e^(-y/theta), y>0:

a) Let theta hat=n*Y_min. Is theta hat unbiased for theta?

For this question, I'm not sure how to generate the pdf for Y_min. Once I get that, I assume I multiply by n, multiply by the other pdf, and solve the integral.

b) Is theta hat=(1/n)summation(from i=1 to n)Y_i unbiased for theta?
(a) If the random variable Y has pdf f(y) then the pdf of $\displaystyle Y_{(1)}= \text{min} \{ Y_1, \, Y_2, \, .... \, Y_n\}$ is found as follows:

The cdf of $\displaystyle Y_{(1)}$ is $\displaystyle G(y) = \Pr(Y_{(1)} \leq y) = 1 - \Pr(Y_{(1)} > y)$.

Since $\displaystyle Y_{(1)}$ is the minimum of $\displaystyle Y_1, \, Y_2, \, .... \, Y_n$ it follows that the event $\displaystyle \Pr(Y_{(1)} > y)$ occurs if and only if the events $\displaystyle \Pr(Y_i > y)$ occur for $\displaystyle i = 1, 2, \, .... \, n$. Since the $\displaystyle Y_i$ are independent and $\displaystyle \Pr(Y_i > y) = 1 - F(y)$ it follows that

$\displaystyle G(y) = \Pr(Y_{(1)} \leq y) = 1 - \Pr(Y_{(1)} > y) = 1 - \Pr(Y_1 > y, \, Y_2 > y, \, .... \, Y_n > y)$

$\displaystyle = 1 - \Pr(Y_1 > y) \cdot \Pr(Y_2 > y) \cdot \, .... \, \cdot \Pr(Y_n > y) = 1 - [1 - F(y)]^n$.

The pdf of $\displaystyle Y_{(1)}$ is given by $\displaystyle g(y) = \frac{dG}{dy}$: $\displaystyle g(y) = n [1 - F(y)]^{n-1} f(y)$.

Now you have to calculate $\displaystyle E(n Y_{(1)})$ and see if it's equal to $\displaystyle \theta$.
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(b) Calculate the expected value of the estimator and see whether or not you get $\displaystyle \theta$.