Question

suppose y1, . . . , yn have the following distribution function, fy (y) = ( e −y θ θ < x < [infinity], 0 o.w. prove that y(1) = min(y1, . . . , yn) is sufficient.

Answer 1

To prove that y(1) = min(y1, . . . , yn) is sufficient, we need to show that the conditional distribution function of the random variables y1, . . . , yn given y(1) does not depend on the parameter theta.

Step-by-step explanation:

To prove that y(1) = min(y1, . . . , yn) is sufficient, we need to show that the conditional distribution function of the random variables y1, . . . , yn given y(1) does not depend on the parameter theta.

1. First, let's find the conditional distribution function of y1, . . . , yn given y(1). The probability that y1, . . . , yn are all greater than some constant c, given that y(1) is also greater than c, is:
2. P(y1 > c, . . . , yn > c | y(1) > c) = P(y1 > c, . . . , yn > c)
3. This probability is equal to the joint distribution function of y1, . . . , yn evaluated at c, which is:
4. P(y1 > c, . . . , yn > c) = (e^(-c*theta))^n = e^(-n*c*theta)
5. Since this probability does not depend on theta, we have shown that the conditional distribution function of y1, . . . , yn given y(1) does not depend on the parameter theta. Therefore, we can conclude that y(1) = min(y1, . . . , yn) is a sufficient statistic.