baileyjanet
New member
- Joined
- Feb 6, 2007
- Messages
- 2
I have Y1, Y2, ..., Yn, a r.s. of size n from a uniformly distributed population with
f(y) = 1/theta, on 0 <=y<=theta
f(y) = 0, elsewhere.
so F(y) = def. integral from 0 to y of 1/theta dtheta = y/theta, on 0<=y<=theta.
I also have Y(n), the max of the Yis, and
U, a function of Y(n), with U = Y(n)/theta.
Using the method of distribution functions, I found the distribution function of U
F(u) = P(U<=u) = P(Y(n)/theta <=u) = P(Y(n) <= theta*u) = Fymax(theta*u)
I know that the distribution function of Y(n) is Fymax(y) = [F(y)]^n
and the density function is the derivative with respect to y of [F(y)]^n
here's where I'm having trouble...
d/dy[F(y)]^n = n[F(y)]^(n-1)*f(y) = n(y/theta)^(n-1)*(1/theta)
this is supposed to simplify to (y/theta)^n.
I'm not seeing it.
Could you please help?
Thanks if you can see what I'm doing!
Janet
f(y) = 1/theta, on 0 <=y<=theta
f(y) = 0, elsewhere.
so F(y) = def. integral from 0 to y of 1/theta dtheta = y/theta, on 0<=y<=theta.
I also have Y(n), the max of the Yis, and
U, a function of Y(n), with U = Y(n)/theta.
Using the method of distribution functions, I found the distribution function of U
F(u) = P(U<=u) = P(Y(n)/theta <=u) = P(Y(n) <= theta*u) = Fymax(theta*u)
I know that the distribution function of Y(n) is Fymax(y) = [F(y)]^n
and the density function is the derivative with respect to y of [F(y)]^n
here's where I'm having trouble...
d/dy[F(y)]^n = n[F(y)]^(n-1)*f(y) = n(y/theta)^(n-1)*(1/theta)
this is supposed to simplify to (y/theta)^n.
I'm not seeing it.
Could you please help?
Thanks if you can see what I'm doing!
Janet