Factorization Theorem for Sufficient Statistics & Indicator Function

[kingwinner]

Problem:
Let Y1,Y2,...,Yn denote a random sample from the uniform distribution over the interval (0,theta). Show that Y(n)=max(Y1,Y2,...,Yn) is a sufficient statistic for theta by the factorization theorem.

Solution:
http://www.geocities.com/asdfasdf23135/stat10.JPG
1) While I understand that I_A (x)I_B (x)=I_{A intersect B} (x), I don't understand the equality circled in red above.

In the solutions, they say that I_0,theta (y1)...I_0,theta (yn)=I_0,theta (y_(n)). Is this really correct?
Shouldn't the right hand side be I_0,theta (y_(n))I_0,infinity (y₍₁₎) ? I believe that the second factor is necessary because the largest observation is greater than zero does not guarantee that the smallest observation is greater than zero.
Which one is correct?


2) Also, is I_0,theta (y_(n)) a function of y_(n), a function of theta, or a function of both y_(n) and theta?
If it is a function of both y_(n) and theta, then there is something that I don't understand. Following the definition of indicator function that I_A (x) is a function of x alone (it is a function of only the stuff in the parenthesis), shouldn't I_0,theta (y_(n)) be a function of only y_(n) alone?


Thank you for explaining! I've been confused with these ideas for at least a week.

 

[statdad]

The left side of

$$\prod_{i=1}^n I_{[0,\theta)} (y_i) = I_{[0,\theta)} (y_{(n)})$$

means that all the $$y_i$$ values are in the interval $$[0,\theta)$$.

This is true if, and only if, the maximum of the y's is in the same interval, and that is the meaning of the right-side.

The indicator $$I_{[0,\theta)} (y_{(n)})$$ is a function of $$y_{(n)}$$ only, since $$\theta$$ is fixed (it's a parameter).

 

[kingwinner]

I understand that
0<X_1,..., X_n<theta here these are the unordered data
is the same as (iff)
0<X_(1)<X_(2)<...<X_(n)<theta

But I don't think
0<X_(1)<X_(2)<...<X_(n)<theta
is EQUIVALENT to (iff)
0<X_(n)<theta
The => direction is true but <= is not. (the fact that the largest observation x(n) is greater than zero does not guarantee that the smallest observation x(1) is greater than zero.)

So that's why I think we should have
I_0,theta(y1)...I_0,theta(yn) = I_0,theta(y(n))I_0,infinity(y(1))
instead of I_0,theta(y1)...I_0,theta(yn)=I_0,theta(y(n)).

Right?

 

[kingwinner]

The indicator $$I_{[0,\theta)} (y_{(n)})$$ is a function of $$y_{(n)}$$ only, since $$\theta$$ is fixed (it's a parameter).

But we can also write it as I _{y(n), inf} (theta), in this case theta would be in the parenthesis, so in this case, would it be a function of theta alone? (in the gerenal case, f(x) means a function of x, f(y) means a function of y, the stuff in the parenthesis)


When we talk about functions, is it always only a function of the stuff in the parenthesis? It looks like that the restrictions/constraints are also important, so shouldn't it be a function also of the variables in the restrictions/constraints?

e.g.)
f(x)=x if x>y
f(x)=x^3 if x<y
Here not only the value of x controls f, the value of y also controls f, so is f a function of BOTH x and y here?

Thanks!