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Jyan

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I previously posted this related thread https://www.physicsforums.com/showthread.php?p=4663806&posted=1#post4663806

I am working on some MIT OCW for probability theory (http://ocw.mit.edu/courses/electric...y-fall-2008/assignments/MIT6_436JF08_hw01.pdf) I attempted exercise 3a and wrote it out here, but quickly realized it was wrong because you can't necessarily find an ordering that is so convenient. I've been working on this problem for hours though, anyone able to provide some assistance?

My previous attempt:

Given [tex] f : A\times B \rightarrow \mathbb{R}[/tex]

Where A,B are finite and non-empty.

Prove: [tex] \max_{x\in A} \min_{y\in B} f(x,y) \le \min_{y\in B} \max_{x\in A} f(x,y) [/tex]

proof:

Let N be the number of elements in A, and M the number of elements in B.

Construct an ordered list of tuples,

[tex] C = \big[(x_0, y_0), ... , (x_0, y_M), (x_1, y_0), ... ,(x_1, y_M), ... , (x_N, y_0), ... , (x_N, y_M)\big] [/tex]

According to

[tex] (x,y) \le (x', y') \Longleftrightarrow f(x,y) \le f(x', y') [/tex]

That is, C is ordered such that the smallest elements of C produce the smallest values of f, and the largest elements of C produce the largest values of f.

Now construct two new sets A' and B' where

[tex] A' = Range \big\{\min_{y\in B} f(x,y)\big\} = \big\{f(x,y_0): x\in A\big\}[/tex]

[tex] B' = Range \big\{\max_{x\in a} f(x,y)\big\} = \big\{f(x_N, y): y\in B\big\}[/tex]

By the ordering on C

[tex] \forall a,b \in A', B' \space a \le b [/tex]

[tex] \blacksquare [/tex]

I feel like I'm pretty close... I just need to fix the problem with creating the ordering for C.

I am working on some MIT OCW for probability theory (http://ocw.mit.edu/courses/electric...y-fall-2008/assignments/MIT6_436JF08_hw01.pdf) I attempted exercise 3a and wrote it out here, but quickly realized it was wrong because you can't necessarily find an ordering that is so convenient. I've been working on this problem for hours though, anyone able to provide some assistance?

My previous attempt:

Given [tex] f : A\times B \rightarrow \mathbb{R}[/tex]

Where A,B are finite and non-empty.

Prove: [tex] \max_{x\in A} \min_{y\in B} f(x,y) \le \min_{y\in B} \max_{x\in A} f(x,y) [/tex]

proof:

Let N be the number of elements in A, and M the number of elements in B.

Construct an ordered list of tuples,

[tex] C = \big[(x_0, y_0), ... , (x_0, y_M), (x_1, y_0), ... ,(x_1, y_M), ... , (x_N, y_0), ... , (x_N, y_M)\big] [/tex]

According to

[tex] (x,y) \le (x', y') \Longleftrightarrow f(x,y) \le f(x', y') [/tex]

That is, C is ordered such that the smallest elements of C produce the smallest values of f, and the largest elements of C produce the largest values of f.

Now construct two new sets A' and B' where

[tex] A' = Range \big\{\min_{y\in B} f(x,y)\big\} = \big\{f(x,y_0): x\in A\big\}[/tex]

[tex] B' = Range \big\{\max_{x\in a} f(x,y)\big\} = \big\{f(x_N, y): y\in B\big\}[/tex]

By the ordering on C

[tex] \forall a,b \in A', B' \space a \le b [/tex]

[tex] \blacksquare [/tex]

I feel like I'm pretty close... I just need to fix the problem with creating the ordering for C.

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