Mircea

P(X=x and Y=y) = P(X = x)P(Y = y) for any x,y

lol

but I'm not sure if that's enough to count as a "proof"

Intuitively they're obviously independent since the 2 PDFs are completely separate from each other

how do I prove that my 2 variables are independent?

Suppose I have 2 probability distribution functions on two sets, A and B. Now suppose that I have variables taking up values from each of these probability spaces

I have a seemingly extremely easy problem but I can't seem to find the way to formally prove it...

Hey, guys!

;)

that's cause you are one

lol

will do

see ya later\

I gtg now

I'll keep an eye on it :)

yeah, MirceaS

gotcha

just send me an email at osebe2 (at) illinois (dot) edu

interesting. Can you let me know when it's done?

haha, goodluck

I'm not sure what's supposed to happen though

nice, looks interesting

I was half expecting a rick roll

here are the 2 operations you can do on the trees

app.ziteboard.com/team/410959f7-cd9a-4637-8ca7-07614aa1dde9

what's that?

fair enough

a whiteboard would certainly help :-)

sorry, I guess I made things harder

because you have n-1 internal nodes that you can swap at, and n-2 internal edges that you can rotate the tree along

if you represent all such trees with n numbered leaves by a node in a huge graph then each node will have degree n-1+n-2

yeah, but for suffficiently large n it breaks down it seems

and the conjecture is that you can turn any such term into an equivalent term in O(n) steps, where a step is either an application of a commutativity law or one of an assoc law

and each step is applying commutativity once or associativity once

it's actually better to think of this in terms of ways to combine n terms via this one binary operation

sorry, I meant full as in each node has either 2 or 0 children

that is associative and commutative\

you can think of it as combining all numbers from 1 to n via some binary operation

only in the leaves

where each step is either a tree rotation, or swapping one node's left and right subtrees

O(n) steps, sorry

and I turned this into a grapg theory problem by noticing that if every tree is a node then it has exactly degree 2n-3

His conjecture was that you can turn a full binary tree with leaves numbered from 1 to n to any other full binary tree with leaves from 1 to n using only tree rotations and swaps in O(n) time

and I think I just succeded :-)

disprove*

well, I was trying to prove a combinatorial conjecture of one of my professors

haha

yeah, thanks!

I will do!

and yeah, good idea taking log on both sides