God damn math

[Lul Thyme]

Originally posted by Jon Miller
You have edited your stuff like 3 times..

JM

So?
It's getting better everytime!!

[Addled Platypus]

dont worry about it

X....marks the spot

[Lul Thyme]

Originally posted by Jon Miller
Well, I was suggesting to come up with E(X^(k+1)) = L^(K+1)+Function(L,k), which would mean you wouldn't have to 'guess' what the sum is. This is just saying that the other direction might be easier.

JM
:lol, and I now am not sure what I was replying to. Either you DanSed me or I am getting sleepy.

Well then you would need to show that
E[X...(X-k)]=E[X^(k+1)]-Function(L,k) which doesnt seem any easier.

[Kuciwalker]

Originally posted by Eli

I'm assuming all the X's are independent,

Eh? How on earth the Xs can be independent? They are the same and therefore as dependent as you can get.

I thought each represented a separate sample.

[Kuciwalker]

Thanks. I'll read all this after I go to the class (and maybe get some more sleep).

[KrazyHorse]

Re: Re: God damn math

Originally posted by Lul Thyme



Right , because only KH knows math...

No, but I'm usually the most helpful one...

[KrazyHorse]

Anyway, I'll assume that Lul got the right answer, but this is sort of a fun problem so I'll do it anyway. A major mistake, though:

I'm assuming all the X's are independent

How the **** can you assume that X and X-1 are independent? They have a correlation coefficient of 1! Don't assume that any deviates are independent without a damn good reason. Now, to the proof:

E[X*(X-1)*...*(X-k)] = sumX=0X=inf(X*(X-1)*...*(X-k)*exp(-L)*(L^X)/X!) (definition)
=exp(-L)*sumX=(k+1)X=inf(X*(X-1)*...*(X-k)*(L^X)/X!) (pulled out constant and dropped 0 terms from start of summation)
=exp(-L)*sumX=(k+1)X=inf((L^X)/(X-(k+1))!) (pulled out common factors from top and bottom)
=exp(-L)*(L^(k+1))sumX=(k+1)X=inf((L^(X -(k+1)))/(X-(k+1))!) (the clever step; pulled out constant for reasons which will be obvious later)
=exp(-L)*(L^(k+1))sumn=0n=inf((L^n)/n!) (substituted n = X-(k+1))
=exp(-L)*(L^(k+1))*exp(L) (definition of exp)
= L^(k+1) (do I need to explain this?)
QED

[LordShiva]

Kuci is teh cheat0r

[alva]

Originally posted by KrazyHorse
...but this is sort of a fun problem so I'll do it anyway.

[KrazyHorse]

Originally posted by Kuciwalker


I thought each represented a separate sample.

If that was the case they would each have been represented by a different symbol and he would have told you they were independent

[KrazyHorse]

Originally posted by alva

?

It took all of 30 seconds to see the answer and about 4 minutes to write it up.

There were no messy or complicated expressions, and everything worked out nicely. Those types of problems are fun...

[Kuciwalker]

It's done. I've transferred from Probability to Intro to Prob and Stats I, a 200-level versus a 300-level course. I just don't have the background in set theory or formal proof to handle this class right now. I'm going to take Real Analysis next semester, which should teach me most of the skills I need, then retake this class sophomore year. The professor agreed this would be a good idea.

[MRT144]

Originally posted by KrazyHorse


?

It took all of 30 seconds to see the answer and about 4 minutes to write it up.

There were no messy or complicated expressions, and everything worked out nicely. Those types of problems are fun...

this is why youre my hero. smart as a whip and witty to boot.

[pchang]

But

he was never a porn star like you!

[Asher]

Not until the tapes leak, anyway.