/sci/ - Science & Math

[Anonymous No.5900733]

If (Xn) is a martingale, then

1) for all integer n (time n), we have E(Xn)=E(X0).

2) according to the Optional Stopping Theorem (Doob's Theorem), for all stopping time T, E(XT)=E(X0).
What is the difference between 1) and 2) ?

Why doesn't 1) imply 2) ?

[Anonymous No.5901153]

BUMP

[Oilerscup20014 !S0ooHAHDTU No.5901236]

isn't that NOT the definition of a martingale?

Xn is a martingale if E(Xn) = Xn-1... Just off my memory.. What you wrote is not the same..

[Anonymous No.5901266]

>>5901236

what i wrote is a consequence of the definition of a martingale

[Anonymous No.5901410]

>>5900733
E(Xn)=E(X0)

this is not what martingale is.

it's E(Xn) = Xn-1

It means, after playing a bunch of games, you have 10 bucks, and on average you win a bit and the lose a bit, expectation of your winnings after next game is again 10 bucks. But this value changes as you play, it's not constant.

What you wrote is: expectation is always the same as before even playing, which means you are a dirty faggot.

[Oilerscup20014 !S0ooHAHDTU No.5901491]

>>5901266

That does ABSOLUTELY NOT follow the def of martingale.

You are dealing with the ACTUAL Xn-1, which is, in general, not equal to E(Xn-1).

If E(Xn-1) = Xn-1 (the expectation is ALWAYS the actual outcome) then you'd be correct. But that is, I hope you see why, totally ridiculous on a few different levels.

[Anonymous No.5902198]

>>5901410
>>5901491

You can look up online, it is written everywhere that Martingales have constant mean.

Just use an induction :

E(Xn+1) = E(E(Xn+1|Fn)) = E(Xn)=...=E(X0).