Need help for Ito Isometry proof

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    Isometry Proof
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The discussion centers on the proof of Ito Isometry, specifically the expectation of the square of the difference of two positions in Brownian motion, denoted as E[(Btj+1(w)-Btj(w))²]. It is established that this expectation equals the time difference tj+1 - tj due to the properties of Brownian motion. The increments of Brownian motion, B_t(w) - B_s(w), are normally distributed with mean 0 and variance t-s, which directly supports the conclusion of the proof.

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Sa7oru
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Hi,

There is one step in Ito Isometry proof I don't understand.

Bt(w) represents the position of w at time t, then E[(Btj+1(w)-Btj(w))2] = tj+1 - tj.

Why is the expectation of the square of the difference of 2 positions equal to their time difference?

Any hint, please.

Thank you.
 
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Could you clarify what Bt(w) is & whether it's normally distributed?
 
I'm guessing B_t(w) is Brownian motion, and yes it's normally distributed because its increments, B_t(w) - B_s(w) are normally distributed (~ N(0, t-s)) if s < t. Indeed, this is the precise property that gives the result the OP wants to understand.
 

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