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- 1/19/22

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Hi all,

I am trying understand a heuristic derivation of Ito's lemma ([imath]B[/imath] follows Brownian motion):

[math]df(t,B) = \frac{\partial f}{\partial t}dt + \frac{\partial f}{\partial B}dB + \frac{1}{2} \frac{\partial^2 f}{\partial^2 B}dB^2,[/math]

but I do not understand why the derivation says that [imath]dB^2[/imath] does not vanish. I've read/watched a few sources (below) to figure out why it doesn't vanish and the one I'm currently stuck with is here. To be specific, I'm stuck on the line after equation (9):

[math]E[(B(t_{i+1}) - B(t_i))^2] = E[B(t_{i+1} - t_i)^2] = t_{i+1} - t_i.[/math]

How could the expected value of the difference squared be at all related to the time interval? Unless we

Almost all of my effort has been trying to understand the link between [imath]dt[/imath] and [imath]dB[/imath]; the intuition that I get is that [imath]dB[/imath] is in a loose sense some kind of random variable whose variance is explicitly made to be related to [imath]dt[/imath] so that when we square it we get nice relations. For example:

[math] B(t_{i+1}) - B(t_i) = \Delta B_i = \sigma \sqrt{\Delta t} X_i, \, \Delta t = t_{i+1} - t_i \\ P(X_i = 1) = P(X_i = -1) = 0.5 \\ E[X_i] = (1) \cdot 0.5 + (-1) \cdot 0.5 = 0 \\ E[X_i^2] = V[X_i], \text{because } V[X_i] = E[X_i^2 - E[X_i]] \text{ but } E[X_i] = 0 \\ V[X_i] = (1)^2 \cdot 0.5 + (-1)^2 \cdot 0.5 = 1 \\ E[(B(t_{i+1}) - B(t_i))^2] = E[\Delta B_i^2] \\ = E[(\sigma \sqrt{\Delta t} X_i)^2] \\ = \sigma^2 \Delta t E[X_i^2] \\ = \sigma^2 (t_{i+1} - t_i) \\ = t_{i+1} - t_i \text{ if } \sigma = 1 [/math]

Really not sure. Also sorry my equations look bad I can't get \begin{align}\end{align} to work. Thanks a lot.

Sources:

I am trying understand a heuristic derivation of Ito's lemma ([imath]B[/imath] follows Brownian motion):

[math]df(t,B) = \frac{\partial f}{\partial t}dt + \frac{\partial f}{\partial B}dB + \frac{1}{2} \frac{\partial^2 f}{\partial^2 B}dB^2,[/math]

but I do not understand why the derivation says that [imath]dB^2[/imath] does not vanish. I've read/watched a few sources (below) to figure out why it doesn't vanish and the one I'm currently stuck with is here. To be specific, I'm stuck on the line after equation (9):

[math]E[(B(t_{i+1}) - B(t_i))^2] = E[B(t_{i+1} - t_i)^2] = t_{i+1} - t_i.[/math]

How could the expected value of the difference squared be at all related to the time interval? Unless we

*explicitly*make the "jump size" proportional to the time intervals i.e. [imath]B(t_{i+1}) - B(t_i) \propto t_{i+1} - t_i[/imath].Almost all of my effort has been trying to understand the link between [imath]dt[/imath] and [imath]dB[/imath]; the intuition that I get is that [imath]dB[/imath] is in a loose sense some kind of random variable whose variance is explicitly made to be related to [imath]dt[/imath] so that when we square it we get nice relations. For example:

[math] B(t_{i+1}) - B(t_i) = \Delta B_i = \sigma \sqrt{\Delta t} X_i, \, \Delta t = t_{i+1} - t_i \\ P(X_i = 1) = P(X_i = -1) = 0.5 \\ E[X_i] = (1) \cdot 0.5 + (-1) \cdot 0.5 = 0 \\ E[X_i^2] = V[X_i], \text{because } V[X_i] = E[X_i^2 - E[X_i]] \text{ but } E[X_i] = 0 \\ V[X_i] = (1)^2 \cdot 0.5 + (-1)^2 \cdot 0.5 = 1 \\ E[(B(t_{i+1}) - B(t_i))^2] = E[\Delta B_i^2] \\ = E[(\sigma \sqrt{\Delta t} X_i)^2] \\ = \sigma^2 \Delta t E[X_i^2] \\ = \sigma^2 (t_{i+1} - t_i) \\ = t_{i+1} - t_i \text{ if } \sigma = 1 [/math]

Really not sure. Also sorry my equations look bad I can't get \begin{align}\end{align} to work. Thanks a lot.

Sources:

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