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Taking the derivative, [itex]d[/itex], of both sides of the equation.

Suppose that [itex]B(t)[/itex] is a Wiener process. Suppose that the following equation is true:

[itex]B(t)(t + \frac13 B(t)) = B(t)^{0.5}[/itex].

I've conjured this equation out of thin air (it's probably not true) to ask the following question. Does the above identity (assuming it's correct) enable us to write the following:

[itex]dB(t)dt + \frac13 dB(t) = dB(t)^{0.5}[/itex]

?
 

SammyS

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Suppose that [itex]B(t)[/itex] is a Wiener process. Suppose that the following equation is true:

[itex]B(t)(t + \frac13 B(t)) = B(t)^{0.5}[/itex].

I've conjured this equation out of thin air (it's probably not true) to ask the following question. Does the above identity (assuming it's correct) enable us to write the following:

[itex]dB(t)dt + \frac13 dB(t) = dB(t)^{0.5}[/itex]

?
In a word, no.

Use the product rule.
 

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