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I'm looking at this example in this lecture note (page 3, number iv):

Let $f(t,x)=t^2+x^2$, and $X_{t}=\mu t + \sigma B_{t}$, where $B_{t}$ is the standard brownian motion. We want to compute $df(t,X_{t})$. Using Ito's lemma, I got $(2t + 2\mu X_{t} + \sigma^2)dt + 2\sigma X_{t} dB_{t}$.

However, the answer stated in that note is $(2t + 2\mu + \sigma^2)dt + 2\sigma dB_{t}$, which basically differs from my solution by setting $X_{t}=1$.

I wonder why do we set $X_{t}=1$? Thanks.

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    $\begingroup$ Your solution is fine, it must be a typo in the note. Best regards. $\endgroup$ – TheBridge Nov 29 '15 at 22:42

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