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As the title says, I need to transform Stratonovich SDEs to Ito form. I get similar results for some, but very different results in others. How do I do this? Thanks a lot!

A) Stratonovich $dX=-X^2dt+2X\circ dW$ to Ito.

Attempt:

I believe that for a Stratonovich SDE $$dX=\left(a-\frac{1}{2}b'b\right)dt+b\circ dW$$ its Ito form will be $$dX=a dt+b dW$$

So I got that $b=2X$ and $a+\frac{1}{2}\times 2\times2X=-X^2$, so $a=-X^2-2X$ and the SDE in Ito form would be $$dX=(-X^2-2X)dt+2XdW$$

However, the actual result is: $$dX=(-X^2+X)dt+2XdW$$

I have the same problem with other conversions so I must be doing it wrong?

s1047857
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2 Answers2

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using your transformation formula (there is a $-$ instead of a $+$) (could you please give a reference?), then you just get:

$$ dX=\left(a-\frac{1}{2}b'b\right)dt+b\circ dW=dX=-X^2dt+2X\circ dW $$ and therefore we get $$ a+\frac{1}{2}b'b=-X^2 \text{ with } b=2X,b'=2 $$ which gives us $$ a=-X^2+\frac{1}{2}b'b=-X^2+2X $$ which again gives then the Ito representation $$ dX=a dt+b dW=dX=(-X^2+2X)dt+2XdW $$

user190080
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  • The transformation formula wasn't in my lecture notes, but I googled it and found one here: https://www.pims.math.ca/files/monahan_2d.pdf (on the second page)

    OK, thanks a lot! One more questoion if you don't mind... What is the formula for differentiating $b=2X$ to obtain $b'=-1$ (in general)?

     – s1047857 Jul 14 '15 at 21:25
  • @user133971 sorry, I just see my mistake, I will edit it right now. It seems your result is almost correct (sign error) and your prof did the mistake! – user190080 Jul 14 '15 at 21:32
  • OK, thanks a lot! :) – s1047857 Jul 14 '15 at 21:35
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    @user133971 I checked the slides and indeed it is $a-1/2b'b$ instead of $a+1/2b'b$ which is still not the desired result – user190080 Jul 14 '15 at 21:39
  • Thanks! I've checked the answers for the other SDEs in the exercises and they would be correct with the new formula... Sorry, I copied it wrong. However, my professor made a mistake in the SDE in part A. – s1047857 Jul 14 '15 at 21:48
  • @user133971 I am sure he did, since there is actually no way that we both calculated the wrong transformation, it is just plug in (thanks for the reference anyway) – user190080 Jul 14 '15 at 21:55
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Let $X$ be a stochastic process such as: $$dX = udt + \sigma(t,\omega)\circ dW$$

to convert to its Ito form, the formula is: $$dX = udt + \sigma(t,\omega)dW + \frac{1}{2}\sigma'(t, \omega)\sigma(t,\omega)dt$$

Your stochastic process is: $$dX = -X^2dt+2X\circ dW$$

So $\sigma(t,\omega) = 2X$

Then, its Ito derivative would be: $$dX = -X^2dt+2XdW + \frac{1}{2}(2)(2X)dt$$ $$dX = -X^2dt+2XdW + 2Xdt$$ $$dX = (-X^2 + 2X)dt+2XdW$$

Broken_Window
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