0

I need to calculate for Brownian motion $dBt^2$ and $dBt^4$. For $dBt^2$, I think that it is correct to use Ito's lemma.

So we get: $$ dBt^2=(1/2)\cdot 2 \cdot dt +2\cdot Bt\cdot dBt$$

Is this correct?

What about $dBt^4$?

Sebastiano
  • 7,649
what_456
  • 147
  • 1
    Your formula is correct. What about using Ito's lemma with the function $f(x) = x^4$? – Rik93 Jun 21 '20 at 11:09
  • So it is $ dBt^4 = 4B_t^3 dBt+1/2 \cdot 12 \cdot t^2 dt ? $ – what_456 Jun 21 '20 at 11:30
  • Almost! It should be $dB_t^4 = 4B_t^3dB_t + \frac12\cdot12\cdot B_t^2dt$. You have to evaluate the second derivative in $B_t$. – Rik93 Jun 21 '20 at 12:56
  • @Rik93 I need to calculate $dSt^4$, where St is geometric Brownian motion. How about that? – what_456 Jun 21 '20 at 15:11
  • Have you also seen Ito's lemma for general processes? And what definition do you use for geometric Brownian motion? – Rik93 Jun 22 '20 at 07:36

0 Answers0