1

Let $X = (X_t)_{t\geq 0}$ be a stochastic process defined by $$ d X_t = 3dt + 2 \sqrt X_t d B_t, $$ where $B$ is a standard Brownian motion. What is its quadratic variation? I have a subtle intuition that is defined by $d\langle X,X\rangle_t = 2\sqrt X_t d t$, but I do not know how to justify it correctly.

R__
  • 301
  • 1
  • 7
  • How is the process $X$ defined? It seems that there is missing an "$=$" somewhere. – mag Jan 15 '21 at 22:15
  • completely right, so sorry. Now is correct – R__ Jan 15 '21 at 22:18
  • 1
    The quadratic variation process of the finite variation part is zero. For the quadratic variation of the Ito integral you can see a proof here: https://math.stackexchange.com/questions/2818913/quadratic-variation-of-ito-integral/3937161#3937161 – UBM Jan 15 '21 at 22:20

1 Answers1

0

The quadratic variation of the process is given by $$\mathrm d \langle X,X\rangle_t=4 |X_t|\mathrm dt. $$ In general, for processes $Z_t=\int_0^tH_s\mathrm dX_t$ and $Q_t=\int_0^tK_s\mathrm dY_t$ holds $$\mathrm d \langle Z,Q\rangle_t=H_tK_t\mathrm d\langle X,Y\rangle_t, $$ for $X,\ Y$ semimartingales and $H,\ K$ adapted càglàd processes.

mag
  • 2,424