Timeline for Change of time or change of measure
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 8, 2011 at 14:58 | vote | accept | SBF | ||
| Jan 6, 2011 at 8:07 | comment | added | SBF | Yes, it's obvious, you're right, thank you. | |
| Jan 5, 2011 at 11:13 | comment | added | Alekk | let $[Y]_t$ denotes the quadratic variation of the process $Y$ between $0$ and $t$. Then the event $A=\{ \omega : [\omega]_T = a \sigma^2 T\}$ satisfies $\mathbb{Q}_Y(A)=1$ while $\mathbb{Q}_X(A)=0$. | |
| Jan 5, 2011 at 9:46 | comment | added | SBF | Why is it obvious? | |
| Jan 4, 2011 at 16:20 | comment | added | Alekk | @Tim van Beek: Y is solution of $dY = \sqrt{a} \sigma dW_t$, not $dY = \sqrt{a} \sigma dY$. Because the quadratic variation of Y on [0;T] is almost surely equal to $a \sigma^2$, and the quadratic variation of X is almost surely equal to $\sigma^2$, this is obvious that the measure are mutually singular. | |
| Jan 4, 2011 at 13:02 | comment | added | SBF | Yes, it is extremely strange result that $$ \frac{dQ_Y}{dQ_X}=0 $$ if $|a|\neq 1$. You can mention that if we take 1a instead of a we should have that $$ \frac{dQ_Y}{dQ_X} =1/\frac{dQ_X}{dQ_Y} = \infty $$ - under the condition that your result is true of course. | |
| Jan 4, 2011 at 12:16 | comment | added | Tim van Beek | The measures of $dX_t = \sigma dW_t$ and $dY_t = \sqrt\alpha \sigma dY_t$ for nonzero constants $\alpha$ and $\sigma$ are absolutely continuous according to the Girsanov theorem... | |
| Jan 4, 2011 at 11:23 | history | answered | Alekk | CC BY-SA 2.5 |