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Consider simple diffusion $dX_t = \sigma dw_t$ and a parameter $a>0$ and $X_0=x$. Let us denote $Y_t = X_{at}$ - thus we made a change of time. Let us denote an original measure as $P$. How to find measure $Q$ such that the process $Y$ is obtained through the process $X$ not with the change of time but with a change of measure, i.e. $$ \mathcal{Law}(Y\text{ under }P) = \mathcal{Law}(X\text{ under } Q). $$ Maybe it is possible to find the density process $$ Z_t = \frac{dQ_t}{dP_t} $$ like for the case when we're looking for the martingale measure? Unfortunately, now it seems the Girsanov theorem is useless for my case.

flag	asked Jan 4 2011 at 8:00 Ilya 470●2●8

Can't one simply calculate the law of $Y_t$ using the Itô symmetry? You have $Y_t = X_0 + \sigma int_0^{\alpha t}$, therefore the increments $Y_t - Y_s$ are Gaussian with variance $\sigma^2 \alpha (t-s)$. When your original measure was Brownian motion $B_t$, X has the same law under Brownian motion as $Y$ has under $\frac{1}{\sqrt(\alpha)} B_t$. – Tim van Beek Jan 4 2011 at 10:03

Alekk · Accepted Answer · 2011-01-04 11:23:46Z UTC

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for example, if $\sigma$ is a constant, then $Y$ satisfies $dY_t = \sqrt{a} \sigma dW_t$ so that the law $\mathbb{Q}_Y$ and $\mathbb{Q}_X$ of the processes $Y$ and $X$ on the Wiener space $C([0;T],\mathbb{R})$ are generaly singular: in other words $\frac{d \mathbb{Q}_Y}{d \mathbb{Q}_X} = 0$ if $|a| \neq 1$.

answered Jan 4 2011 at 11:23

Alekk
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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... – Tim van Beek Jan 4 2011 at 12:16

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. – Ilya Jan 4 2011 at 13:02

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@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. – Alekk Jan 4 2011 at 16:20

Why is it obvious? – Ilya Jan 5 2011 at 9:46

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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 $\mathbb{Q}_X(A)=0$. – Alekk Jan 5 2011 at 11:13

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The Bridge · Answer 2 · 2011-01-04 09:54:38Z UTC

Hi,

I think what you are looking for is not attainable (at least I can't see how) because the time change changes the filtration of the processes you are studying.

Nevertheless you can get by the transform below an equivalent process that the one you get by time change. (from X_t=\sigma W_t getting $Y_t=dW_t$)

Under some conditions on the volatility of the diffusion you can (by so-called Lamperti's Transformation) get a process with unit volatility + drift term, then use change of measure technique (i.e. Girsanov) to get back to pure diffusion term and you are done.

NB: You can adapt Lamperti's transform (+ Girsanov) to get a process equal in law to your time change in more general case.

with this operation you still don't get the same law under different measure (because you have to transform the process first) but at least you stay within the same filtation.

Regards,

Reference for Lamperti's Transform : Theorem 2 in

http://www.imm.dtu.dk/English/Research/Scientific_Computing/People.aspx?lg=showcommon&id=271164

Change of time or change of measure

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