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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.

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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 '11 at 10:03

2 Answers 2

up vote 3 down vote accepted

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 $|a| \neq 1$.

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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 '11 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 '11 at 13:02
@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 '11 at 16:20
Why is it obvious? –  Ilya Jan 5 '11 at 9:46
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 '11 at 11:13


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.


Reference for Lamperti's Transform : Theorem 2 in

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I think, my question is not correct, I will reask it. –  Ilya Jan 4 '11 at 10:09
Does a deterministic time change change the filtration of Brownian motion? I think only a stochastic time change by a stopping time does that. –  Tim van Beek Jan 4 '11 at 10:13
Well what I meant was that $Y_t$ is a $\mathcal{F}_{at}$-measurable random variable as $X_t$ is $\mathcal{F}_{t}$-measurable (where $\mathcal{F}_t$ is the usual augmentation of the Brownian filtration generated by $(W_t)_{t>=0}$). So, I don't see how in a strong sense a change of measure can address the problem, but maybe there are others ways to treat the problem (for example in a weak sense of equality in law but I just can't see it). Regards –  The Bridge Jan 4 '11 at 12:29
Sure, I understand - it's a very nice correction, thanks. I mean that I need to find change of measure which makes the same change of distributions, i.e. $Q_X\to Q_Y$ such that $$ dX_t = \sigma dw_t\to dY_t = \sqrt{a}\sigma dw_t. $$ I ask this question as a new question… so I will be happy if can forward our discussion to the new question. –  Ilya Jan 4 '11 at 13:05

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