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Let $(\Omega,\mathcal{F},\mathbb{F},P)$ be a filtered probability space and $X=(X_t)_{t\geq 0}$ be a Brownian motion with respect to the natural filtration $(\mathcal{F}_t)_{t\geq 0}$ generated by $X$.

Define $$Y_t=X_t+\mu t$$ for $t\geq 0$ and a measure $Q_t$ on $\mathcal{F}_t$ by $$Q_t(A)=E[exp(\mu X_t-\mu^2\frac{t}{2}),A]$$ for $t\geq 0$.

I have a proof that $Q\circ Y^{-1}=P\circ X^{-1}$ holds, but have a few questions about it.

The aim is to show that $$E_Q[f_1(X_{t_1})\ldots f_n(X_{t_n})]=E_P[f(X_{t_1}+\mu t_1)\ldots f_n(X_{t_n}+\mu t_n)]$$ for every bounded and continuous $f_1,\ldots,f_n$ and $t_1\leq\ldots\leq t_n$.

We have $E_Q[f(X_t)|\mathcal{F_s}]=E_P[f(X_t)exp(\mu X_t-\mu^2\frac{t}{2})|\mathcal{F}_s]=\ldots=E_P[f(X_t+\mu(t-s))|\mathcal{F}_s]exp(\mu X_s-\mu^2\frac{s}{2})$ (the first equation is not perfectly clear, obviously we are using the the formula how conditional expectations behave under change of measure as can be seen here, but this would result in: $E_Q[f(X_t)|\mathcal{F}_s]=\frac{1}{exp(-\mu X_s+\mu^2\frac{s}{2})}E[f(X_t)exp(\mu X_t-\mu^2\frac{t}{2})|\mathcal{F}_s]=E[f(X_t)exp(\mu(X_t+X_s)-\mu^2\frac{t-s}{2})|\mathcal{F}_s]$)

How can we use this to compute $$E_Q[f_1(X_{t_1})\ldots f_n(X_{t_n})]=E_P[f(X_{t_1}+\mu t_1)\ldots f_n(X_{t_n}+\mu t_n)]$$?

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  • $\begingroup$ should it be $Y_t = X_t+\mu t$? $\endgroup$ – Gordon Oct 19 '16 at 15:38
  • $\begingroup$ Yes, sorry. We get a drift term which depends on t. $\endgroup$ – peer Oct 19 '16 at 15:41
  • $\begingroup$ In addition, is $X_t=B_t$? $\endgroup$ – Gordon Oct 19 '16 at 15:44
  • $\begingroup$ Yes, i changed it. $\endgroup$ – peer Oct 19 '16 at 15:48
  • $\begingroup$ I was able to verify the equation with the conditional expectations by using independence of $X_t-X_s$ from $\mathcal{F}_s$ and momemt generating functions for normal distributed random variables. $\endgroup$ – peer Oct 19 '16 at 15:55
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Note that, for $0 \le s < t$, \begin{align*} E_P\left(f(X_t) e^{\mu (X_t-X_s) -\frac{\mu^2}{2}(t-s)} \mid \mathcal{F}_s \right) &= E_P\left(f(X_s + X_t-X_s) e^{\mu (X_t-X_s) -\frac{\mu^2}{2}(t-s)} \mid \mathcal{F}_s \right)\\ &=\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}} f(X_s + x\sqrt{t-s})e^{\mu\sqrt{t-s}x -\frac{\mu^2}{2}(t-s)}e^{-\frac{x^2}{2}}dx\\ &=\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}} f(X_s + x\sqrt{t-s})e^{-\frac{(x-\mu\sqrt{t-s})^2}{2}}dx\\ &=\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}} f(X_s + \mu(t-s) + x\sqrt{t-s})e^{-\frac{x^2}{2}}dx\\ &=E_P\Big(f(X_s +\mu(t-s) + X_t-X_s) \mid \mathcal{F}_s \Big). \end{align*} Let $t_0=0$. Then, \begin{align*} &\ E_Q\left(\prod_{i=1}^n f_i(X_{t_i})\right)\\ =&\ E_P\left(\prod_{i=1}^n f_i(X_{t_i}) e^{-\frac{\mu^2}{2}t_n + \mu X_{t_n}} \right)\\ =&\ E_P\left(\prod_{i=1}^{n-1} f_i(X_{t_i})e^{-\frac{\mu^2}{2}t_{n-1} + \mu X_{t_{n-1}}} E_P\left( f_n(X_{t_n}) e^{-\frac{\mu^2}{2}(t_n-t_{n-1}) + \mu (X_{t_n}-X_{t_{n-1}}) } \mid \mathcal{F}_{t_{n-1}}\right)\right)\\ =&\ E_P\left(\prod_{i=1}^{n-1} f_i(X_{t_i})e^{-\frac{\mu^2}{2}t_{n-1}+ \mu X_{t_{n-1}}} E_P\left( f_{n}\Big(X_{t_{n-1}} + \mu (t_n-t_{n-1})+ X_{t_n}-X_{t_{n-1}}\Big)\mid \mathcal{F}_{t_{n-1}}\right) \right)\\ =&\ E_P\left(\prod_{i=1}^{n-1} f_i(X_{t_i})e^{-\frac{\mu^2}{2}t_{n-1}+ \mu X_{t_{n-1}}}f_{n}\Big(X_{t_{n-1}} + \mu (t_n-t_{n-1})+ X_{t_n}-X_{t_{n-1}}\Big)\right)\\ =&\ E_P\Bigg(\prod_{i=1}^{n-2} f_i(X_{t_i})e^{-\frac{\mu^2}{2}t_{n-2}+ \mu X_{t_{n-2}}} E_P\bigg(f_{n-1}(X_{t_{n-1}})f_{n}\Big(X_{t_{n-1}} + \mu (t_n-t_{n-1})+ X_{t_n}-X_{t_{n-1}}\Big)\\ &\qquad e^{-\frac{\mu^2}{2}(t_{n-1}-t_{n-2}) + \mu (X_{t_{n-1}}-X_{t_{n-2}}) } \mid \mathcal{F}_{t_{n-2}}\bigg)\Bigg). \end{align*} Moreover, \begin{align*} &\ E_P\bigg(f_{n-1}(X_{t_{n-1}})f_{n}\Big(X_{t_{n-1}} + \mu (t_n-t_{n-1})+ X_{t_n}-X_{t_{n-1}}\Big) e^{-\frac{\mu^2}{2}(t_{n-1}-t_{n-2}) + \mu (X_{t_{n-1}}-X_{t_{n-2}}) } \mid \mathcal{F}_{t_{n-2}}\bigg)\\ =&\ E_P\bigg(f_{n-1}(X_{t_{n-2}}+X_{t_{n-1}}-X_{t_{n-2}})\\ &\ f_{n}\Big(X_{t_{n-2}}+X_{t_{n-1}}-X_{t_{n-2}} + \mu (t_n-t_{n-1})+ X_{t_n}-X_{t_{n-1}}\Big)e^{-\frac{\mu^2}{2}(t_{n-1}-t_{n-2}) + \mu (X_{t_{n-1}}-X_{t_{n-2}}) } \mid \mathcal{F}_{t_{n-2}}\bigg)\\ =&\ \frac{1}{2\pi}\iint_{\mathbb{R}^2}f_{n-1}(X_{t_{n-2}}+\sqrt{t_{n-1}-t_{n-2}}x) f_n(X_{t_{n-2}}+\sqrt{t_{n-1}-t_{n-2}}x +\mu (t_n-t_{n-1})\\ &\qquad + \sqrt{t_n-t_{n-1}}y)e^{-\frac{\mu^2}{2}(t_{n-1}-t_{n-2}) +\mu \sqrt{t_{n-1}-t_{n-2}}x} e^{-\frac{x^2+y^2}{2}} dx dy\\ =&\ \frac{1}{2\pi}\iint_{\mathbb{R}^2}f_{n-1}(X_{t_{n-2}}+ \mu(t_{n-1}-t_{n-2}) +\sqrt{t_{n-1}-t_{n-2}}x)\\ & \quad f_n(X_{t_{n-2}}+ \mu(t_{n-1}-t_{n-2})+\sqrt{t_{n-1}-t_{n-2}}x +\mu (t_n-t_{n-1})+ \sqrt{t_n-t_{n-1}}y)e^{-\frac{x^2+y^2}{2}} dx dy\\ &=E_P\bigg(f_{n-1}(X_{t_{n-2}} + \mu(t_{n-1}-t_{n-2}) + X_{t_{n-1}}-X_{t_{n-2}})\\ &\qquad f_{n}\Big(X_{t_{n-2}} + \mu(t_{n-1}-t_{n-2}) + \mu (t_n-t_{n-1})+ X_{t_{n-1}}-X_{t_{n-2}}+ X_{t_n}-X_{t_{n-1}}\Big) \mid \mathcal{F}_{t_{n-2}}\bigg). \end{align*} Therefore, \begin{align*} &\ E_Q\left(\prod_{i=1}^n f_i(X_{t_i})\right)\\ =&\ E_P\Bigg(\prod_{i=1}^{n-2} f_i(X_{t_i})e^{-\frac{\mu^2}{2}t_{n-2}+ \mu X_{t_{n-2}}} E_P\bigg(f_{n-1}(X_{t_{n-2}} + \mu(t_{n-1}-t_{n-2}) + X_{t_{n-1}}-X_{t_{n-2}})\\ &\qquad f_{n}\Big(X_{t_{n-2}} + \mu(t_{n-1}-t_{n-2}) + \mu (t_n-t_{n-1})+ X_{t_{n-1}}-X_{t_{n-2}}+ X_{t_n}-X_{t_{n-1}}\Big) \mid \mathcal{F}_{t_{n-2}}\bigg)\Bigg)\\ =&\ E_P\Bigg(\prod_{i=1}^{n-2} f_i(X_{t_i})e^{-\frac{\mu^2}{2}t_{n-2}+ \mu X_{t_{n-2}}} f_{n-1}(X_{t_{n-2}} + \mu(t_{n-1}-t_{n-2}) + X_{t_{n-1}}-X_{t_{n-2}})\\ &\qquad f_{n}\Big(X_{t_{n-2}} + \mu(t_{n-1}-t_{n-2}) + \mu (t_n-t_{n-1})+ X_{t_{n-1}}-X_{t_{n-2}}+ X_{t_n}-X_{t_{n-1}}\Big) \Bigg)\\ =& \ \cdots\cdots\\ =&\ E_P\left(\prod_{i=1}^n f_i\left(X_{t_{0}} + \sum_{j=1}^i\mu\left(t_j-t_{j-1}\right) + \sum_{j=1}^i \left(X_{t_j}-X_{t_{j-1}}\right)\right) \right)\\ =&\ E_P\left(\prod_{i=1}^n f_i(X_{t_i} + \mu t_i) \right). \end{align*}

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  • $\begingroup$ And how can i proof the martingale property out of this? $\endgroup$ – peer Oct 25 '16 at 1:08
  • $\begingroup$ I mean the martingale property of $Y$ w.r.t. $Q$? $\endgroup$ – peer Oct 25 '16 at 7:18
  • $\begingroup$ I would treat that as another question. $\endgroup$ – Gordon Oct 25 '16 at 12:41
  • $\begingroup$ I added a question here: math.stackexchange.com/questions/1984847/… $\endgroup$ – peer Oct 25 '16 at 19:20

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