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Let $X_{t}:=(1-t)\int_{0}^{t}\frac{1}{1-s}dB_{s}$. This satisfies SDE:

$$dX_{t}=-\frac{X_{t}}{(1-t)}+dB_{t}$$

So the generator will be $A(f)=\frac{-x}{1-t}f'+\frac{1}{2}f''$ and so I think the pde for the transition density will be:

$$-\frac{x}{1-t}\frac{\partial }{\partial x}p(t,x)+\frac{1}{2}\frac{\partial^{2} }{\partial^{2} x}p(t,x)=\frac{\partial }{\partial t}p(t,x)$$

Is this any known equation? It looks like wave equation. Any ideas on how to solve it.

Thomas Kojar
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    Note that $(X_t)_{t \geq 0}$ is not a time-homogeneous process; therefore, the transition density does not only depend on $t$, but another (time) variable $s$ (i.e. $p(s,t,x)$.) – saz Oct 25 '14 at 06:31
  • Thanks. How would the pde for the transition density look like? – Thomas Kojar Oct 25 '14 at 17:48

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Note that $X_t=(1-t)Y_t$ for every $t$ in $(0,1)$, where $\mathrm dY_t=\mathrm dB_t/(1-t)$ hence $(Y_t)_t$ is a Brownian martingale such that $\mathrm d\langle Y\rangle_t=\mathrm dt/(1-t)^2$, that is, $\langle Y\rangle_t=t/(1-t)$. Thus, there exists a Brownian motion $(\beta_t)_t$ such that, for every $t$, $$Y_t=\beta_{A(t)}\qquad A(t)=\frac{t}{1-t}.$$ This proves that the transition probabilities of $(Y_t)_t$ are $$q_{s,t}(y,y')=g(A(t)-A(s),y'-y),$$ for every $s\lt t$, where $(g(t,y))_{t,y}$ is the standard Brownian semi-group of transition probabilities, that is, $$g(t,y)=\frac{1}{\sqrt{2\pi t}}\mathrm e^{-y^2/(2t)}.$$ Finally, the (inhomogenous) transition probabilities $(p_{s,t})_{s\lt t}$ of the Markov process $(X_t)_t$ are $$p_{s,t}(x,x')=\frac1{1-t}q_{s,t}\left(\frac{x}{1-s},\frac{x'}{1-t}\right)=\frac1{1-t}g\left(\frac{t}{1-t}-\frac{s}{1-s},\frac{x'}{1-t}-\frac{x}{1-s}\right).$$

Did
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