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Let $B = (B_t)_{t\geq 0}$ be a standard brownian motion started at $0$. Consider the two following stochastic equations: \begin{equation} \begin{split} dX_t &=& (13 + 2X_t)\,dt + (6 + X_t)\,dB_t \\ dY_t &=& 2Y_t\,dt + Y_t\,dB_t \end{split} \end{equation} with $X_0 = Y_0 = 1$.

I am trying to show that the following identity in law holds: $$ X_t - Y_t\left(1 + 6\int_0^t\frac{1}{Y_s}\,dB_s \right) \stackrel{law}{=} 7\int_0^tY_s\,ds $$

Combining $X$ and $Y$, it can be shown through Ito's formula that the solution for $X$ is given by $$ X_t = Y_t\left(1 + 7\int_0^t\frac{1}{Y_s}\,ds + 6\int_0^t\frac{1}{Y_s}\,dB_s \right) $$

But I am having trouble operating with the first equation to work out the law identity, and the solution for $X$ is not really helping me. Any hint is most welcome, as I am a bit lost now. I suspect the solution might not be trivial.

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    $\begingroup$ So, basically your claim boils down to $$Y_t \int_0^t \frac{1}{Y_s} \, ds \stackrel{\text{law}}{=} \int_0^t Y_s \, ds, $$ right? $\endgroup$
    – saz
    Feb 13, 2015 at 11:56
  • $\begingroup$ yes, with a 7 multiplying the integral on the left of your identity also. Maybe now applying integration by parts? $\endgroup$
    – Adam
    Feb 13, 2015 at 13:27

1 Answer 1

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Note that the solution of the SDE, $\mathrm dY_t = 2Y_t\mathrm dt + \mathrm dB_t\ (\mbox{for }Y_0 = 1)$, is a collection of log-normal random variables,

$$ Y_t = e^{\frac{3}{2}t + B_t}~(\mbox{for }t\geqslant 0)\,. $$

Therefore, for $0\leqslant s\leqslant t$,

$$ \left\{\frac{Y_t}{Y_s}\right\}_{0\leqslant s\leqslant t}\quad \stackrel{\text{law}}{=}\quad \left\{Y_{s}\right\}_{0\leqslant s\leqslant t}\,.\tag{1} $$

proof:

For $n\in\mathbb{N}$ and $0\leqslant t_1<t_2<\ldots<t_n\leqslant t$, the time-reversal properties of Brownian motion justify the following:

$ \,\,\,P\left(\frac{Y_t}{Y_{t_1}}\leqslant y_1,\ \ldots,\ \frac{Y_t}{Y_{t_n}}\leqslant y_n\right)\\ \begin{eqnarray*} &=& P\left(\frac{3}{2}(t-t_1) + B_t - B_{t_1}\leqslant \log y_1,\ \ldots,\ \frac{3}{2}(t-t_n) + B_t - B_{t_n}\leqslant \log y_n \right) \\ &=& P\left( B_t - B_{t_1}\leqslant \log y_1 - \frac{3}{2}(t-t_1),\ \ldots,\ B_t - B_{t_n}\leqslant \log y_n - \frac{3}{2}(t-t_n) \right) \\ &=& P\left( -(B_{t-r_1}-B_t)\leqslant \log y_1 - \frac{3}{2}r_1,\ \ldots,\ -(B_{t-r_n}-B_t)\leqslant \log y_n - \frac{3}{2}r_n \right) \\ &=& P\left( B_{t-r_1}-B_t\leqslant \log y_1 - \frac{3}{2}r_1, \ldots, B_{t-r_n}-B_t\leqslant \log y_n - \frac{3}{2}r_n \right) \\ &=& P\left( B_{r_1}\leqslant \log y_1 - \frac{3}{2}r_1,\ \ldots,\ B_{r_n}\leqslant \log y_n - \frac{3}{2}r_n \right) \\ &=& P\left( \frac{3}{2}r_1 + B_{r_1}\leqslant \log y_1,\ \ldots,\ \frac{3}{2}r_n + B_{r_n}\leqslant \log y_n \right) \\ &=& P\left(Y_{r_1}\leqslant y_1,\ \ldots,\ Y_{r_n}\leqslant y_n\right), \end{eqnarray*} $

where we defined $r_i:=t-t_i\ .$

The equality in law, $(1)$, implies $$ \int_0^t \frac{Y_t}{Y_s} \, ds \stackrel{\text{law}}{=} \int_0^t Y_s \, ds\,, $$

which, from your application of Ito's lemma and as deduced by @saz, is the relationship to be proved.

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