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Consider a linear SDE of the form:

$$dX_t = \alpha X_tdt + \beta dW_t,\\ X_t = x$$ I would like to determine an explicit solution, as well as that solution's distribution if possible.

My issue with the first task is that we are given $X_t = x$, and not, as usual, $X_0 = x$. With regards to the second task, I do believe such solutions will be .... normal? Log-normal?

EDIT: I found the solution to the first question, but only for $X_0 = x$. So I am wondering, do we just replace $t$ in the formula with $T-t$?

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  • $\begingroup$ Avoid writing $dX_t$ and $X_t$ with the same subscript, that is causing confusion. And you could write it as $T-t$ sure, note that in this example the drift and diffusion coefficients do not depend on time, and therefore this problem is equivalent to solving up to time $S = T - t$ the SDE with $X_0 = x$. $\endgroup$
    – Nadiels
    Feb 24, 2017 at 16:59
  • $\begingroup$ So you would say the solution is $exp( \alpha (T-t)) + \int_0^{T-t} exp( \alpha (T - t - s)) \beta dW_s$? Seems a but clunky. $\endgroup$
    – Jaood
    Feb 24, 2017 at 19:54
  • $\begingroup$ It does look clunky! My point was to, as you first suggested, replace any mention of $T-t $ with $S $ and solve from $s=0$ $\endgroup$
    – Nadiels
    Feb 24, 2017 at 20:18

1 Answer 1

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We follow Integrating factor and homogeneous equation for SDEs., but write the details for this specific case \begin{align*} \mathrm{d}X_t = a(t)X_t \mathrm{d}t + d(t)\mathrm{d}B_t. \end{align*}

We start with integrating factor

$$Y_t=\exp\left(\int_0^t a(s)ds\right).$$ For $Z_t:=X_tY_t^{-1}$ we use the product rule to get \begin{align} dZ_t=& X_t d(Y_t^{-1})+Y^{-1}dX_t+d\langle X, Y^{-1}\rangle_t\\ &= \frac{-X_t}{Y_t}a(t)dt+\frac{1}{Y_t}\big[a(t)X_t dt + d(t)dB_t\big]\\ &= d(t)Y_t^{-1}dB_t. \end{align}

Now integrate to obtain

$$Z_t=X_0\int_0^t d(s)Y_s^{-1}dB_s,$$

and so

$$X_t=Y_t\left(X_0\int_0^t d(s)Y_s^{-1}dB_s\right)$$

$$=\exp\left(\int_0^t a(s)ds\right).\left(X_0\int_0^t d(s)\exp\left(-\int_0^t a(s)ds\right).dB_s\right)$$

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