I am stuck in solving the following SDE, which looks suspiciously simple, but I am unable to get any answer! Any suggestions would be very helpful! It is:

$dX(t) = \frac{1}{X(t)^3} dW(t)$,

where $W(t)$ is the standard Wiener process.


Let the process $Y_t$ defined $Y_t = X_t^8$, then we have that

$$ dY_t=8 X_t^7dX_t+\frac{56}{2} X_t^6 d\langle X\rangle_t $$

Knowing that $d\langle X\rangle_t = \frac{1}{X_t^6}dt$ and $X_t^7dX_t = X_t^4 dW_t = \sqrt{Y_t}dW_t$

$$ dY_t=8 \sqrt{Y_t} dW_t+28 dt $$

So that $Y_t$ follows a Cox-Ingersoll-Ross diffusion of the general form $dY_t=(a-kY_t)dt+\sigma \sqrt{Y_t}dW_t$ and with parameters $a = 28, k = 0, \sigma =8$.

Since $0 \leq 2a\leq \sigma^2 $, the process $Y_t$ is non-negative as long as $Y_0 \geq 0$ and we can define two possible solutions for $X_t$ as

$$ X_t = Y_t^{\frac{1}{4}} \quad\text{or}\quad X_t = -Y_t^{\frac{1}{4}} $$

  • $\begingroup$ Hi. Why have you defined the process to be Y_t = X_t^8? $\endgroup$ – Thomas Moore Oct 10 '17 at 23:37
  • $\begingroup$ Hello, because it's the only power of X such that you can get rid of X terms in the drift part. $\endgroup$ – Tom Mike Oct 11 '17 at 9:37

You can use Ito's lemma. Write $X_t=f(t,W_t)$ and get some PDEs for $f$.


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