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What is the solution of the SDE $$dX_t = 3dt + 2\sqrt{X_t} dB_t$$ when $X_0=0$ and B is a one dimensional brownian motion?

I know what the solution is when the dimension of the brownian motion is the same as the dimension of the bessel process but how do you find the solution when the dimensions are not equal?

1 Answers 1

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Skipping all technicalities.

Ito's formula gives:

$d f(X_t) = f'(X_t) d X_t + \frac 12 f''(X_t) d \langle X\rangle_t$

Here $$dX_t = adt + b \sqrt{X_t} d B_t,$$

(with $a=3,b=2$).

Here $d \langle X\rangle_t =b^2 X_t dt$. Now let's take $f(x) =\sqrt{t}$, and let $Y_t = f(X_t) = \sqrt{X_t}$. Then

\begin{align*} d Y_t &=d f(X_t)\\ & = \frac{1}{2\sqrt{X_t}}(a dt + 2\sqrt{X_t} d B_t) - \frac 18 \frac{1}{\sqrt{X_t^{3}}} b^2 X_t dt\\ & = d B_t -(\frac a2 - \frac {b^2}{ 8})\frac{1}{\sqrt{X_t}} dt\\ & = d B_t + \left (a - \frac{b^2}{4}\right) \frac{1}{2Y_t} dt. \end{align*}

Since Bessel process of dimension $r$ satisfies

$$ d \rho_t = d B_t + (r-1) \frac{1}{2\rho_t} dt,$$

It follows that $Y$ is Bessel process of dimension $a-\frac{b^2}{4}+1$. When $a=3$ and $b=2$, this means $Y$ is a Bessel process of dimension $3$, hence $X$ is the square of a Bessel process of dimension $3$.

  • 0
    But what is the closed form solution of the above SDE? – 2017-01-12
  • 0
    Xt is the square of the norm of 3d Bm – 2017-01-12
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    Sorry, I do not understand, the Brownian motion is one dimensional – 2017-01-12
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    The norm of d-dimensional BM is a one dimensional process that solves an SDE driven by 1d BM. Take a look at revuz and yor 3rd edition p. 439 (chapter xi, section 1). – 2017-01-12
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    (+1 ) , typo : $f(x)=\sqrt{x}$ – 2017-01-12