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Consider the stochastic differential equation $\frac{dX_t}{X_t}=adt+bdW_t$ for the diffusion $X_t$ . The parameters $a,b$ are constant.Using Ito's lemma and suitable integration over $[0,T]$, show that $X_T=X_oexp[(a-\frac{1}{2}b^2)T+b\phi\sqrt T]$ where $\phi$~$N(0,1)$

My solution:

$X_t$ follows geometric brownian motion with $\mu=a$ and $\sigma=b$, so I used substitution the $F(X_t)=log(X_t)$ and used Ito's lemma for a geometric brownian and then integrated the result over $[0.T]$ to get $log(\frac{X_T}{X_0})=(a-\frac{1}{2}b^2)T+b(W_T-W_0)$ which gives me $X_T=X_oexp[(a-\frac{1}{2}b^2)T+b(W_T-W_0)]$ when taking exponentials on both sides.

Now I know that $(W_T-W_0)$ is normally distributed with mean $0$ and variance $T$ by definition of Brownian motion, can you explain how/why $W_T-W_0=b\phi\sqrt T$?

Vladimir Nabokov
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1 Answers1

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The solution of the SDE is $$X_T = X_0 \exp\left\{\left(a-\frac{1}{2}b^2\right)T+bW_T\right\}.$$ Since $W_T \sim N(0,T)$, we have for any $x$ \begin{align} \mathbb{P}(W_T \leq x) &= \mathbb{P}\left(\phi \leq \frac{x}{\sqrt{T}}\right) \\ &= \mathbb{P}\left(\sqrt{T} \phi \leq x\right), \end{align} where $\phi \sim N(0,1)$. Hence, $W_T$ and $\sqrt{T} \phi$ have the same distribution.

Cavents
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