Consider the Itô process $\{X_t,t\ge0\}$ described by the stochastic differential equation $$dX_t = 0.10X_t dB_t + 0.25X_t dt.$$ Calculate the probability that $X_t$ is at least $5\%$ higher than $X_0$
- at time $t = 0.01$, and
- at time $t = 1$.
Source: University of Regina, Fall 2014, 441 (assignment #4) pg. 1
My Work So Far
We note the price of the stock follows a Brownian motion given by $$dX_t=0.25X_tdt+0.10X_tdB_t,$$ therefore we can calculate \begin{align*} X_t&=X_0e^{0.10B_t+(0.25-\frac{0.10^2}{2})t}\\ &=X_0e^{0.19B_t+0.245t}\\ \end{align*} If $Z\sim N(0,1)$, then \begin{align*} \mathbb{P}(X_t\gt1.05X_0)&=\mathbb{P}\left(B_t \geq\frac{\log(1.05)-0.245t}{0.10}\right)\\ &=\mathbb{P}\left(Z\ge\frac{\log(1.05)-0.245t}{0.10\sqrt{t}}\right)\\ \end{align*} We can conclude with
- if $t=0.01$, then \begin{align*} \mathbb{P}(X_{0.01}\geq1.05X_0)&=\mathbb{P}\left(Z\ge\frac{\log(1.05)-0.245(0.01)}{0.10\sqrt{0.01}}\right)\\ &=\mathbb{P}(Z\geq4.634)\\ &=1-\Phi(4.634)\\ &\approx0.000002.\\ \end{align*}
1-pnorm(4.634)
[1] 1.793335e-06
- if $t=1$, then \begin{align*} \mathbb{P}(X_{1}\geq1.05X_0)&=\mathbb{P}\left(Z\ge\frac{\log(1.05)-0.245(1)}{0.10\sqrt{1}}\right)\\ &=\mathbb{P}(Z\geq-1.962)\\ &=\Phi(1.962)\\ &\approx0.9751.\\ \end{align*}
pnorm(1.962)
[1] 0.9751188
Would this be correct? Is there a better way to calculus this?