Probability of stock price modeled by brownian motion with drift

Question

Given a stock price modeled by Brownian motion with drift with $u=1$ and $\sigma=1.5$. Assume the stock price is $40$ at $t=0$ and $60$ at $t=10$, find the probability that the stock price is between $40$ and $60$ at $t=5$?

My attempt: I computed $P(X(5)>40\ |\ X(0)=40) = 1 - \phi(\frac{-5}{\sqrt{11.25}})$ and $P(X(5)>60\ |\ X(0)=40) = 1 - \phi(\frac{15}{\sqrt{11.25}})$, so $P(X(5)\in (40,60)\ |\ X(0) = 40) = \phi(\frac{15}{\sqrt{11.25}}) - \phi(\frac{-5}{\sqrt{11.25}})$.

Now, using Brownian bridge, I also could compute $P(X(5)> 60\ |\ X(10)=60) = 1-\phi(\frac{59}{3\sqrt{2.5}})$, as well as $P(X(5)>40\ |\ X(10) = 60) = 1-\phi(\frac{19}{3\sqrt{2.5}})$, thus $P(X(5)\in (40,60)\ |\ X(10) = 60) = \phi(\frac{59}{3\sqrt{2.5}}) - \phi(\frac{19}{3\sqrt{2.5}})$.

Finally, since $X(10)$ and $X(0)$ are independent, $P(X(5)\in (40, 60)\ |\ X(0)=40, X(10) = 60) = P(X(5)\in (40, 60)\ |\ X(0)=40)\ P(X(5)\in (40, 60)\ |\ X(10)=60) = [\phi(\frac{59}{3\sqrt{2.5}}) - \phi(\frac{19}{3\sqrt{2.5}})][\phi(\frac{15}{\sqrt{11.25}}) - \phi(\frac{-5}{\sqrt{11.25}})]$

My question: Is the last step in my solution correct? On the other hand, I could not really combine the two given informations at $t=0$ and $t=10$, as we actually only need $1$ of them to figure out the required probability. I am confused why we need two given information rather than one. Could anyone please help explain the reason? Any thoughts on the solution above would be appreciated as well.

Answer 1

Let study $X_5$ and $X_{10}$. We have a condition on $X_0$(backward) and $X_{10}(forward)$.

We have that $X_t=X_0+ut+\sigma W_t$, and therefore

$E(X_5)=45$

$E(X_{10})=50$

$var(X_5)=\sigma^25$

$var(X_{10})=\sigma^210$ and finally $cov(X_5,X_{10})=\sigma^25$, or $corr(X_5,X_{10})=\sqrt{\frac{1}{2}}$

We want to know the distribution U defined as $X_5|{X_{10}=60}$ ; it is Gaussian too (https://en.wikipedia.org/wiki/Multivariate_normal_distribution) with mean m, and variance $v^2$ such as

$m=45+\frac{\sigma\sqrt{5}}{\sigma\sqrt{10}}\sqrt{\frac{1}{2}}(60-50)$

$v^2=(1-{\sqrt{\frac{1}{2}}}^2)\sigma^25$

Finally ,

$P(X(5)\in (40, 60)\ |\ X(0)=40, X(10) = 60)=P(U\leq60)-P(U\leq40)$

or

$P(m+vY\leq60)-P(m+vY\leq40)=\phi(\frac{60-m}{v})-\phi(\frac{40-m}{v})$

with Y normally distributed mean 0, variance 1