# Help integrating the transition probability of the Brownian Motion density function.

1. Problem:

Given the Brownian Motion with Drift: $$dx = \mu \, dt+\sigma \, dW$$ It can be shown that the transition density function is the following: $$p(x, t) = \frac{e^{-\frac{(x-\mu t)^2}{2t\sigma^2}}}{\sqrt{2\pi} \sqrt{t\sigma^2}}$$

Therefore, If I want to find the cumulative probability for some range:

Example: $$\mu=0.05, \sigma=0.5$$ $$Pr(T \leq 10, \ -\infty \leq X \leq 5) = \int_0^{10}\int_{-\infty}^5 \frac{e^{-\frac{(x-\mu t)^2}{2t\sigma^2}}}{\sqrt{2\pi} \sqrt{t\sigma^2}}\,dx\,dt = 9.997$$

• I get values that are greater than 1; therefore, I am definitely wrong. This is not a probability.

So, I have some questions:

1. What is the value that I am getting?
2. How do I read $Pr(T \leq t, \ X \leq x)$?

Simulated Paths for the Diffusion (Equation 1, above):

Contour Plot of the Transition Probability Function:

What basic probability questions can be answered by inferring from the transition probability density?

2. Follow up question:

What if there was a threshold where the paths of the diffusion are being killed - doesn't the time become a random variable? i.e. A book I am reading provides the following formula for the probability of trajectories that are killed:

$$Pr(T < t \mid y) = \int_{0}^{\infty }\int_{\Omega }k(x)p(t, x\mid y)\,dx\,dt \\$$

Example:

$$k(x)=\lambda x^2, \ where \ \lambda=0.001 \\$$

Diffusion paths crossing the killing function $k(x)$ (Red line)

This is where my confusion about having a random variable $T$ comes from.

So how can I relate this conclusion with the information below about the transition probability density and the double integral on time?

3. Lessons Learned (Needs Verification)

Given the Brownian Motion with Drift:

$$dx = \mu dt+\sigma dW \\$$

i. Diffusion without Killing:

The Transition Probability Function defined as: $$p(x,t)dx=P(x(t)\in(x,x+dx)) \\$$ ... is obtained by solving the Forward Kolmogorov PDE: $$\frac{\partial p(t, x)}{\partial t} = -\frac{\partial\mu p(x, t)}{\partial x}+\frac{1}{2}\frac{\partial^{2}\sigma p(x, t)}{\partial x^{2}} \\ p(x, 0) = \delta (x-x_{0}) \\$$

... Inference: $$Pr(X \leq x) = \int_{-\infty }^{x}p(x, t)\,dx \\$$

ii. Diffusion with Killing:

The Transition Probability Function defined as: $$p(x,t)dx=P(x(t)\in(x,x+dx),T>t)$$

... is obtained by solving the Backward Kolmogorov PDE: $$\frac{\partial p(x, t)}{\partial t} = -k(x)p(x, t) + \frac{\partial\mu p(x, t)}{\partial x} +\frac{1}{2}\frac{\partial^{2}\sigma p(x, t)}{\partial x^{2}} \\ p(x, 0) = \delta (x-x_{0})$$ $$and \ BCs$$ ... Inference: $$Pr(T \leq t, X \leq x) = \int_{0}^{t}\int_{-\infty }^{x}k(x)p(t, x)\,dx\,dt$$

• "the cumulative probability for some range" Ambiguous expression, please explain what you mean. "$Pr(T \leq 10, \ -\infty \leq X \leq 5)$" ?? $T$ and $X$ are not even defined, please explain what you mean. The integral you are computing is $E(L)$, where $$L=\int_0^{10}\mathbf 1_{X_t<5}dt$$ is the time spent by the process $(X_t)$ below $5$ between times $0$ and $10$. – Did Nov 26 '15 at 7:57
• What is your question in the end? The formula in my first comment yields $$E(L)=\int_0^{10}\int_{-\infty}^5p_t(x)dxdt,$$ and, obviously, $E(L)$ is not constrained to be $\leqslant1$, only $$E(L)\leqslant\int_0^{10}\int_{-\infty}^\infty p_t(x)dxdt=\int_0^{10}1\cdot dt=10.$$ – Did Nov 27 '15 at 6:32
• Yes E stands for expectation. Re your edit, please indicate your source for the formula with $P(T<t\mid y)$ (at present the RHS does not depend on $t$). – Did Nov 28 '15 at 8:59
• You misread: the LHS is $P(T<\tau)$ where $T$ and $\tau$ are both random (killing time and time of absorption by the boundary, respectively), as explained on page 23. – Did Nov 28 '15 at 14:27
• First the easy part: $P(T<\tau)$ is also $P((T-\tau)<0)$ hence this fits the notations you learned (but I should mention this is the first time I meet somebody to whom $P(X<Y)$ with $X$ and $Y$ random variables, is a problem). Now the real beef: you misled everybody by mentioning the Brownian density $p(x,t)$ at the beginning of your post since the paper you take this from, define $$p(x,t)dx=P(x(t)\in(x,x+dx),T>t,\tau>t),$$ while the PDF you wrote is $$\gamma(x,t)dx=P(x(t)\in(x,x+dx)).$$ To see the difference, note that, for every $t$, $$\int_\Omega\gamma(x,t)dx=1$$ while ... – Did Nov 28 '15 at 15:08

What you write doesn't make sense and I think you have a lot of misunderstandings. Your double integral makes no sense, I don't know where you got that. You don't define $T$ or $X$. I have no idea where you get the idea of "temporal" or "spatial" random variable. Let me try to clear things up for you.

$X_t$ is a collection of random variables. Each $t$ represents a single random variable. $t$ is a parameter. The probability density function is as you say, $p(x, t) = \frac{e^{-\frac{(x-\mu t)^{2}}{2t\sigma ^{2}}}}{\sqrt{2\pi }\sqrt{t\sigma ^{2}}}$. However this doesn't mean what you think it means. $t$ just indexes a collection of random variables. There is no such thing as the "temporal part" or "spatial part". This is NOT a bivariate density function. $t$ is a parameter just like $\sigma$ and $\mu$.

If you want to find $P(X_t \leq x)$, then just compute:

$$P(X_t \leq x)=\int_{-\infty}^x \frac{e^{-\frac{(y-\mu t)^{2}}{2t\sigma ^{2}}}}{\sqrt{2\pi }\sqrt{t\sigma ^{2}}} dy$$

Your outcome will depend on $t$, because $X_t$ depends on $t$.

Edit, for another example consider the Poisson random variable $X_{\lambda}$ indexed by $\lambda$. What $P(X_{\lambda}=1)$? Well, $\lambda e^{-\lambda}$. Of course this depends on $\lambda$ as $X_{\lambda}$ depends on $\lambda$.

The main difference between the Brownian motion (with drift) and this Poisson example, is typically with Poisson random variables you pick a $\lambda$ and stay with it, where with Brownian motion you care about the evolution over $t$. However $t$ is still just some parameter.

If you really wanted to be specific, you could write $X_{t,\mu,\sigma}$ although you don't do this because $\mu$ and $\sigma$ are fixed throughout the problem. This is why you don't see $X_{\lambda}$. That's because $\lambda$ is typically fixed.

• This post answered my original question. – Edv Beq Dec 29 '15 at 0:47