# Stochasitc Processes

1. Feb 18, 2008

### wildman

Could someone give me a hint for this problem? I have no idea how to attack it.

The random variables A and B are independent $$N(0;\sigma)$$ and p is the probability that the process X(t) = A - Bt crosses the t axis in the interval (0,T). Show that $$\pi p = arctan T$$. Hint: p = P{0 <= A/B <= T}.

Well I know that independent means:

$$f(x,y) = f_x(x) f_y(y)$$

Normal is equal to:

$$\frac{1}{\sqrt{2\pi {\sigma}^2}} e^{\frac{-x^2}{2 {\sigma}^2}}$$

The hint says that the answer is the probability that A/B is between 0 and T.

Last edited: Feb 18, 2008
2. Feb 18, 2008

### John Creighto

Why not start by writing the probability distribution for x(t) or trying to show the hit the question gave you.

3. Feb 20, 2008

### wildman

Ok. Here is the answer. It was actually pretty much on page 187 of the textbook (Papoulis and Pillai, 4th edition):

The example is to show the ratio Z = X/Y has a Cauchy density centered at
$$\frac {r\sigma_1}{\sigma_2}$$ and X and Y are random variables with zero mean and a normal density.
After doing some math (which I am too lazy to copy), the book ends up with this formula (6-63) (of which I simplified since the random variables are independent and the variances are the same:)

$$F_z(z) = \frac{1}{2} + \frac{1}{\pi}arctan Z$$ where Z = A/B

From the hint we know that the probabilities are between 0 and T so

$$p = P[0<z<T] = F_z(T) - F_z(0)$$

so
$$p = \frac{1}{2} + \frac{1}{\pi}arctan T - (\frac{1}{2} + \frac{1}{\pi}arctan (0))$$

and from that the answer is obvious.

4. Feb 23, 2008

### uart

Hi wildman. I dont have the text you reference so I don't know what the "math which you're too lazy to copy" is, but there is a way to do this that doesn't require very difficult maths.

The best way to approach this type of problem is to identify the region (R) in the "BA" plane where A-Bt will have a zero-crossing. Once you've identified this region then the probability you seek is nothing more than the integral over the region R of the joint pdf. This integral can be difficult in some cases but fortunately in this case it's pretty trivial.

Since you want A-tB to have a zero-crossing for t in [0..T] then it follows that A and B must be of the same sign (since T and hence t is non-negative). This means that the regions can only be in quadrants I and III.

It doesn't take much to find that the region we require in quadrant-I is A<BT. This is the area between the straight line A=BT and the B axis. Since A=BT is a straight line with gradient T then from basic geometry this region is a sector in the "BA plane" of angle $$\theta = \tan^{-1}(T)$$. Obviously there is a similar region in quadrant-III where |A|<|B|T and again this second region is a sector of angle $$\,\tan^{-1}(T)$$.

So the probablity we require is the integral of the joint pdf over the union of these regions. Since the joint pdf has radial symmetry and it's total integral over 0..2Pi is unity, then the integral over each sector of angle theta is simply $$\theta / 2 \pi$$ and the total integral over both (quadrant I and III) sectors is $$\theta / \pi$$.

That is,

$$p = \frac{\theta}{\pi} = \frac{\tan^{-1}(T)}{\pi}$$

Last edited: Feb 23, 2008