Brownian Motions

Random walk: X(t) denotes the position of the r.v. at time t.
The Brownian motion (BM) is a symmetric random walk with the step-size approaches zero.
Properties:
X(0) = 0
{X(t), t 0} has stationary and independent increments.
X(t) is normally distributed with mean = 0 and Variance = 2 t,
where X(t)
X(t) is everywhere continuous but nowhere differentiable

Standard BM:
B(t) = X(t) / 
 = 0 ,  = t
Conditional probability of BM:
The conditional distribution of X(s) given that X(t) = B is normal.
E[X(s) | X(t) = B] = (s/t)B
Var[X(s) | X(t) = B] = (s/t)(t – s)
Hitting times:
Ta = The first time the BM hits a.

The maximum value in [0,t]:

P{up A before down B} =

P{up B before down A} =

Examples:

P{Win $7 before lose $5} = 5/(5+7) = 5/12

P{Win $5 before lose $7} = 7/(5+7) = 7/12

Problem 10.5

P{T1T-1T2} = P{T1T-1} P{T-1T2 | T1T-1} = (1/2) (1/3) = 1/6

Arbitrage Theorem

Example: Pricing stock options (P. 608)
Assuming a stock is $100 per share now (t = 0), and it will worth either $200 or $50 after 1 time period (t = 1).
At t = 0 one can purchase the option of buying y shares at t = 1 at $150/share. The cost of this option is C  y and must be paid in advance at t = 0.
Goal: Set the price C so that you can make money either way.
x = units purchased at t = 0; y = options purchased at t = 1
x and y can be positive or negative (or 0):
x < 0: sell x units at t = 0; x > 0: buy x units at t = 0
y < 0: sell y option units at t = 0; y > 0 buy x option units at t = 0
At t = 0, you buy x units of stock and buy y units of options.
At t = 1
If price = $200, each option will worth ($200 – $150) = $50: Value = 200x + 50y
If price = $50, options are worthless: Value = 50x
Choose y such that the value is the same regardless of the stock price at t = 1. Equating the two values we have: y = -3x
The value at t = 1 is 50x for both scenarios
Now consider the gain:
Cost at t = 0: cost = 100x + Cy = 100x – 3xC
Gain at t = 1: gain = value – cost = 50 x – 100x + 3xC = x(3C – 50)
If x > 0, gain > 0 if x(3C – 50) > 0, or 3C > 50  C > 50/3 = 16 2/3.
If x < 0, gain > 0 if x(3C – 50) > 0, or 3C < 50  C < 50/3 = 16 2/3.
Let C = 20. If you buy 1 unit and sell 3 options at t = 0 (x = 1, y = -3):
Cost at t = 0: cost = 100x + Cy = 100x – 3xC = 40
If unit goes up to $200 at t = 1:

Value = 200x + 50y = 50 and gain = 50 – 40 = 10

If unit goes down to $50 at t = 1:

Value = 50x = 50 and gain = 50 – 40 = 10

Let C = 15. If you sell 1 unit and buy 3 options at t = 0 (x = -1, y = 3):
Cost at t = 0: cost = 100x + Cy = 100x – 3xC = -55 (gain)
If unit = $200 at t = 1: value = 200x + 50y = -50 (loss) and gain = -50 – (-55) = 5
If unit = $50 at t = 1: value = 50x = -50 and gain = -50 – (-55) = 5
This is the so-called sure-win strategy.

The arbitrage theorem – One of the following will happen:
There exists a P that  Ep[ri(x)] = 0 (sure-win is not possible)
There exists a P that (sure-win is possible)
The general solution is very difficult to solve (linear programming).
For the special case: Only one outcome can be chosen, sure-win is possible if

where Oi is the odd for outcome i.

Example: O1 = 1, O2 = 2, O3 = 3:  Sure-win is possible. For no sure-win: O1 = 1, O2 = 2, O3 = 1/ (1 – 1/2 – 1/3) – 1 = 5

White Noise

Introduction: A white noise process is a random process that generates a time series with a constant spectral density across the entire frequency band. It is a non-physical process since it requires infinite energy.
White noise transformation: If {X(t), t 0} is a standard Brownian motion and f(t) is a smooth function, the following integral is called white noise transformation:

The result of this integration can be used to represent the output g(t) of a physical process (voltage, EM wave, force, …) f(t) passing through a white noise medium.

Question: Why would the output always average to zero? Answer: The white noise process has infinite energy, which can overwhelm any DC energy of the input signal.

Stationary and Weakly Stationary Systems

Introduction: One of the most important applications of probability theory is the analysis of random signals and systems. Here we will introduce the basic concept of analyzing the response of an LTI system if the input is a random process.
Stationary processes: A process is stationary iff its statistical properties remain unchanged under any time shift (time-invariant).
The requirement for a process to be stationary is rather stringent, which often renders the study of such process impractical. To relax the requirement so that the results can be applied to a wider range of physical systems and processes, we define a process called weakly stationary or second-order stationary process.
Weakly stationary processes: A process {X(t),  0} is weakly stationary if its first two moments remain unchanged under any time shift (time-invariant) and the covariance between X(t) and X(s) depends only on |t – s|.
If the input of an LTI system has a random part and the random portion is a stationary or weakly stationary process, we can use most of the regular techniques such as the impulse response, Fourier transform, and Laplace transform to solve the problems. Only minor modifications are required.
I will not go through the details in sections 10.7 and 10.8 since the notations used in the text are vastly different from the notations used in most EE textbooks. Let’s depart from the text at this point. I will provide handout for the subsequent lectures.