By Kardi Teknomo, PhD.

stochastic process

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Wiener Process

Random walk model has no parameter and it has no trend. How if the underlying process actually has trend (either increasing or decreasing) and some kind of volatility? Wiener process includes two parameters called drift rate \( \alpha \)  that represent trend and degree of volatility \( \sigma \). Similar to Random Walk, Wiener process also follow Markov property, thus we can still use equation \() x_t = x_{t-1} + \Delta x \). The change \( \Delta x \) is based on the following two properties:

1.       Small change in the variable occurs within a small time-interval according to \( \Delta x = N(0, \sigma \sqrt{t}) \)

2.       The changes in the variable \( \Delta x \) of any two time-periods \( \Delta t \)  are independent (it follow Markov process or random walk). That is \( x_t = x_0 + \sum_{i=1}^{n}N(0,\sigma \sqrt{\Delta t}) = x_0 + N(0,\sigma \sqrt{n\Delta t}) = x_0 + N(0,\sigma \sqrt{\Delta t}) \) .

The change equation is

$$ \Delta x_t = \alpha t + N(0, \sigma \sqrt{t}) = N(\alpha t , \sigma \sqrt{t}) $$

Thus, we have

$$ x_t = x_0 + \alpha t + N(0, \sigma \sqrt{t}) $$

Very small positive drift rate produce higher variation and higher drift rate produce smaller variation. Positive drift rate produce positive trend and negative drift rate produce negative trend. Similarly, higher degree of volatility produces higher variation. There two numbers are determined by the data to produce minimum sum of square error.

Figure above shows the comparison of Wiener process and random walk. The Wiener process is using zero drift and the unit standard deviation to make it the same parameter with the random walk. In general, Wiener process has much higher spread because of the factor of the square root of time.

We can set the parameters of Wiener Process such that it will resemble random walk. For instance, setting drift rate to 0.0001 and volatility to 0.2 in the following chart produces Wiener process that almost similar to random walk. Thus, we can see Wiener process as generalization of random walk.

Example

The data of the last many years show that average increase (trend) in its yearly demand of a supermarket is 10 thousand customers with standard deviation of 5 thousand customers. Suppose this year demand is 750 thousand customers, what would be the prediction of customers in the next year and next two years?

Answer:

If we use Wiener process, we have \( x_0=750, \alpha=10, \sigma=5 \) and next year prediction would be \( x_1=x_0+\alpha t+N(0,\sigma \sqrt{t})=750+10+N(0,5\sqrt{1})=760+N(0,5)=N(760,5) \) and the next two years prediction of demand would be \( x_2=x_1+\alpha t+N(0,\sigma \sqrt{t}) = 750+2\cdot 10+N(0,5\sqrt{2})=770+N(0,5\sqrt{2})=N(770,7.07) \).

What does it mean when the prediction is still in random value of Normal distribution with certain mean and standard deviation? A Normal distribution has certain characteristics that the 99% of the values will not go beyond \( \mu \pm 3 \sigma \). Thus, we can have a 99% confidence that the prediction value of \( N(770,7.07) \) will be in the range of \( 770 \pm 3\cdot 7.07 = [748.79,791.21] \). The chance to get demand of 700 or 800 are very low.

In the stock market, we often have initial stock price \( x_0 \) and the change of the price after time period \( \Delta t \) is written as \( \frac{\Delta x}{x_0} = N(\alpha \Delta t,\sigma \sqrt{t}) \) where the drift rate \( \alpha \)  is called the expected rate of return and the standard deviation is called volatility.  Integrating both side of the equation over time will get \( \ln{x_t} = \ln{x_0} +N(\alpha t,\sigma \sqrt{t}) \) or \( \Delta x = x_0 (e^{N(\alpha t, \sigma \sqrt{t})}-1) \). Thus, \( x_t \) is distributed as lognormal distribution.

Example

Suppose a stock has volatility of 30% per year with expected rate of return of 6%. If the current stock price is $20 what is the expected price change in the next three months?

Answer:

$$ \Delta x_t = x_0 (e^{N(\alpha t,\sigma \sqrt{t})}-1) =$20 (e^{N(0.06 \cdot \frac{1}{4}, 0.3 \cdot \sqrt{\frac{1}{4}})} - 1) =$20(e^{N(0.015,0.15)}-1) $$

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Preferable reference for this tutorial is

Teknomo, Kardi. (2019) Stochastic Process Tutorial .
http://people.revoledu.com/kardi/tutorial/StochasticProcess/