**ST412 Stochastic Processes Assignment Sample NUIG Ireland**

ST412 Stochastic Processes is a probabilistic study of random phenomena. The course is an introduction to the theory and applications of stochastic processes. Processes studied in the course include basic discrete and continuous time models; birth-death processes; Poisson process and exponential distribution; Markov chains; renewal theory; storage models; reliability theory; queuing models.

The emphasis in the course is placed on learning how to model real systems using stochastic processes, how to obtain results for these models using analytical or simulation techniques, and how to interpret these results.

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In this section, we are describing some assigned tasks. These are:

**Assignment Task 1: Use probability and moment generating functions to calculate corresponding distributional properties.**

To calculate the corresponding distributional properties using probability and moment generating functions, you need to determine the following:

First, what is the underlying distribution of the random variable? This will give you the functional form of the moment-generating function. Second, what are the values of the parameters in this distribution? This will allow you to calculate specific values for (e.g., in a uniform distribution, all Taylor coefficients are equal to 1/2). Lastly, what is your expectation or result? This will be some function of one or more variables (e.g., find if E(X)=3).

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With these three pieces of information in hand, we can use probability and moment-generating functions to calculate corresponding distributional properties.

For example, let’s say we have a random variable X~Uniform(0,1) with the moment-generating function:

$$ M_X(t)=\frac{e^{tx}-1}{t}=\sum_{n=0}^{\infty}\frac{t^n}{n!}x^n$$

Now, let’s say we want to calculate E(X), the expectation of X. We can do this by taking the first derivative of the moment-generating function and setting it equal to 0:

$$\frac{dM_X(t)}{dt}=\sum_{n=0}^{\infty}\frac{t^{n-1}}{(n-1)!}x^n=xe^{tx}$$

Now, setting this derivative equal to 0 and solving for x gives us:

$$xe^{tx}=0 \Rightarrow x=0$$

Therefore, E(X)=0.

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**Assignment Task 2: Derive properties of branching processes such as expectation, variance, and probability of extinction.**

**Expectation:** The expected value of a branching process is the sum of the probabilities of all the possible outcomes at the end of the process.

**Variance:** The variance of a branching process is the sum of the squared differences between each outcome and its expected value.

**Probability of extinction: **The probability of extinction is equal to the probability that no descendants remain after a particular number of steps.

Assuming that each branch produces k offspring with probability p, the expected number of offspring at the end of the process is:

$$E = \sum_{i=0}^{\infty}i^kp^i(1-p)^{n-i}$$

The variance of the process is:

$$V = \sum_{i=0}^{\infty}(i-E)^2p^i(1-p)^{n-i}$$

And the probability of extinction is:

$$P(extinction) = \lim_{n\to\infty}\sum_{i=0}^np^i(1-p)^n$$

In each of these cases, the summation is over all possible outcomes at the end of the process.

**Assignment Task 3: Calculate relevant probabilities in random walks with and without barriers.**

There are many different ways to calculate probabilities in random walks, and it can get quite complex. But here is a very basic explanation of how it works.

A random walk with barriers is like a game of hopscotch. You can only move forwards or sideways, and you can’t go back to where you’ve already been. This makes the path much more predictable since once you’ve reached a certain point, you know that you won’t be going back there.

A random walk without barriers is like walking around your neighborhood without any restrictions. You can go anywhere you want, and there’s no telling which way you’ll end up going next. This makes the path much more unpredictable since anything could happen.

The probabilities in a random walk with barriers are calculated by dividing the number of ways you can reach a certain point by the total number of possible paths.

For example, let’s say you’re starting at point A and you want to calculate the probability of reaching point B. There are two possible paths:

Path 1: A-B

Path 2: A-C-B

Since there are two paths, the probability of reaching point B is 1/2.

The probabilities in a random walk without barriers are calculated by dividing the number of ways you can reach a certain point by the total number of possible paths.

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**Assignment Task 4: Use Markov property to prove various probabilistic statements about the Markov chain.**

The Markov chain is a type of probabilistic model that is used to describe a sequence of random variables. The Markov chain can be used to model a wide variety of phenomena, including but not limited to the evolution of words in a language, the movement of financial assets, and the states of a physical system.

One useful property of the Markov chain is the Markov property. The Markov property states that the conditional probability distribution for any given variable X depends only on the history up to and including time t-1. In other words, the probability that X takes on a particular value at time t does not depend on any information about X at time t+1 or later. This property is what makes the Markov chain a powerful tool for modelling random processes.

The Markov property can be used to prove various probabilistic statements about the Markov chain. For example, suppose we have a Markov chain with two states, A and B. We can use the Markov property to calculate the probability that the chain will be in state B at time t, given that it is in state A at time t-1.

We know that the probability of being in state B at time t is equal to the sum of the probabilities of all the paths that lead from state A to state B. Since there are only two states in our Markov chain, there are only two paths that lead from state A to state B:

**Path 1:** A-B

**Path 2: **A-A-B

The probability of being in state B at time t is therefore equal to the sum of the probabilities of these two paths. We can use the Markov property to calculate the probability of each path:

**Path 1:** A-B

The probability of this path is equal to the probability of being in state A at time t-1, multiplied by the transition probability from state A to state B.

**Path 2: **A-A-B

The probability of this path is equal to the probability of being in state A at time t-2, multiplied by the transition probability from state A to state A, multiplied by the transition probability from state A to state B.

Since there are only two paths that lead from state A to state B, the probability of being in state B at time t is equal to the sum of these two probabilities. This proves that the Markov property can be used to calculate the probability of being in a particular state at a particular time.

**Assignment Task 5: Classify states of Markov chains and determine stationarity properties.**

There are three main stages that a Markov chain can be in:

- Stationary
- Nonstationary with no trend
- Nonstationary with a trend

A state is stationary if the expected value of the next state only depends on the current state and not on the past. A state is nonstationary with no trend if the expected value of the next state depends on the current state and the past, but there is no consistent direction to the changes. A state is nonstationary with a trend if the expected value of the next state depends on the current state and the past, and there is a consistent direction to the changes.

The stationary states of a Markov chain are important because they can be used to calculate the long-term behaviour of the chain. The nonstationary states, on the other hand, cannot be used to calculate the long-term behaviour and are therefore less useful for modelling purposes.

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**Assignment Task 6: Calculate limiting and stationary distributions.**

There are a few different ways to calculate limiting and stationary distributions. One way is to use a mathematical formula, such as the one below:

**Limiting Distribution:** p(x) = C * e^(-λx)

**Stationary Distribution:** P* = [p(0), p(1), …, p(n)]

where C is a normalizing constant, λ is the Parzen windows parameter, and n is the number of data points. The approach you take depends on what information you have available and what you want to achieve. If you have complete information about all possible states (such as in a system with only finitely many potential outcomes), then it might be possible to calculate the exact stationary distribution. However, if you have incomplete information or want to approximate the stationary distribution, then you will need to use a numerical method, such as the one described above.

**Assignment Task 7: Prove and calculate various properties of the Poisson process.**

Proving and calculating various properties of the Poisson process can be a complex undertaking. However, with a clear understanding of the underlying principles, it is relatively straightforward to derive the results you need.

At its most basic, the Poisson process is a way of modelling the arrival of events over time. It is commonly used in situations where events arrive at random and there is no underlying pattern to their arrival. Examples of situations where the Poisson process could be used include modelling customer arrivals at a store or calls arriving at a call centre.

To derive results for the Poisson process, we first need to define some key terms and concepts. The rate parameter, λ, governs how fast events occur on average. The time interval, t, is the length of time over which we are measuring the arrivals. The probability of an event occurring, P(t), is a function of both λ and t.

The Poisson distribution is a discrete probability distribution that describes the probability of k events occurring in a given time interval. It is governed by two parameters, λ, and t.

The Poisson process is a way of modelling the arrival of events over time. It is commonly used in situations where events arrive at random and there is no underlying pattern to their arrival. Examples of situations where the Poisson process could be used include modelling customer arrivals at a store or calls arriving at a call centre.

The Poisson process has several properties that make it a useful tool for modelling random events. First, the Poisson process is memoryless, which means that the probability of an event occurring is independent of when the last event occurred. Second, the Poisson process is stationary, which means that the rate at which events occur does not change over time. Finally, the Poisson process is independent, which means that the arrival of one event does not affect the arrival of another event.

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**Assignment Task 8: Build and describe Markov chains to represent simplified real-world problems, for example, such as those used to model credit mobility.**

A Markov chain is a mathematical model that allows you to predict the next state in a sequence of events, based on the current state and the previous event. In other words, it’s a model that helps you understand how something changes over time.

For example, you could use a Markov chain to model credit mobility. This would involve tracking the movement of people’s credit scores over time, and then using the Markov chain to predict how likely it is that someone with a low credit score will move up to a higher credit score (or down to a lower one). You could also use it to predict how likely it is that someone with no credit history will suddenly start building up good credit.

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