Stochastics includes the two sub-areas of statistics and probability theory. The goal of stochastics is, in short, to examine chance. This means that random processes are categorized into uniform models and then examined for their potential outcomes.
The central question is: How probable are the different outcomes of these models? In this post we will cover the following topics:
Random experiments in stochastics
Ever since people no longer held exclusively higher forces responsible for the outcome of the most varied of situations, the term “coincidence” has existed. Coincidence means that the outcome of an event cannot be fully predicted. When it comes to examining this coincidence, it is necessary to set up models for it.
These models enable us to make consistent statements about chance using the tools of mathematics.
A random experiment describes the modeling of a process that can be repeated as often as required, subject to certain rules, and the outcome of which cannot be predicted with certainty. There must be several possible outcomes.
But don’t be fooled! A soccer game can also be a random experiment, because there are different outcomes and even the victory of a favorite team cannot be predicted with 100% certainty.
results and events
If you want to understand stochastics, or in the narrower sense the probability calculation, it is important that you can distinguish results from events. These two terms form the cornerstone of probability theory.
Results
Each possible outcome of a random experiment represents an outcome. The set of all possible outcomes is called the outcome space (or outcome set) and is denoted by the Greek letter Ω (omega).
- Example: If a die is rolled, the result set is: Ω = {1,2,3,4,5,6}, if a coin is tossed, it is: Ω = {heads, tails}
events
An event represents a subset (or subset) of the result space of a random experiment (ZE). This means that several results of a ZE are grouped on the basis of certain characteristics, so that after a result has occurred, it is clearly recognizable whether the event has occurred or not.
- Example: When a dice is rolled, an event might be: «The number rolled is odd». The subset A of Ω is therefore: A = {1,3,5}. Another event could be: «The number of points is > 2». The subset B of Ω is thus: B = {3,4,5,6}
More definitions
- The set of all events is called the event space
- Events that can only be brought about by a single result (e.g. “number of points = 6”) are called elementary events
- An event that occurs through all results in Ω, i.e. A = Ω, is called the certain event, since it will definitely occur
- If the subset B is empty, i.e. contains no result, there is an impossible event. Example: The event “number over 6” cannot occur with an ordinary die.
More on the topic: random experiments, results and events
Stochastics – combinatorics and urn model
The goal of combinatorics is to find out how many different ways there are to choose exactly k elements from a set with n elements. This consideration is often explained using the example of an urn from which balls are drawn.
In general, such experiments can be divided into different types of models:
- from one variation(also “ordered random sample”) is always relevant when:
- k < n and
- the order in which the balls are drawn from the urn is important
- One combination(also «unordered random sample») is always present when:
- k < n and
- the order is unimportant
- at permutationsis:
- n = k and
- The order matters
It is therefore a variation in which all the balls are drawn from the urn.
With and Without «Repeat»
Now that you know the rough model of your experiment, all you have to do is answer the following question: After a ball is drawn, is it set aside so that it is no longer an option for the next draw? Or is it put back and can be drawn again next time?
The two variants can be described as follows:
- With replacement = with repetition
- Without replacement = without repetition
In order to be able to go through the individual models, there is only one thing you should have understood beforehand: the product rule of combinatorics or the «general counting principle»!
Stochastic – The General Counting Principle
At the General counting principle is about how to find out the number of combinations of different elements.
Example:
Let’s say you want to blend three different ingredients into a smoothie. A smoothie should always contain exactly 1x fruit (pink), 1x nuts (blue) and 1x liquid (orange). Now you have 3 different types of fruit, 4 different types of nuts and 2 different liquids to choose from at home.
The question is:
How many different smoothies can you mix with these ingredients?
The answer is easier than you think!
Because you could add 4 different nuts to each of the 3 types of fruit. Thus there are 3 × 4 = 12 different fruit-nut combinations. If you can now combine each of these 12 combinations with either milk or water, you get 3 × 4 × 2 = 24 different smoothies.
In general this means:
- There are k quantities (in this case 3: fruit, nuts, liquid) M1-Mk
- There are n elements in each of these sets, where M1 has n1 elements (e.g. fruit is M1 and has 3 elements 🡪 n1 = 3), M2 has n2 elements, etc.
- n1 · n2 · … · nk combinations of the elements in the sets can be generated
- These combinations are called k-tuples (x1, x2, …, xk).
- In the example, 24 different k-tuples are created, one of which could be: (fruit 1, nut 3, water)
Permutation without repetition
Permutation: n = k and the order is important
- Imagine there are 30 chairs in a row on a soccer field and there are a total of 30 students in your class. Now the question is: In how many different orders can your class arrange itself on the chairs?
- We can approach the whole thing quite logically! The first person to sit down on one of the chairs is free to choose. So there are n = 30 different ways for him to sit down. If the second person wants to sit down, he or she can only choose from (n – 1) = (30 – 1) = 29 different chairs. If he or she sits down, only (n – 2) = 28 different options remain for the next one and so on, until the last one has to sit on the last free chair.
- So to get the total number of possible combinations, all you have to do is multiply the choices you and your classmates have: n · (n – 1) · (n – 2) · … · 1 = n!so the factorial of n. In In this example, there would be many different ways of sitting down in a row, namely: 30! = 30 * 29 * 28 * … * 1 = 2.653 * 1032 = 265,300,000,000,000,000,000,000,000,000,000
permutation with repetition
You now know that there n! There are ways to arrange n things in n places in a different order. But what if some of the things we place are the same, i.e. repeat? Imagine we want to arrange all 10 of a delivery of 10 colored plastic buckets in a row.
There are three different situations:
1. All buckets are distinguishable except three:
This could be the case if, for example, the first 7 buckets are numbered consecutively, but there are 3 buckets numbered 8. These 3 buckets are then identical. In this case, there are k = 3 objects that you could swap places in the row without anyone noticing. As described above, there are k! = 3! Ways to arrange the 3 buckets differently (if you could tell them apart). So there is also k! Arrangements in which the row remains «the same». This k! So arrangements are not multiplied by the other possibilities, so we end up with the following formula:
2. There are several indistinguishable groups:
This situation could arise if the buckets were different colors, for example, and there were 3 green, 2 red, and 5 blue buckets. That would give k1! = 3! Ways to switch the green buckets unnoticed, k2! = 2! Ways to place the red buckets and k3! = 5! Possibilities of arranging the blue buckets differently.
The total number of different arrangement options for s = 3 groups can therefore be calculated as follows:
3. All buckets are indistinguishable from each other:
Since you can’t see any change in the arrangement, no matter how you swap the buckets in such a row, there is only one way to distribute n indistinguishable buckets to n places. Mathematically, this also results from:
Excellent! Now you’re really familiar with permutations! Let’s continue with the variations with and without repetition.
variation without repetition
Variation: k < n and the order is important
The only difference between variations and permutations: In variations, only one sample k < n is drawn from the urn instead of all n elements. Since we are considering a variation without repetition here, each of these k elements can only be selected once.
Let’s go back to the example with the chairs on the soccer field. This time, let’s imagine that only 10 designated students are to be distributed among the 30 chairs. n is then still 30 and k = 10.
As with the permutation n = 30, the first has a choice of free chairs, one of which he can occupy. The second has only (n – 1) = 29 free chairs, etc. The last one is number 10 in this example, so he still has (n – k + 1) = 21 chairs to choose from. If you now multiply all these possibilities, it looks like this:
→ There are significantly fewer different possibilities than with permutation!
Variation with repetition
One Variation with repetition turns out to be much simpler in comparison. Because if an element from the group can be selected several times, i.e. does not «take up» space from the other elements, there is the same number of possibilities for the next element in the row after each drag, namely n. The number k of elements corresponds the sample size, so the formula for calculating the possibilities is:
n n n n n … =
combination without repetition
Combination: k < n and the order is unimportant
With a combination (i.e. a random sample) without repetition, it is almost like with a variation without repetition, only this time the order doesn’t matter either! Using the example, this means that you and your classmates no longer have to sit on chairs in…