Problem 7 - Basiswissen Statistik: Kompaktkurs für Anwender aus Wirtschaft, Informatik und Technik

Question

\(X\) sei eine Zufallsvariable mit den möglichen Werten \(1,2,3\) und \(Y\) eine Zufallsvariable mit Werten in \(\\{A, B, C\\}\) für drei verschiedene Zahlen \(A, B, C . X\) sei diskret gleichverteilt und für \(Y\) gelte: $$ P(Y=A)=0.1, P(Y=B)=0.5, P(Y=C)=0.4 $$ Stellen Sie die zugehörige Tafel der gemeinsamen Verteilung auf, wenn \(X\) und \(Y\) unabhängig sind. Geben Sie für alle \(x \in\\{1,2,3\\}\) und \(y \in\\{A, B, C\\}\) die bedingten Wahrscheinlichkeiten \(P(X=x \mid Y=y)\) an.

Step-by-Step Solution

Answer

All joint probabilities are products of individual probabilities, and conditional probabilities are all \(\frac{1}{3}\).

Step 1: Identify Probability Distributions

First, we determine the probability distribution for the random variables \(X\) and \(Y\). \(X\) is uniformly distributed, so \(P(X=1) = P(X=2) = P(X=3) = \frac{1}{3}\). For \(Y\), we have the given probabilities: \(P(Y=A) = 0.1\), \(P(Y=B) = 0.5\), and \(P(Y=C) = 0.4\).

Step 2: Set Up Joint Distribution

Since \(X\) and \(Y\) are independent, the joint probability \(P(X=x, Y=y)\) is simply the product of the individual probabilities: \(P(X=x) \times P(Y=y)\).

Step 3: Calculate Joint Probabilities

For each combination of \(x\) and \(y\), calculate the joint probability:- \(P(X=1, Y=A) = \frac{1}{3} \times 0.1 = \frac{1}{30}\)- \(P(X=1, Y=B) = \frac{1}{3} \times 0.5 = \frac{5}{30} = \frac{1}{6}\)- \(P(X=1, Y=C) = \frac{1}{3} \times 0.4 = \frac{4}{30} = \frac{2}{15}\)- Repeat for \(X=2, X=3\) with \(Y=A, B, C\) using similar steps.

Step 4: Create Joint Probability Table

Tabulate the joint probabilities for each combination of \(X\) and \(Y\) in a table: \[ \begin{array}{c|c|c|c} & Y=A & Y=B & Y=C \ \hline X=1 & \frac{1}{30} & \frac{1}{6} & \frac{2}{15} \ X=2 & \frac{1}{30} & \frac{1}{6} & \frac{2}{15} \ X=3 & \frac{1}{30} & \frac{1}{6} & \frac{2}{15} \ \end{array} \]

Step 5: Calculate Conditional Probabilities

Since \(X\) and \(Y\) are independent, the conditional probability \(P(X=x \mid Y=y)\) is simply \(P(X=x)\). Therefore, for each \(y\), \(P(X=1 \mid Y=y) = \frac{1}{3}\), \(P(X=2 \mid Y=y) = \frac{1}{3}\), and \(P(X=3 \mid Y=y) = \frac{1}{3}\), for \(y = A, B, C\).

Key Concepts

Random Variables

In probability and statistics, a random variable is a variable that takes on different numerical values, each associated with a probability. It's a way to quantitatively describe a random process. There are two types of random variables:

• Discrete Random Variables: These take on a finite or countably infinite set of values. An example is the roll of a die, where the outcomes are discrete integers from 1 to 6.
• Continuous Random Variables: These take on a continuum of possible values. An example is the amount of time it takes for a bus to arrive, which can take any positive value.

In our original problem, both \(X\) and \(Y\) are discrete random variables. \(X\) can be 1, 2, or 3, each with equal probability, making it uniformly distributed. \(Y\) assumes values from \(\{A, B, C\}\), each with predefined probabilities.

Probability Distributions

Probability distributions describe how probabilities are distributed over the various values that a random variable can take. They serve as models to predict the behavior of random variables.
For a discrete random variable, a probability mass function (PMF) is used. The PMF assigns probabilities to each possible value of the random variable.
For example, the PMF of the random variable \(X\) in the exercise is:

• \(P(X=1) = \frac{1}{3}\)
• \(P(X=2) = \frac{1}{3}\)
• \(P(X=3) = \frac{1}{3}\)

Similarly, for \(Y\), the distribution is described with distinct values:

• \(P(Y=A) = 0.1\)
• \(P(Y=B) = 0.5\)
• \(P(Y=C) = 0.4\)

Understanding these distributions helps in predicting outcomes and understanding the likelihood of potential events.

Conditional Probability

Conditional probability is pivotal when analyzing the probability of an event given that another event has occurred. It's mathematically represented as \(P(A \mid B)\), which denotes the probability of event \(A\) occurring given event \(B\) is true.
This concept doesn't play a major role in our discussed exercise because we are dealing with independent events. However, it is crucial in real-world applications where events are often dependent.

• For example, if you're trying to determine the probability of rain given there's a storm, you'll use conditional probability.
• It helps in scenarios involving sequences of events, where outcomes of past events can affect future outcomes.

In the exercise, since \(X\) and \(Y\) are independent, the conditional probabilities \(P(X=x \mid Y=y)\) are simply \(P(X=x)\), meaning knowing \(Y\)'s outcome doesn't alter the probability of \(X\).

Independent Events

Two events are considered independent if the occurrence or non-occurrence of one event doesn't influence the probability of the other event occurring. Mathematically, events \(A\) and \(B\) are independent if:

• \(P(A \cap B) = P(A) \times P(B)\)

For the random variables \(X\) and \(Y\) in the exercise, their independence means:

• Joint probability \(P(X=x, Y=y)\) is the product \(P(X=x) \times P(Y=y)\).
• Knowing the value of \(Y\) doesn't provide any information about \(X\), and vice versa.

This independence simplifies calculations, as seen in the exercise where joint and conditional probabilities could be easily computed.