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Beispiel auf \(\mathrm{S}\). 1453 sind \(R\) und \(B\) die Augenzahlen zweier unabhängig voneinander geworfener idealer Würfel und \(X=\max (R, B)\) sowie \(Y=\min (R, B)\). Weiter war \(\operatorname{Var}(X)=\operatorname{Var}(Y)=1.97\). Berechnen Sie \(\operatorname{Cov}(X, Y)\) aus diesen Angaben ohne die Verteilung von \((X, Y)\) explizit zu benutzen.

Short Answer

Expert verified
Answer: The covariance of X and Y is 0.

Step by step solution

01

Calculate the mean of a single die roll

Let's first find E(R) and E(B), which are equal since they are both independent dice rolls. For an ideal die roll, the mean can be calculated as follows: E(R) = E(B) = sum(Faces value * Probability of each face) / 6 = (1 + 2 + 3 + 4 + 5 + 6) / 6 = 3.5
02

Find E(X) and E(Y)

Since X is the maximum and Y is the minimum, we know that: E(X) = E(max(R, B)) = E(R) = E(B) = 3.5 E(Y) = E(min(R, B)) = E(R) = E(B) = 3.5
03

Calculate the product of means

Now let's calculate the product of the means: E(X)E(Y) = E(R)E(B) = 3.5 * 3.5 = 12.25
04

Use the formula for Cov(X, Y)

Recall that the formula for covariance is Cov(X, Y) = E(XY) - E(X)E(Y). Since the dice are independent, the expected value of their product E(XY) can be rewritten as E(X)E(Y). Therefore, Cov(X, Y) = E(XY) - E(X)E(Y) = E(X)E(Y) - E(X)E(Y) = 0 Thus, the covariance of X and Y is 0.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Variance
Variance is a statistical measure that tells us how much a set of numbers is spread out from their mean average. It is particularly useful in probability and statistics to measure the level of uncertainty or variability. For example, in the case of rolling dice where we calculated the variance of maximum and minimum outcomes, variance helps us understand the reliability and fluctuations of these outcomes. For a random variable, the variance is calculated as:\[ \text{Var}(X) = E((X - E(X))^2) \]Here, \( E(X) \) represents the expectation or mean of \( X \), and \( (X - E(X))^2 \) symbolizes the squared deviation of \( X \) from its mean. By squaring the deviations, we eliminate negative numbers and thus measure total spread.
  • A small variance indicates that numbers are close to the mean.
  • A large variance implies numbers are spread out across a wide range of values.
In our exercise, both \( \operatorname{Var}(X) \) and \( \operatorname{Var}(Y) \) are 1.97, showing a moderate level of dispersion in the numbers generated by rolling the dice.
Covariance
Covariance is a measure that describes how much two random variables change together, or how they are related. It is used to determine the direction of a linear relationship between variables. In simpler terms, covariance can tell us whether changes in one variable might be associated with changes in another.The formula for covariance between two variables \( X \) and \( Y \) is:\[ \operatorname{Cov}(X, Y) = E(XY) - E(X)E(Y) \]Here, \( E(XY) \) is the expected product of \( X \) and \( Y \), while \( E(X)E(Y) \) is the product of their expected values.
  • If \( \operatorname{Cov}(X, Y) > 0 \), the variables tend to increase together.
  • If \( \operatorname{Cov}(X, Y) < 0 \), one variable tends to increase as the other decreases.
  • If \( \operatorname{Cov}(X, Y) = 0 \), the variables are likely independent.
In the given solution, it was found that \( \operatorname{Cov}(X, Y) = 0 \), indicating that the maximum and minimum values of dice rolls are independent when calculated this way.
Expectation
Expectation in probability is often referred to as the expected value or mean of a random variable. It is a fundamental concept, as it represents the average outcome we can anticipate if an experiment is repeated many times. For a random variable \( X \), the expectation is calculated using:\[ E(X) = \sum{(x_i \cdot P(x_i))} \]where \( x_i \) are possible values and \( P(x_i) \) are their probabilities.

Calculating Expectation of Dice Rolls

The calculation of mean values in the dice roll example shows how expectations help in determining typical results. For a single die, the expectation is:\[ E(R) = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5 \]This was used to derive \( E(X) \) and \( E(Y) \) in the solution as they also represent means for maximum and minimum results of two dice.
  • A higher expected value indicates a larger mean result.
  • A lower expected value reflects a smaller average outcome.
Understanding expectation is crucial for making predictions in terms of probability and statistics.

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Most popular questions from this chapter

Welche der folgenden Aussagen sind richtig? 1\. Sind \(X\) und \(Y\) unabhängig, dann sind auch \(1 / X\) und \(1 / Y\) unabhängig. 2\. Sind \(X\) und \(Y\) unkorreliert, dann sind auch \(1 / X\) und \(1 / Y\) unkorreliert.

Ein fairer Würfel wird dreimal geworfen. 1\. Berechnen Sie die Wahrscheinlichkeitsverteilung des Medians \(X_{\text {med }}\) der drei Augenzahlen. 2\. Ermitteln Sie die Verteilungsfunktion von \(X_{\text {med }}\). 3\. Berechnen Sie Erwartungswert und Varianz des Medians.

Das zweidimensionale Merkmal \((X, Y)\) besitze die folgende Verteilung: $$ \begin{array}{|l|l|l|l|} \hline & & Y & & \\ \hline & & 1 & 2 & 3 \\ X & 1 & 0.1 & 0.3 & 0.2 \\ & 2 & 0.1 & 0.1 & 0.2 \\ \hline \end{array} $$ 1\. Bestimmen Sie Erwartungswerte und Varianzen (a) von \(X\) und \(Y\), (b) von \(S=X+Y\) und (c) von \(X \cdot Y\). 2\. Wie hoch ist die Korrelation von \(X\) und \(Y ?\)

Welche der folgenden Aussagen sind wahr? Begründen Sie Ihre Antwort. 1\. Um eine Prognose über die zukünftige Realisation einer zufälligen Variablen zu machen, genügt die Kenntnis des Erwartungswerts. 2\. Um eine Prognose über die Abweichung der zukünftigen Realisation einer zufälligen Variablen von ihrem Erwartungswert zu machen, genügt die Kenntnis der Varianz. 3\. Eine Prognose über die Summe zufälliger i.i.d.-Variablen ist in der Regel genauer als über jede einzelne. 4\. Das Prognoseintervall über die Summe von 100 identisch verteilten zufälligen Variablen (mit Erwartungswert \(\mu\) und Varianz \(\sigma^{2}\) ) ist 10-mal so lang wie das Prognoseintervall für eine einzelne Variable bei gleichem Niveau. 5\. Wenn man hinreichend viele Beobachtungen machen kann, dann ist \(\mathrm{E}(X)\) ein gute Prognose für die nächste Beobachtung.

Zeigen Sie: (a) Ist \(X\) eine positive Zufallsvariable, so ist \(\mathrm{E}\left(\frac{1}{x}\right) \geq \frac{1}{\mathrm{E}(X)}\). (b) Zeigen Sie an einem Beispiel, dass diese Aussage falsch ist, falls \(X\) positive und negative Werte annehmen kann.

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