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a. Use the rules of expected value to show that \(\operatorname{Cov}(a X+b, c Y+d)=a c \operatorname{Cov}(X, Y)\). b. Use part (a) along with the rules of variance and standard deviation to show that \(\operatorname{Corr}(a X+b, c Y+d)=\operatorname{Corr}(X, Y)\) when \(a\) and \(c\) have the same sign. c. What happens if \(a\) and \(c\) have opposite signs?

Short Answer

Expert verified
Covariance changes by ac, correlation remains the same when a and c have the same sign; reverses if they have opposite signs.

Step by step solution

01

Express the Covariance

To begin, let's express the covariance for any random variables \( X \) and \( Y \). The covariance is defined as:\[\operatorname{Cov}(X, Y) = E[(X - E[X])(Y - E[Y])]\]We must analyze \( \operatorname{Cov}(aX+b, cY+d) \).
02

Apply the Linear Transformation in Covariance

Now apply the definition of covariance to the transformed variables:\[\operatorname{Cov}(aX+b, cY+d) = E[((aX+b) - E[aX+b])((cY+d) - E[cY+d])]\]Using the linearity of expectation, \( E[aX+b] = aE[X] + b \) and \( E[cY+d] = cE[Y] + d \). So,\[((aX+b) - E[aX+b]) = a(X - E[X])\]and\[((cY+d) - E[cY+d]) = c(Y - E[Y])\].

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Key Concepts

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

Expected Value
Understanding expected value is crucial for grasping many statistical concepts. The expected value, often represented as \(E[X]\), is essentially the average or mean you would expect from a random variable if you could repeat the random process an infinite number of times.
For a discrete random variable \(X\), it is calculated as:
\[E[X] = \sum_{i}x_i P(x_i)\]
where \(x_i\) is each possible outcome and \(P(x_i)\) is the probability of that outcome.
For continuous random variables, we use integrals:
\[E[X] = \int_{-fty}^{\infty} x f(x) \mathrm{d}x\]
where \(f(x)\) is the probability density function of \(X\).
This concept helps us understand distributions and make predictions based on probabilities. By applying linear transformations, such as \( aX + b \), the expected value transforms linearly:
\[E[aX+b] = aE[X] + b\]
This simplicity is what makes expected value a powerful tool in statistical analysis.
Linearity of Expectation
The linearity of expectation is an incredibly useful property in probability theory and helps simplify calculations significantly. It states that the expected value of a sum of random variables is equal to the sum of their expected values.
This means:
  • \(E[X+Y] = E[X] + E[Y]\)
  • \(E[cX] = cE[X]\)\ where \(c\) is any constant.
This rule applies even if the variables \(X\) and \(Y\) are not independent, which is a major benefit compared to other statistical measures.
It's particularly useful in scenarios involving many random variables, allowing us to split complex problems into smaller, manageable parts.
For linear transformations, you use linearity to express transformations like \(aX + b\). In our problem, applying linear transformations helps us reorganize and simplify the expressions for covariance and correlation.
Variance and Standard Deviation
Variance and standard deviation are fundamental concepts for measuring the spread or dispersion of a set of values. Understanding these helps you grasp how much variability exists within your data.

**Variance:** Measures the average squared deviation from the mean. It's given by:
\[\operatorname{Var}(X) = E[(X - E[X])^2]\]
The variance gives insight into the distribution width. High variance means data points are more spread out.

**Standard Deviation:** Is simply the square root of the variance and provides a measure of dispersion in the same units as the original random variable:
\[\operatorname{SD}(X) = \sqrt{\operatorname{Var}(X)}\]
Standard deviation is preferred because it is easier to interpret compared to variance.
When dealing with transformations of variables, understanding how variance changes is important:
  • \(\operatorname{Var}(aX + b) = a^2 \operatorname{Var}(X)\)
This is useful in complex calculations, as seen in the context of covariance and correlation, ensuring you adjust correctly for scaling factors.

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Most popular questions from this chapter

Use the proposition on the expected product to show that when \(X\) and \(Y\) are independent, \(\operatorname{Cov}(X, Y)=\operatorname{Corr}(X, Y)=0\)

A market has both an express checkout line and a superexpress checkout line. Let \(X_{1}\) denote the number of customers in line at the express checkout at a particular time of day, and let \(X_{2}\) denote the number of customers in line at the superexpress checkout at the same time. Suppose the joint pmf of \(X_{1}\) and \(X_{2}\) is as given in the accompanying table. $$ \begin{array}{cc|cccc} & & & &{x_{2}} & \\ & & 0 & 1 & 2 & 3 \\ \hline & 0 & .08 & .07 & .04 & .00 \\ & 1 & .06 & .15 & .05 & .04 \\ x_{1} & 2 & .05 & .04 & .10 & .06 \\ & 3 & .00 & .03 & .04 & .07 \\ & 4 & .00 & .01 & .05 & .06 \end{array} $$ a. What is \(P\left(X_{1}=1, X_{2}=1\right)\), that is, the probability that there is exactly one customer in each line? b. What is \(P\left(X_{1}=X_{2}\right)\), that is, the probability that the numbers of customers in the two lines are identical? c. Let \(A\) denote the event that there are at least two more customers in one line than in the other line. Express \(A\) in terms of \(X_{1}\) and \(X_{2}\), and calculate the probability of this event. d. What is the probability that the total number of customers in the two lines is exactly four? At least four? e. Determine the marginal pmf of \(X_{1}\), and then calculate the expected number of customers in line at the express checkout. f. Determine the marginal pmf of \(X_{2}\). g. By inspection of the probabilities \(P\left(X_{1}=4\right)\), \(P\left(X_{2}=0\right)\), and \(P\left(X_{1}=4, X_{2}=0\right)\), are \(X_{1}\) and \(X_{2}\) independent random variables? Explain.

Annie and Alvie have agreed to meet for lunch between noon \((0: 00\) p.m. \()\) and 1:00 p.m. Denote Annie's arrival time by \(X\), Alvie's by \(Y\), and suppose \(X\) and \(Y\) are independent with pdf's $$ \begin{aligned} &f_{X}(x)=\left\\{\begin{array}{cc} 3 x^{2} & 0 \leq x \leq 1 \\ 0 & \text { otherwise } \end{array}\right. \\ &f_{Y}(y)=\left\\{\begin{array}{cl} 2 y & 0 \leq y \leq 1 \\ 0 & \text { otherwise } \end{array}\right. \end{aligned} $$ What is the expected amount of time that the one who arrives first must wait for the other person?

Suppose two identical components are connected in parallel, so the system continues to function as long as at least one of the components does so. The two lifetimes are independent of each other, each having an exponential distribution with mean \(1000 \mathrm{~h}\). Let \(W\) denote system lifetime. Obtain the moment generating function of \(W\), and use it to calculate the expected lifetime.

Suppose the amount of rainfall in one region during a particular month has an exponential distribution with mean value 3 in., the amount of rainfall in a second region during that same month has an exponential distribution with mean value 2 in., and the two amounts are independent of each other. What is the probability that the second region gets more rainfall during this month than does the first region?

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