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If \(X_{1}, X_{2}, X_{3}, X_{4}\) are (pairwise) uncortelated random variables each having mean 0 and variance 1 , compute the correlations of (a) \(X_{1}+X_{2}\) and \(X_{2}+X_{3}\); (b) \(X_{1}+X_{2}\) and \(X_{3}+X_{4}\)

Short Answer

Expert verified
The correlations are: (a) \(0.5\) (between \(X_{1}+X_{2}\) and \(X_{2}+X_{3}\)) (b) \(0\) (between \(X_{1}+X_{2}\) and \(X_{3}+X_{4}\))

Step by step solution

01

Calculate Variances

First, we need to calculate the variances of the given expressions using the properties of variances: Var(A+B) = Var(A) + Var(B) + 2Cov(A, B), where A and B are random variables. For (a): \( \text{Var}(X_{1}+X_{2}) = \text{Var}(X_{1}) + \text{Var}(X_{2}) + 2\text{Cov}(X_{1}, X_{2}) = 1 + 1 + 0 = 2\) \( \text{Var}(X_{2}+X_{3}) = \text{Var}(X_{2}) + \text{Var}(X_{3}) + 2\text{Cov}(X_{2}, X_{3}) = 1 + 1 + 0 = 2\) For (b): \( \text{Var}(X_{1}+X_{2}) = 2 \) (already calculated) \( \text{Var}(X_{3}+X_{4}) = \text{Var}(X_{3}) + \text{Var}(X_{4}) + 2\text{Cov}(X_{3}, X_{4}) = 1 + 1 + 0 = 2\) Step 2: Calculate the covariances of given expressions
02

Calculate Covariances

Now we need to calculate the covariances of the given expressions: For (a): \( \text{Cov}(X_{1}+X_{2}, X_{2}+X_{3}) = \text{Cov}(X_{1}, X_{2}) + \text{Cov}(X_{1}, X_{3}) + \text{Cov}(X_{2}, X_{2}) + \text{Cov}(X_{2}, X_{3}) = 0 + 0 + \text{Var}(X_{2}) + 0 = 1\) For (b): \( \text{Cov}(X_{1}+X_{2}, X_{3}+X_{4}) = \text{Cov}(X_{1}, X_{3}) + \text{Cov}(X_{1}, X_{4}) + \text{Cov}(X_{2}, X_{3}) + \text{Cov}(X_{2}, X_{4}) = 0 + 0 + 0 + 0 = 0\) Step 3: Calculate the correlations of given expressions
03

Calculate Correlations

Now we can calculate the correlations of the given expressions using the formula mentioned above: For (a): \( \text{Corr}(X_{1}+X_{2}, X_{2}+X_{3}) = \frac{\text{Cov}(X_{1}+X_{2}, X_{2}+X_{3})}{\sqrt{\text{Var}(X_{1}+X_{2}) \cdot \text{Var}(X_{2}+X_{3})}} = \frac{1}{\sqrt{2 \cdot 2}} = \frac{1}{2}\) For (b): \( \text{Corr}(X_{1}+X_{2}, X_{3}+X_{4}) = \frac{\text{Cov}(X_{1}+X_{2}, X_{3}+X_{4})}{\sqrt{\text{Var}(X_{1}+X_{2}) \cdot \text{Var}(X_{3}+X_{4})}} = \frac{0}{\sqrt{2 \cdot 2}} = 0\) So, the correlations are: (a) 0.5 (between \(X_{1}+X_{2}\) and \(X_{2}+X_{3}\)) (b) 0 (between \(X_{1}+X_{2}\) and \(X_{3}+X_{4}\))

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Key Concepts

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

Random Variables
In probability theory, random variables are a fundamental concept that underpins much of the analysis in fields such as statistics, economics, and engineering. A random variable is a numerical quantity whose value depends on the outcomes of a random phenomenon.

For example, if we roll a fair six-sided die, the outcome is a random variable because it can be any number between 1 and 6. Typically, random variables are divided into two types: discrete, which can take on a countable number of values (like the roll of a die), and continuous, which can take any value within a given range (like the temperature on a given day).

In the exercise, the random variables \(X_{1}\), \(X_{2}\), \(X_{3}\), and \(X_{4}\) are assumed to have a mean of 0 and a variance of 1, which suggests that they are standardized, with their outcomes centered around the mean (average) value with a consistent spread or dispersion represented by the variance.
Variance
The variance is a measure of how much the values of a random variable differ from the mean value of the random variable. It is one of the most widely used measures of variability and gives us an indication of the 'spread' of a set of values.

In formal terms, the variance of a random variable \(X\) is the expected value of the squared deviation of \(X\) from its mean \(\mu\), often denoted as \(\text{Var}(X)\) or \(\sigma^2\). The calculation is \(\text{Var}(X) = E[(X - \mu)^2]\), where \(E\) represents the expected value.

The workout in the original exercise incorporated the property that the variance of the sum of two independent random variables is equal to the sum of their variances. This is why we add the variances of \(X_1\) and \(X_2\) directly to find the variance of their sum. However, if the variables are not independent, their covariance, which is assumed to be zero in the given exercise, would also play a role.
Covariance
The concept of covariance is crucial when analyzing the relationship between two random variables. It measures the extent to which two random variables change together. If the two variables tend to increase and decrease together, the covariance is positive. If one tends to increase when the other decreases, the covariance is negative.

The formula to compute covariance between two random variables \(X\) and \(Y\) is given by \(\text{Cov}(X, Y) = E[(X - E[X])(Y - E[Y])]\). In simpler terms, it's the expected value of the product of their deviations from their respective means.

Using this concept, the solution to the exercise derives the covariance values for the pairs of expressions. Since the variables were given to be pairwise uncorrelated, meaning their covariance is zero, only when the same random variable is paired with itself—like \(X_2\) with \(X_2\) in part (a)—does the covariance contribute to the calculation, being equal to its variance.
Correlation
While covariance indicates the direction of a linear relationship between two variables, it does not provide a standardized measure of the strength of that relationship. That’s where correlation comes in. It is a dimensionless measure that provides both the direction and the strength of a linear relationship between two random variables.

The correlation is often denoted by \(\rho\) or \(r\) and calculated as \(\text{Corr}(X,Y) = \frac{\text{Cov}(X,Y)}{\sqrt{\text{Var}(X) \text{Var}(Y)}}\). It has a value between -1 and 1, where 1 means a perfect positive linear relationship, -1 means a perfect negative linear relationship, and 0 indicates no linear relationship.

In the solution provided, correlation calculations apply this definition to find the relationship between the sums of different pairs of random variables. The exercise illustrates that knowing the variances and covariances of individual variables allows us to determine the correlation between their sums, showcasing the interconnected nature of these statistical concepts.

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Most popular questions from this chapter

Cards from an ordinary deck of 52 playing cards are turned face up one at a time. If the first card is an ace, or the second a deuce, or the third a three, or \(\ldots\), or the thirteenth a king, or the fourteenth an ace, and so on, we say that a match occurs. Note that we do not require that the \((13 n+1)\) th card be any particular ace for a match to occur but only that it be an ace. Compute, the expected number of matches that occur.

Show that \(X\) is stochastically larger than \(Y\) if and only if $$ E[f(X)] \geq E[f(Y)] $$ for all increasing functions \(f\). HINT: If \(X \geq_{s t} Y\), show that \(E[f(X)] \geq E[f(Y)]\) by showing that \(f(X) \geq_{\text {st }}\) \(f(Y)\) and then using Theoretical Exercise 7. To show that \(E[f(X)] \geq E[f(Y)]\) for all increasing functions \(f\) implies that \(P\\{X>t\\} \geq P\\{Y>t\\}\), define an appropriate increasing function \(f\).

In Example \(3 \mathrm{~g}\) we showed that the covariance of the multinomial random variables \(N_{i}\) and \(N_{j}\) is equal to \(-m P_{l} P_{j}\) by expressing \(N_{l}\) and \(N_{j}\) as the sum of indicator variables. This result could also have been obtained by using the formula $$ \operatorname{Var}\left(N_{i}+N_{j}\right)=\operatorname{Var}\left(N_{i}\right)+\operatorname{Var}\left(N_{j}\right)+2 \operatorname{Cov}\left(N_{i}, N_{j}\right) $$ (a) What is the distribution of \(N_{i}+N_{j} ?\) (b) Use the identity above to show that \(\operatorname{Cov}\left(N_{i}, N_{j}\right)=-m P_{i} P_{j}\)

Let \(X\) be a random variable having finite expectation \(\mu\) and variance \(\sigma^{2}\), and let \(g(\cdot)\) be a twice differentiable function. Show that $$ E[g(X)] \approx g(\mu)+\frac{g^{\prime \prime}(\mu)}{2} \sigma^{2} $$ Hint: Expand \(g(\cdot)\) in a Taylor series about \(\mu\). Use the first three terms and ignore the remainder.

Consider a population consisting of individuals able to produce offspring of the same kind. Suppose that each individual will, by the end of its lifetime, have produced \(j\) new offspring with probability \(P_{j}, j \geq 0\), independently of the number produced by any other individual. The number of individuals initially present, denoted by \(X_{0}\), is called the size of the zeroth generation. All offspring of the zeroth generation constitute the first generation, and their number is denoted by \(X_{1} .\) In general, let \(X_{n}\) denote the size of the \(n\)th generation. Let \(\mu=\sum_{j=0}^{x} j P_{j}\) and \(\sigma^{2}=\sum_{j=0}^{x}(j-\mu)^{2} P_{j}\) denote, respectively, the mean and the variance of the number of offspring produced by a single individual. Suppose that \(X_{0}=1\) - that is, initially there is a single individual in the population. (a) Show that $$ E\left[X_{n}\right]=\mu E\left[X_{n-1}\right] $$ (b) Use part (a) to conclude that $$ E\left[X_{n}\right]=\mu^{n} $$ (c) Show that $$ \operatorname{Var}\left(X_{n}\right)=\sigma^{2} \mu^{n-1}+\mu^{2} \operatorname{Var}\left(X_{n-1}\right) $$ (d) Use part (c) to conclude that $$ \operatorname{Var}\left(X_{n}\right)= \begin{cases}\sigma^{2} \mu^{n-1}\left(\frac{\mu^{n}-1}{\mu-1}\right) & \text { if } \mu \neq 1 \\ n \sigma^{2} & \text { if } \mu=1\end{cases} $$ The case described above is known as a branching process, and an important question for a population that evolves along such lines is the probability that the population will eventually die out. Let \(\pi\) denote this probability when the population starts with a single individual. That is, $$ \pi=P\left\\{\text { population eventually dies out } \mid X_{0}=1\right. \text { ) } $$ (e) Argue that \(\pi\) satisfies $$ \pi=\sum_{j=0}^{\alpha} P_{j} \pi^{j} $$ HINT: Condition on the number of offspring of the initial member of the population.

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