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Chapter 7: Problem 52

Show how to compute \(\operatorname{Cov}(X, Y)\) from the joint moment generating function of \(X\) and \(Y\).

Short Answer

Expert verified
To compute the covariance of two random variables \(X\) and \(Y\) from their joint moment generating function \(M_{X,Y}(t_1, t_2)\), follow these steps: 1. Find the means of \(X\) and \(Y\): \[ \mu_X = \frac{\partial M_{X,Y}}{\partial t_1} (0,0) \quad \text{and} \quad \mu_Y = \frac{\partial M_{X,Y}}{\partial t_2} (0,0) \] 2. Find the mixed second moment of \(X\) and \(Y\): \[ E[XY] = \frac{\partial^2 M_{X,Y}}{\partial t_1 \partial t_2} (0,0) \] 3. Compute the covariance \(\operatorname{Cov}(X, Y)\) as: \[ \operatorname{Cov}(X, Y) = E[XY] - \mu_X \mu_Y \]

Step by step solution

01

1.Definition of Moment Generating Function (MGF) and Joint MGF

The moment generating function (MGF) of a random variable X is defined as \( M(t) = E[e^{tX}] \), where \(t\) is any real number and \(E[ ]\) represents the expected value. Similarly, the joint moment generating function (JMGF) of two random variables X and Y can be defined as \( M_{X,Y}(t_1, t_2) = E[e^{t_1X+t_2Y}] \), where \(t_1\) and \(t_2\) are any real numbers.
02

2. Derive the Mean of X and Y from Joint MGF

We can find the mean of X (\(\mu_X\)) and Y (\(\mu_Y\)) using the joint MGF by computing the first derivatives of the MGF with respect to \(t_1\) and \(t_2\), and evaluating them at \(t_1 = 0\) and \(t_2 = 0\), respectively: \[ \mu_X = \frac{\partial M_{X,Y}}{\partial t_1} (0,0) = E[X] \] \[ \mu_Y = \frac{\partial M_{X,Y}}{\partial t_2} (0,0) = E[Y] \]
03

3. Derive the Second Moments of X and Y from Joint MGF

Next, we can find the second moments of X and Y: \[ E[X^2] = \frac{\partial^2 M_{X,Y}}{\partial^2 t_1} (0,0) \] \[ E[Y^2] = \frac{\partial^2 M_{X,Y}}{\partial^2 t_2} (0,0) \]
04

4. Derive the Mixed Second Moment of X and Y from Joint MGF

Now we can find the mixed second moment, which is required to compute the covariance: \[ E[XY] = \frac{\partial^2 M_{X,Y}}{\partial t_1 \partial t_2} (0,0) \]
05

5. Compute Covariance of X and Y

Finally, we can compute the covariance of X and Y as follows: \[ \operatorname{Cov}(X, Y) = E[XY] - \mu_X \mu_Y \] Using the values obtained from the previous steps, we can compute the covariance. Remember that the joint MGF contains all the information needed to find these moments.

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Key Concepts

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

Moment Generating Function (MGF)
The Moment Generating Function (MGF) is a powerful tool in the field of probability and statistics, especially when dealing with random variables. It provides a way to summarize all the moments of a random variable.

The MGF is defined for a random variable \( X \) as \( M(t) = E[e^{tX}] \), where \( t \) is any real number and \( E[ ] \) represents the expected value. The purpose of the MGF is to encode information about the entire distribution of the random variable. From the MGF, if it exists for \( t \) in a neighborhood of zero, you can derive all the moments of the distribution:
  • The first derivative of the MGF, evaluated at zero, gives the mean \( \mu_X \) of the random variable.
  • The second derivative, evaluated at zero, gives the second moment, which can be used to calculate the variance.
This makes the MGF a handy function as it consolidates the entire distribution information in a single expression.
Joint MGF
The Joint Moment Generating Function (JMGF) extends the concept of the MGF to multiple random variables. If you have two random variables \( X \) and \( Y \), their joint MGF is given by:
  • \( M_{X,Y}(t_1, t_2) = E[e^{t_1 X + t_2 Y}] \)
The joint MGF captures the relationship between \( X \) and \( Y \), encoding not just their individual distributions, but also their combined behavior.

The JMGF can be used to compute various moments by taking specific partial derivatives:
  • The first partial derivatives (w.r.t. \( t_1 \) and \( t_2 \)) evaluated at \( t_1 = 0 \) and \( t_2 = 0 \) will give the means \( E[X] \) and \( E[Y] \).
  • The second mixed partial derivative evaluated at \( t_1 = 0 \) and \( t_2 = 0 \) provides the mixed moment \( E[XY] \), which is essential for computing covariance.
This function greatly simplifies the process of discovering the relationship between two random variables, allowing the computation of complex quantities like covariance.
Random Variables
Random variables represent numerical outcomes from a random phenomenon. They are a fundamental concept in probability theory, and they are categorized into discrete and continuous types, based on the nature of the outcomes they represent.
  • Discrete Random Variables: These take on a countable number of distinct values. An example could be the number of heads in a series of coin tosses.
  • Continuous Random Variables: These can take on any value within a certain range or interval. An example could be the temperature on a given day.
Both types of random variables can be analyzed through their probability distributions, which describe the likelihood of each possible outcome.

In terms of MGFs, while discrete random variables use probabilities for each specific outcome, continuous random variables use a probability density function (PDF). Analyzing these variables often involves calculating expected values and variances, which provide a summary of the distribution's central tendency and spread, respectively. Understanding random variables is crucial for effectively using MGFs and JMGFs in statistical analysis.

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Most popular questions from this chapter

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.

One ball at a time is randomly selected from an um containing \(a\) white and \(b\) black balls until all of the remaining balls are of the same color. Let \(M_{a, b}\) denote the expected number of balls left in the urn when the experiment ends. Compute a recursive formula for \(M_{a, b}\) and solve when \(a=3, b=5\).

A coin having probability \(p\) of landing heads is flipped \(n\) ti es. Compute the expected number of runs of heads of size 1 , of size 2, of \(:\) e \(k, 1 \leq k \leq n\).

Um 1 contains 5 white and 6 black balls, while um 2 contains 8 white and 10 black balls. Two balls are randomly selected from um 1 and are then put. in urn 2 . If 3 balls are then randomly selected from urn 2 , compute the expected number of white balls in the trio. HNT: Let \(X_{i}=1\) if the \(i\) th white ball initially in urn 1 is one of the three selected, and let \(X_{i}=0\) otherwise. Similarly, let \(Y_{i}=1\) if the ith white ball from urn 2 is one of the three selected, and let \(Y_{i}=0\) otherwise. The number of white balls in the trio can now be written as \(\sum_{1}^{5} X_{i}+\sum_{1}^{8} Y_{i}\).

Let \(X_{(i)}, i=1, \ldots, n\), denote the order statistics from a set of \(n\) uniform. \((0,1)\) random variables and note that the density function of \(X_{(i)}\) is given by $$ f(x)=\frac{n !}{(i-1) !(n-i) !} x^{i-1}(1-x)^{n-i} \quad 0
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