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Show that (a) \(\quad E[X Y \mid Y=y]=y E[X \mid Y=y]\) (b) \(E[g(X, Y) \mid Y=y]=E[g(X, y) \mid Y=y]\) (c) \(E[X Y]=E[Y E[X \mid Y]]\)

Short Answer

Expert verified
In summary, we have proved the following properties of conditional expectation: (a) \(E[XY \mid Y = y] = y E[X \mid Y = y]\) (b) \(E[g(X, Y) \mid Y = y] = E[g(X, y) \mid Y = y]\) (c) \(E[XY] = E[Y E[X \mid Y]]\)

Step by step solution

01

Proof for Part (a)

: We want to show that \(E[XY \mid Y = y] = y E[X \mid Y = y]\). By definition of conditional expectation, we have: \(E[XY \mid Y = y] = \sum_x x\cdot y \cdot P(X = x \mid Y = y)\) But, since the value of y is constant in this expression, we can write: \(E[XY \mid Y = y] = y \sum_x x \cdot P(X = x \mid Y = y) = y E[X \mid Y = y]\) This proves (a).
02

Proof for Part (b)

: We want to show that \(E[g(X, Y) \mid Y = y] = E[g(X, y) \mid Y = y]\). By definition of conditional expectation, we have: \(E[g(X, Y) \mid Y = y] = \sum_x g(x, y) \cdot P(X = x \mid Y = y)\) Notice that the function g(X, Y) is now independent of Y because we replace Y with y when we calculate the particular value g(x, y). Thus, we can write: \(E[g(X, Y) \mid Y = y] = \sum_x g(x, y) \cdot P(X = x \mid Y = y) = E[g(X, y) \mid Y = y]\) This proves (b).
03

Proof for Part (c)

: We want to show that \(E[XY] = E[Y E[X \mid Y]]\). First, let's find the value of \(E[XY | Y]\). By definition of conditional expectation, we have: \(E[XY \mid Y] = \sum_x x P(X = x \mid Y = y) \cdot Y = Y \sum_x x \cdot P(X = x \mid Y = y)\) Now, using the Law of Iterated Expectation, we have: \(E[XY] = E[E[XY \mid Y]]\) \(E[XY] = E[Y \sum_x x \cdot P(X = x \mid Y = y)] = E[Y E[X \mid Y]]\) This proves (c).

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Key Concepts

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

Law of Iterated Expectation
The Law of Iterated Expectation is a powerful tool in probability theory. It connects the concept of conditional expectation with regular expectation. Essentially, it states that the expectation of a random variable can be expressed through an iterative process of taking expectations.

The formula for the Law of Iterated Expectation is:
  • \( E[X] = E[E[X \mid Y]] \)
This means that the overall expectation of \( X \) can be computed by first calculating the expectation of \( X \) given another variable \( Y \), and then taking the expectation of that result across all possible values of \( Y \).

This law is particularly useful when dealing with joint distributions or complex dependencies between random variables. It helps us break down the problem into simpler parts. For example, in part (c) of the exercise, we used the law to express \( E[XY] \) in terms of the expectation of \( Y \cdot \) an inner expectation \(E[X \mid Y]\).

In practice, the Law of Iterated Expectation helps in managing uncertainties and conditional dynamics in various fields such as finance, risk management, and statistics.
Joint Distribution
When working with probability, understanding joint distributions is crucial for analyzing multiple random variables simultaneously. A joint distribution gives us the probabilities of different combinations of outcomes for two or more random variables.

For example, if \( X \) and \( Y \) are two random variables, the joint distribution \( P(X = x, Y = y) \) tells us the probability that \( X \) equals \( x \) while \( Y \) equals \( y \).

In the context of conditional expectations, as shown in parts (a) and (b) of the exercise, joint distributions are essential. They allow us to derive expressions like \( E[XY \mid Y = y] \), using the probabilities of \( X \) given \( Y \).

Furthermore, joint distributions help in calculating conditional probabilities and expectations, simplifying complex multivariable problems into manageable tasks. By understanding the behavior and dependencies between variables, we can accurately compute their joint effects and relationships, thus providing a deeper insight into their interactions.
Probability Theory
Probability theory is the mathematical framework for quantifying uncertainty. It's foundational to understanding random processes and is the language in which we express concepts like conditional expectation and joint distribution.

Key elements of probability theory include:
  • *Random Variables*: Functions that assign numerical values to outcomes of random phenomena.
  • *Probability Distributions*: Functions that provide the probabilities of outcomes of random variables.
  • *Expected Values*: The long-term average or mean of random variables.
  • *Conditional Probability*: The probability of an event given that another event has occurred.
In the context of the exercise, we use probability theory to understand the calculations and transformations involved. It's what allows us to transition smoothly between joint distributions, conditional expectations, and overall expectations.

With probability theory, we can model real-world phenomena, making predictions and decisions based on uncertain data. It's what enables us to handle risks, manage uncertainties, and optimize outcomes in many fields including science, engineering, economics, and beyond.

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Most popular questions from this chapter

Polya's urn model supposes that an urn initially contains \(r\) red and \(b\) blue balls. At each stage a ball is randomly selected from the urn and is then returned along with \(m\) other balls of the same color. Let \(X_{k}\) be the number of red balls drawn in the first \(k\) selections. (a) Find \(E\left[X_{1}\right]\) (b) Find \(E\left[X_{2}\right]\). (c) Find \(E\left[X_{3}\right]\). (d) Conjecture the value of \(E\left[X_{k}\right]\), and then verify your conjecture by a conditioning argument. (e) Give an intuitive proof for your conjecture. Hint: Number the initial \(r\) red and \(b\) blue balls, so the urn contains one type \(i\) red ball, for each \(i=1, \ldots, r ;\) as well as one type \(j\) blue ball, for each \(j=1, \ldots, b\). Now suppose that whenever a red ball is chosen it is returned along with \(m\) others of the same type, and similarly whenever a blue ball is chosen it is returned along with \(m\) others of the same type. Now, use a symmetry argument to determine the probability that any given selection is red.

Let \(X\) be exponential with mean \(1 / \lambda ;\) that is, $$ f_{X}(x)=\lambda e^{-\lambda x}, \quad 01]\)

\(A\) and \(B\) roll a pair of dice in turn, with \(A\) rolling first. A's objective is to obtain a sum of 6 , and \(B\) 's is to obtain a sum of 7 . The game ends when either player reaches his or her objective, and that player is declared the winner. (a) Find the probability that \(A\) is the winner. (b) Find the expected number of rolls of the dice. (c) Find the variance of the number of rolls of the dice.

The joint density of \(X\) and \(Y\) is $$ f(x, y)=\frac{\left(y^{2}-x^{2}\right)}{8} e^{-y}, \quad 0

In the list problem, when the \(P_{i}\) are known, show that the best ordering (best in the sense of minimizing the expected position of the element requested) is to place the elements in decreasing order of their probabilities. That is, if \(P_{1}>P_{2}>\cdots>P_{n}\) show that \(1,2, \ldots, n\) is the best ordering.

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