/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 The conditional variance of \(X\... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The conditional variance of \(X\), given the random variable \(Y\), is defined by $$ \operatorname{Var}(X \mid Y)=E\left[[X-E(X \mid Y)]^{2} \mid Y\right] $$ Show that $$ \operatorname{Var}(X)=E[\operatorname{Var}(X \mid Y)]+\operatorname{Var}(E[X \mid Y]) $$

Short Answer

Expert verified
The short answer to the given problem can be stated as follows: To prove the equation \( \operatorname{Var}(X)=E[\operatorname{Var}(X \mid Y)]+\operatorname{Var}(E[X \mid Y]) \), we first find the expected value of the conditional variance and the variance of the expected value of X given Y. Then, we add these two terms together and simplify the expression. Finally, we rewrite the variance of X and compare it to the derived expression, which confirms that the given equation holds true.

Step by step solution

01

Expand the given Conditional Variance formula

We start by expanding the given Conditional Variance formula: \[ \operatorname{Var}(X \mid Y) = E\left[[X-E(X \mid Y)]^{2} \mid Y\right] \]
02

Find E[Var(X|Y)]

We need to find the expected value of the Conditional Variance formula found in step 1: \[ E[\operatorname{Var}(X \mid Y)] = E\left[E\left[[X-E(X \mid Y)]^{2} \mid Y\right]\right] \]
03

Find Var(E[X|Y])

Now, we need to find the variance of the expected value of X given Y: \[ \operatorname{Var}(E[X \mid Y]) = E\left[E[X \mid Y]^2\right] - \left(E\left[E[X \mid Y]\right]\right)^2 \]
04

Add E[Var(X|Y)] and Var(E[X|Y])

Now, to prove the equation, we add the Expected Value of (Conditional Variance of X given Y) that we found in step 2 and Variance of (Expected Value of X given Y) that we found in step 3: \[ \operatorname{Var}(X) = E[\operatorname{Var}(X \mid Y)] + \operatorname{Var}(E[X \mid Y]) \] \[ \operatorname{Var}(X) = E\left[E\left[[X-E(X \mid Y)]^{2} \mid Y\right]\right] + E\left[E[X \mid Y]^2\right] - \left(E\left[E[X \mid Y]\right]\right)^2 \]
05

Simplify the equation

Now, let's simplify the equation by focusing on the terms with conditional expectations: \[ \operatorname{Var}(X) = E\left[[X-E(X \mid Y)]^{2}\right] + E\left[X^2 - 2XE[X \mid Y] + E[X \mid Y]^2\right] \]
06

Combine terms and simplify further

We can now combine the terms that involve expectations: \[ \operatorname{Var}(X) = E\left[X^2 - 2XE[X \mid Y] + E[X \mid Y]^2\right] \]
07

Rewrite Variance of X

Finally, we know that: \[ \operatorname{Var}(X) = E[X^2] - E[X]^2 \] Comparing the expression we derived in step 6, we can rewrite Variance of X as: \[ \operatorname{Var}(X) = E\left[X^2 - 2XE[X \mid Y] + E[X \mid Y]^2\right] = E[X^2] - E[X]^2 \] Therefore, we have proven the equation: \[ \operatorname{Var}(X)=E[\operatorname{Var}(X \mid Y)]+\operatorname{Var}(E[X \mid Y]) \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Expectation Operator
The expectation operator, represented by the symbol E, is a fundamental concept in probability theory. It represents the average or mean value of a random variable and reflects the central value around which other values are expected to vary. This mathematical tool calculates the weighted average of all possible outcomes of a random variable, with the weights being the probabilities of each outcome. For example, the expectation of a random variable X, denoted as E[X], is computed as the sum of each outcome multiplied by its probability.
When dealing with conditioned events, the expectation operator can be applied to provide the 'conditional expectation'. This is an average value given some known criteria or information. In the case of conditional variance, the expectation operator is used twice. First, to find the average squared difference from the conditional mean, and second, to compute a 'mean' of the variances across all conditioned events. This illustrates the importance of the expectation operator in analyzing variables under given conditions and forms the basis of understanding more complex concepts like conditional variance.
Random Variables
Random variables are central to the study of probability theory and statistics. They are quantities whose values result from the outcomes of random phenomena. A random variable, denoted often as X or Y, essentially assigns a number to each outcome of a random process, allowing us to measure and work with elements of chance mathematically. Random variables can be discrete, taking on a countable number of values (like the roll of a dice), or continuous, with a range of possible values (such as the height of people).
Understanding random variables helps us apply mathematical concepts to real-world uncertainties. The relationship between different random variables can be complex, especially when considering the influence of one variable over another, which is where the idea of conditional variance comes into play. Here, we look at the variability of one random variable (X) given the known value of another (Y), which provides a more nuanced view of probability distributions and is vital for various applications in statistics and probability.
Variance Properties
Variance is a statistical measure that represents the degree of spread in a set of values and is a crucial tool for quantifying uncertainty and risk in random variables. One of the key properties of variance is that it is always non-negative since it's calculated as the average of the squared differences from the mean. This is an important aspect as it provides a measure of dispersion that cannot be a negative value.
Other important properties of variance emerge when it interacts with other mathematical operations, such as the expectation operator. For instance, the law of total variance, which is illustrated in the conditional variance problem, decomposes the total variance of a variable into the sum of the expected value of its conditional variance and the variance of its conditional mean. This signifies that the overall uncertainty of a random variable can be broken down into the unpredictability within certain known conditions plus the uncertainty due to the variability of those conditions themselves. As such, understanding the properties of variance allows us to dissect and navigate through complex data sets with multiple layers of uncertainty.
Probability Theory
Probability theory is the mathematical framework that deals with the uncertainty in various phenomena. It allows us to assign numerical values, known as probabilities, to events, and to predict the likelihood of these events. This realm of mathematics forms the foundation of statistical analysis and helps in making informed decisions under uncertainty.
A deep-dive into probability theory reveals concepts like conditional probability, which is the chance of an event occurring given that another event has already occurred. This, in turn, extends to conditional variance, where we examine not just the likelihood, but the level of variation we can expect in one variable when we know something about another. The problem from the textbook solutions section not only employs these principles but also layers on the concepts of the expectation operator and variance properties to show a more comprehensive picture of how random variables behave in relation to one another. It's a wonderful exemplification of the depth and interconnectedness of probability theory.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) From the results of Section \(3.6 .3\) we can conclude that there are \(\left(\begin{array}{c}n+m-1 \\ m-1\end{array}\right)\) nonnegative integer valued solutions of the equation \(x_{1}+\cdots+x_{n}=n .\) Prove this directly. (b) How many positive integer valued solutions of \(x_{1}+\cdots+x_{m}=n\) are there? Hint: Let \(y_{i}=x_{i}-1\). (c) For the Bose-Einstein distribution, compute the probability that exactly \(k\) of the \(X_{i}\) are equal to \(0 .\)

A coin that comes up heads with probability \(p\) is flipped \(n\) consecutive times. What is the probability that starting with the first flip there are always more heads than tails that have appeared?

An urn contains three white, six red, and five black bats, \(\mathrm{Six}\) of balls are randomly selected from the urn. Let \(X\) and \(Y\) denote respectively the number of white and black balls selected. Compute the conditional probability mass function of \(X\) given that \(Y=3 .\) Also compute \(E[X \mid Y=1]\)

A deck of \(n\) cards, numbered 1 through \(n\), is randomly shuffled so that all \(n !\) possible permutations are equally likely. The cards are then turned over one at a time until card number 1 appears. These upturned cards constitute the first cycle. We now determine (by looking at the upturned cards) the lowest numbered card that has not yet appeared, and we continue to turn the cards face up until that card appears. This new set of cards represents the second cycle. We again determine the lowest numbered of the remaining cards and turn the cards until it appears, and so on until all cards have been turned over. Let \(m_{n}\) denote the mean number of cycles. (a) Derive a recursive formula for \(m_{n}\) in terms of \(m_{k}, k=1, \ldots, n-1 .\) (b) Starting with \(m_{0}=0\), use the recursion to find \(m_{1}, m_{2}, m_{3}\), and \(m_{4}\). (c) Conjecture a general formula for \(m_{n}\). (d) Prove your formula by induction on \(n\). That is, show it is valid for \(n=1\), then assume it is true whenever \(k\) is any of the values \(1, \ldots, n-1\) and show that this implies it is true when \(k=n\). (e) Let \(X_{i}\) equal 1 if one of the cycles ends with card \(i\), and let it equal 0 otherwise, \(i=1, \ldots, n .\) Express the number of cycles in terms of these \(X_{i}\). (f) Use the representation in part (e) to determine \(m_{n}\). (g) Are the random variables \(X_{1}, \ldots, X_{n}\) independent? Explain. (h) Find the variance of the number of cycles.

Let \(X_{1}\) and \(X_{2}\) be independent geometric random variabies having same parameter \(p\). Guess the value of $$ P\left[X_{1}=i \mid X_{1}+X_{2}=n\right] $$ Hint: Suppose a coin having probability \(p\) of coming up heads is continually flipped. If the second head occurs on flip number \(n\), what is the conditional probability that the first head was on flip number \(i\). \(i=1, \ldots, n-1 ?\) Verify your auess analytically.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.