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Chapter 3: Problem 35

If \(E[X \mid Y=y]=c\) for all \(y\), show that \(\operatorname{Cov}(X, Y)=0\).

Short Answer

Expert verified
Given that \(E[X \mid Y = y] = c\), we can find the covariance between X and Y as follows: 1. Define covariance and conditional expectation. 2. Expand the covariance using the given conditional expectation. 3. Apply the properties of conditional expectation. 4. Substitute the given conditional expectation into the equation. 5. Simplify the equation and note that the expected value of the random variable \(Y - E[Y]\) is equal to zero. 6. Multiply and get the result that the covariance, \(\operatorname{Cov}(X, Y) = 0\).

Step by step solution

01

Defining covariance and conditional expectation

The covariance between two random variables X and Y is defined as follows: \[ \operatorname{Cov}(X, Y) = E[(X - E[X])(Y - E[Y])] \] The conditional expectation of a random variable X given another random variable Y is denoted as \(E[X \mid Y]\), and since it is a constant for all possible values of Y in this case, we write it as \(E[X \mid Y = y] = c\).
02

Expanding the covariance using the given conditional expectation

Using the given information, we can expand the covariance using the expected value and the conditional expectation: \begin{align*} \operatorname{Cov}(X, Y) &= E\left[(X - E[X])(Y - E[Y])\right] \\ &= E\left[E\left[(X - E[X])(Y - E[Y]) \mid Y\right]\right] \end{align*}
03

Apply the properties of conditional expectation

Now we can apply the properties of conditional expectation to simplify the above expression: \begin{align*} \operatorname{Cov}(X, Y) &= E\left[E\left[(X - E[X])(Y - E[Y]) \mid Y\right]\right] \\ &= E\left[(E[X \mid Y] - E[X])(Y - E[Y])\right] \end{align*}
04

Substitute the given conditional expectation

Now, we substitute the given conditional expectation \(E[X \mid Y = y] = c\) into the equation: \begin{align*} \operatorname{Cov}(X, Y) &= E\left[(c - E[X])(Y - E[Y])\right] \end{align*}
05

Simplify the equation

Since both c and E[X] are constants, we can simplify the above equation by taking them out of the expected value: \begin{align*} \operatorname{Cov}(X, Y) &= E\left[(c - E[X])(Y - E[Y])\right] \\ &= (c - E[X])E\left[(Y - E[Y])\right] \end{align*} Notice that the expected value of the random variable \(Y - E[Y]\) is equal to zero: \[ E\left[(Y - E[Y])\right] = E[Y] - E[E[Y]] = E[Y] - E[Y] = 0 \]
06

Multiply and get the result

Now, we just need to multiply the expression to get the covariance: \begin{align*} \operatorname{Cov}(X, Y) &= (c - E[X])E\left[(Y - E[Y])\right] \\ &= (c - E[X]) \times 0 \\ &= 0 \end{align*} Therefore, \(\operatorname{Cov}(X, Y) = 0\) when \(E[X \mid Y = y] = c\) for all y.

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Key Concepts

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

Conditional Expectation
Understanding conditional expectation is crucial for various applications in probability and statistics. When we talk about the conditional expectation, denoted as \(E[X | Y]\), we are interested in the expected value of a random variable \(X\) given that another random variable \(Y\) is known to be a certain value. Think of it like adjusting our predictions or expectations based on new information we've received about related events.

Conditional expectation is an essential concept because it adjusts our estimations based on new data, reflecting a more accurate prediction under given circumstances. It's especially useful in scenarios where two variables have some form of dependency. In the case of the exercise, the notation \(E[X | Y = y] = c\) indicates that no matter what value \(Y\) takes, the expected value of \(X\) does not change; it remains as the constant \(c\). This property allows us to simplify and solve more complex problems like determining covariance, as seen in the solution provided.
Expected Value
The expected value is a fundamental concept in probability and statistics that represents the average outcome if an experiment is repeated many times. It's denoted as \(E[X]\) for a random variable \(X\), and can be thought of as the long-term average or the 'center' of the distribution of the random variable.

The expected value is a weighted average, where each possible outcome is weighted by its probability of occurrence. Expected value is not just about one-off expectations; it captures the idea of an average outcome over the long run. When we subtract the expected value from a random variable, as in \(X - E[X]\), we're essentially centering the variable at zero, which is useful in analyzing the variance and covariance. This centering technique aids in extracting insights from the data by removing the baseline level and focusing on deviations from the expected outcome.
Probability Models
Probability models are mathematical representations of random phenomena. They describe the structures and rules that govern the outcome of random processes or events. These models define a set of outcomes and assign probabilities to those outcomes, ensuring that the probabilities sum up to one.

Why are probability models important? They are critical tools for predicting future events, assessing risks, and making informed decisions under uncertainty. A probability model can simplify the complex real-world by providing a framework to calculate probabilities of events and statistical measures like mean, variance, and in our case, covariance.

Building a Model

To build a probability model, statisticians determine the sample space and the associated probabilities of outcomes. This involves identifying all possible results and ascribing a likelihood to each. It's the foundation upon which we rely for further analysis, like calculating conditional expectations or expected values which help in deducing properties like independence or correlation between random variables.

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Most popular questions from this chapter

There are three coins in a barrel. These coins, when flipped, will come up heads with respective probabilities \(0.3,0.5,0.7\). A coin is randomly selected from among these three and is then flipped ten times. Let \(N\) be the number of heads obtained on the ten flips. Find (a) \(P(N=0\\}\). (b) \(P\\{N=n\\}, n=0,1, \ldots, 10\) (c) Does \(N\) have a binomial distribution? (d) If you win \(\$ 1\) each time a head appears and you lose \(\$ 1\) each time a tail appears, is this a fair game? Explain.

Let \(X_{1}\) and \(X_{2}\) be independent geometric random variables having the saime 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 guess analytically.

A set of \(n\) dice is thrown. All those that land on six are put aside, and the others are again thrown. This is repeated until all the dice have landed on six. Let \(N\) denote the number of throws needed. (For instance, suppose that \(n=3\) and that on the initial throw exactly two of the dice land on six. Then the other die will be thrown, and if it lands on six, then \(N=2\).) Let \(m_{n}=E[N]\). (a) Derive a recursive formula for \(m_{n}\) and use it to calculate \(m_{i}, i=2,3,4\) and to show that \(m_{5} \approx 13.024\). (b) Let \(X_{i}\) denote the number of dice rolled on the \(i\) th throw. Find \(E\left[\sum_{i=1}^{N} X_{i}\right]\).

Show in the discrete case that if \(X\) and \(Y\) are independent, then $$ E[X \mid Y=y]=E[X] \quad \text { for all } y $$

Let \(X_{l}, i \geqslant 0\) be independent and identically distributed random variables with probability mass function $$p(j)=P\left[X_{i}=j\right\\}, \quad j=1, \ldots, m \quad \sum_{j=1}^{m} P(j)=1$$ Find \(E[N]\), where \(N=\min \left[n>0: X_{n}=X_{0}\right\\}\).
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