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Let \(X\) and \(Y\) be independent with mean \(\mu\). Explain the error in the foliowing equation: \(\mathrm{E}(X \mid X+Y=z)=\mathrm{E}(X \mid X=z-Y)=E(z-Y)=z-\mu\)

Short Answer

Expert verified
The error is assuming \( \mathrm{E}(X \mid X+Y=z) = z - \mu \) without considering the conditional dependency properly.

Step by step solution

01

Analyze the Given Equation

The given equation is \( \mathrm{E}(X \mid X+Y=z)=\mathrm{E}(X \mid X=z-Y)=\mathrm{E}(z-Y)=z-\mu \). It suggests that the expected value of \( X \) given \( X+Y = z \) can somehow be equated to \( z-Y \) and then to \( z-\mu \). Let's analyze each part.
02

Understanding Independence

Since \( X \) and \( Y \) are independent with mean \( \mu \), this independence affects conditional expectations. Conditional expectation \( \mathrm{E}(X \mid X+Y = z) \) depends on the relationship that any realization of \( X \) and \( Y \) must satisfy \( X+Y = z \), not just \( X = z-Y \).
03

Conditional Expectation Formula

The correct formula for conditional expectation when \( X \) and \( Y \) are independent is based on the law of total expectation. \( \mathrm{E}(X \mid X+Y=z) \) is influenced by the distribution of \( Y \) and the constraint \( X + Y = z \), not by treating \( X \) directly as \( z - Y \).
04

Error Identification

The error lies in equating \( \mathrm{E}(X \mid X+Y=z) \) directly to \( \mathrm{E}(z-Y) \). The expectation \( \mathrm{E}(z-Y) = z - \mathrm{E}(Y) = z - \mu \) incorrectly ignores the condition \( X+Y = z \) which implicitly alters \( X \)'s distribution given \( z \).
05

Correct Formulation

The correct approach involves calculating \( \mathrm{E}(X \mid X+Y=z) \) with the consideration that \( Y = z - X \) influences the distribution of \( X \). This requires detailed probability distribution calculations rather than simple arithmetic manipulation.

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Key Concepts

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

Independence of Random Variables
When we say random variables are independent, it means that the occurrence or value of one does not affect the probability distribution of the other. In simpler terms, knowledge of one variable does not give us information about the other.
For example, consider rolling two separate dice. The result on one die does not influence the result on the other. Mathematically, for independent random variables \(X\) and \(Y\), this is expressed by the equation:
  • \( P(X = x \text{ and } Y = y) = P(X = x) \, P(Y = y) \)
Understanding independence is crucial when calculating expectations and conditional expectations, as independence can often simplify complex probability calculations. In this problem, the independence of \(X\) and \(Y\) should inform how we approach their conditional expectation, ensuring that one variable's effect does not improperly skew the analysis.
Law of Total Expectation
The Law of Total Expectation provides a way to break down expected values into simpler components. This rule states that we can calculate an expected value by considering several conditional expectations and their probabilities.
This law can be very useful in problems involving conditional expectations, especially when dealing with independent random variables. The general formula is:
  • \( \mathrm{E}(X) = \mathrm{E}[ \mathrm{E}(X \mid Y) ] \)
Essentially, it says that the overall expectation of a random variable can be understood as an average of its conditional expectations. When \(X\) and \(Y\) are independent, the total expectation takes into account these conditions, allowing us to calculate \(\mathrm{E}(X \mid X+Y=z)\) more accurately by considering the integral over all possible values of \(Y\).
This understanding prevents errors in assumptions and calculations, as it guides the correct formulation of conditional expectation in the context of independent random variables.
Probability Distribution
Probability distribution describes the likelihood of different outcomes for a random variable. In the context of independent variables, it's crucial to calculate the distribution correctly as it lays the foundation for expected values and other statistical calculations.
Each random variable has its own probability distribution, which can take many forms, such as a normal distribution, binomial distribution, etc. When we talk about conditional distributions, like \( \mathrm{E}(X \mid X+Y=z) \), we are referring to how the probability distribution of one variable changes when influenced by some condition or another variable, though, not directly since they are independent.
  • For example, to find the correct probability distribution for \(X\) when \(X + Y = z\), it might involve using integral or summation methods, depending on the form of the distribution.
  • This knowledge enables us to make correct inferences when calculating expectations, especially in multi-variable cases.
Understanding the probability distribution helps identify where mistakes might arise, such as incorrectly assuming that one variable can be isolated without considering its conditional environment.
Expectation Calculation Error
In probability, an expectation calculation error occurs when premise errors or simplifications lead to incorrect results. In this exercise, such an error arises from improperly equating \(\mathrm{E}(X \mid X+Y=z)\) with \(z - \mu\).
The error lies in capturing the conditionality incorrectly. Instead of simply manipulating variables algebraically, it is important to adhere to probability rules and incorporate actual distributions and dependences.
  • Misapplying law of total expectation or independence principles leads to such errors.
  • Assuming one can subtract the mean directly from a conditional expression ignores new dependencies introduced by conditioning.
To avoid expectation calculation errors, always maintain the integrity of the probability structure. For example, always consider how conditions such as \(X+Y=z\) could affect the distribution of \(X\), making sure all assumptions about distribution and independence are accounted for throughout calculations.

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Most popular questions from this chapter

A secretary drops \(n\) matching pairs of letters and envelopes down the stairs, and then places the letters into the envelopes in a random order. Use indicators to show that the number \(X\) of correctly matched pairs has mean and variance 1 for all \(n \geq 2\). Show that the mass function of \(X\) converges to a Poisson mass function as \(n \rightarrow \infty\).

If one picks a numerical entry at random from an almanac, or the annual accounts of a corporation, the first two significant digits, \(X\), \(Y\), are found to have approximately the joint mass function $$ f(x, y)=\log _{10}\left(1+\frac{1}{10 x+y}\right), \quad 1 \leq x \leq 9,0 \leq y \leq 9 $$ Find the mass function of \(X\) and an approximation to its mean. [A heuristic explanation for this phenomenon may be found in the second of Feller's volumes (1971).]

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