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Let \(X\) be a random digit \((0,1,2, \ldots, 9\) are equally likely) and let \(Y\) be a random digit not equal to \(X\). That is, the nine digits other than \(X\) are equally likely for \(Y\). a. Determine \(p_{X}(x), p_{Y X X}(y \mid x), p_{X, Y}(x, y)\). b. Determine a formula for \(E(Y \mid X=x)\). Is this a linear function of \(x\) ?

Short Answer

Expert verified
a. \(p_X(x) = \frac{1}{10}, p_{Y | X}(y | x) = \frac{1}{9}, p_{X,Y}(x,y) = \frac{1}{90}\). b. \(E(Y | X=x) = \frac{45-x}{9}\) is not a linear function of \(x\).

Step by step solution

01

Determine Probability Mass Function of X

Since each digit from 0 to 9 is equally likely for the random variable \(X\), the probability mass function \(p_X(x)\) assigns the same probability to each digit. Therefore,\[p_X(x) = \frac{1}{10} \text{ for each } x = 0, 1, 2, \ldots, 9.\]
02

Conditional Probability for Y Given X

Given \(X = x\), \(Y\) is any digit other than \(x\), thus there are 9 possible outcomes for \(Y\). Therefore, the conditional probability \(p_{Y | X}(y | x)\) is \[p_{Y | X}(y | x) = \frac{1}{9} \text{ for } y eq x.\]
03

Joint Probability for X and Y

To find \(p_{X,Y}(x,y)\), use the fact that \[p_{X,Y}(x,y) = p_X(x) \cdot p_{Y | X}(y | x).\]Since \(p_X(x) = \frac{1}{10}\) and \(p_{Y | X}(y | x) = \frac{1}{9}\), we get:\[p_{X,Y}(x,y) = \frac{1}{10} \times \frac{1}{9} = \frac{1}{90} \text{ for } y eq x.\]
04

Determine Expected Value E(Y | X=x)

The conditional expected value is given by \[E(Y | X=x) = \sum_{y eq x} y \cdot p_{Y | X}(y | x) = \sum_{y eq x} y \cdot \frac{1}{9}.\]This can be simplified to:\[E(y) = \frac{1}{9} \left(\sum_{y eq x} y\right).\]
05

Evaluate if E(Y | X=x) is a Linear Function

Calculate \(\sum_{y eq x} y\), which is the sum of all digits except \(x\), i.e., sum of 0 to 9 minus \(x\). The sum of digits 0 through 9 is 45.Thus, \[\sum_{y eq x} y = 45 - x\text{, so } E(Y | X=x) = \frac{45-x}{9}.\] This expression is not linear in \(x\) because it doesn't conform to the form \(ax + b\), specifically the term \(\frac{-x}{9}\).

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Key Concepts

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

Probability Mass Function
In probability theory, a Probability Mass Function (PMF) is used to describe the probability of each potential outcome for a discrete random variable. For the exercise involving the random variable \(X\), which represents a digit from 0 to 9, each digit is equally likely to occur because they are all chosen without bias. Thus, the probability assigned to each digit in this scenario is the same.
  • Probability of any specific digit \(x\): \(p_X(x) = \frac{1}{10}\).
  • This equal distribution reflects the uniformity of the selection process.
Such a PMF indicates a uniform distribution over the set of all digits (0 through 9). This united distribution implies that no individual outcome is more likely than another.
Conditional Probability
Conditional probability is a measure that quantifies the probability of an event occurring given that another event has already occurred. Here, we explore the conditional probability of \(Y\) given a specific \(X = x\). Given that \(X\) is a chosen digit, \(Y\) may be any other digit besides \(x\).
The probability space for \(Y\) thus comprises nine digits, making it equally likely for each digit other than \(x\) to be chosen as \(Y\).
  • For \(y eq x\), \(p_{Y | X}(y | x) = \frac{1}{9}\).
This means each permissible outcome for \(Y\) holds an equal chance of occurrence, provided \(X\) is known. Understanding such probabilities is crucial, especially when dealing with events where other factors or prior outcomes influence the probabilities.
Joint Probability
Joint probability considers the probability of simultaneous occurrences of two outcomes or events. In the context of the exercise, it involves evaluating the probability of choosing specific digits \(x\) and \(y\) for \(X\) and \(Y\), respectively. The joint probability \(p_{X,Y}(x,y)\) is computed by multiplying the probability of \(X=x\) by the conditional probability of \(Y=y\) given \(X=x\).
  • Using the formula: \(p_{X,Y}(x,y) = p_X(x) \cdot p_{Y | X}(y | x)\).
  • This becomes \(p_{X,Y}(x,y) = \frac{1}{10} \times \frac{1}{9} = \frac{1}{90}\).
This joint probability reflects the likelihood of two dependent outcomes occurring together, necessary for understanding complex arrangements of probabilistic events.
Expected Value
The expected value is a fundamental concept in probability that predicts a type of average outcome when an experiment is repeated multiple times. For the conditional expected value \(E(Y | X = x)\), it represents the mean value of \(Y\) assuming \(X=x\).
This involves taking a weighted average of all possible digit values for \(Y\), excluding \(x\), using the conditional probability:
  • Expression: \(E(Y | X = x) = \sum_{y eq x} y \cdot p_{Y | X}(y | x)\).
The simplification leads to \(E(Y | X=x) = \frac{45-x}{9}\), where 45 is the sum of digits from 0 to 9. The resulting expression demonstrates that \(E(Y | X = x)\) isn't linear in \(x\), highlighting the potentially non-linear relationships inherent in conditioned expectations, critical for both theoretical and applied settings.

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