91Ó°ÊÓ

Show that the following function satisfies the properties of a joint probability mass function. $$ \begin{array}{ccc} \hline x & y & f_{X Y}(x, y) \\ \hline-1 & -2 & 1 / 8 \\ -0.5 & -1 & 1 / 4 \\ 0.5 & 1 & 1 / 2 \\ 1 & 2 & 1 / 8 \\ \hline \end{array} $$ Determine the following: (a) \(P(X < 0.5, Y < 1.5)\) (b) \(P(X < 0.5)\) (c) \(P(Y < 1.5)\) (d) \(P(X > 0.25, Y < 4.5)\) (e) \(E(X), E(Y), V(X),\) and \(V(Y)\) (f) Marginal probability distribution of the random variable \(X\) (g) Conditional probability distribution of \(Y\) given that \(X=1\) (h) Conditional probability distribution of \(X\) given that \(Y=1\) (i) \(E(X \mid y=1)\) (j) Are \(X\) and \(Y\) independent?

Step by step solution

01

Verify Joint PMF Properties

To verify that the given function is a valid joint PMF, we must check two conditions:1. **Non-negativity**: Each value in the table should be non-negative. All given values \( \frac{1}{8}, \frac{1}{4}, \frac{1}{2}, \frac{1}{8} \) are non-negative.2. **Sum to 1**: The sum of all probabilities should equal 1: \[ \frac{1}{8} + \frac{1}{4} + \frac{1}{2} + \frac{1}{8} = \frac{1}{8} + \frac{2}{8} + \frac{4}{8} + \frac{1}{8} = \frac{8}{8} = 1 \]

02

Determine P(X < 0.5, Y < 1.5)

Identify all events where both \(X < 0.5\) and \(Y < 1.5\): - \(X = -1, Y = -2\) with probability \(\frac{1}{8}\) - \(X = -0.5, Y = -1\) with probability \(\frac{1}{4}\)Add these probabilities:\[P(X < 0.5, Y < 1.5) = \frac{1}{8} + \frac{1}{4} = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}\]

03

Determine P(X < 0.5)

Consider all events where \(X < 0.5\): - \(X = -1\) with probability \(\frac{1}{8}\) - \(X = -0.5\) with probability \(\frac{1}{4}\)Add the probabilities:\[P(X < 0.5) = \frac{1}{8} + \frac{1}{4} = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}\]

04

Determine P(Y < 1.5)

Identify events where \(Y < 1.5\):- \(Y = -2\) with probability \(\frac{1}{8}\) - \(Y = -1\) with probability \(\frac{1}{4}\)Add these probabilities:\[P(Y < 1.5) = \frac{1}{8} + \frac{1}{4} = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}\]

05

Determine P(X > 0.25, Y < 4.5)

All combinations satisfy \(X > 0.25\) and \(Y < 4.5\), as all \(Y\) values are less than 4.5 and selected \(X\) values are greater than 0.25:- \(X = 0.5, Y = 1\) with probability \(\frac{1}{2}\)- \(X = 1, Y = 2\) with probability \(\frac{1}{8}\)Sum the probabilities:\[P(X > 0.25, Y < 4.5) = \frac{1}{2} + \frac{1}{8} = \frac{4}{8} + \frac{1}{8} = \frac{5}{8}\]

06

Calculate Expected Values E(X) and E(Y)

Use the expected value formula \(E(X) = \sum x_i P(x_i)\) and \(E(Y) = \sum y_i P(y_i)\).For \(E(X)\): - \(E(X) = (-1)\frac{1}{8} + (-0.5)\frac{1}{4} + (0.5)\frac{1}{2} + 1\frac{1}{8} = -\frac{1}{8} - \frac{1}{8} + \frac{1}{2} + \frac{1}{8} = \frac{1}{8}\)For \(E(Y)\):- \(E(Y) = (-2)\frac{1}{8} + (-1)\frac{1}{4} + (1)\frac{1}{2} + 2\frac{1}{8} = -\frac{2}{8} - \frac{2}{8} + \frac{4}{8} + \frac{2}{8} = 0\)

07

Calculate Variances V(X) and V(Y)

Use the variance formula \(V(X) = E(X^2) - (E(X))^2\) and \(V(Y) = E(Y^2) - (E(Y))^2\).For \(V(X)\):- \(E(X^2) = (-1)^2 \frac{1}{8} + (-0.5)^2 \frac{1}{4} + (0.5)^2 \frac{1}{2} + 1^2 \frac{1}{8} = \frac{1}{8} + \frac{1}{16} + \frac{1}{8} + \frac{1}{8} = \frac{9}{16}\)- \(V(X) = \frac{9}{16} - \left(\frac{1}{8}\right)^2 = \frac{9}{16} - \frac{1}{64}\)- \(V(X) = \frac{36}{64} - \frac{1}{64} = \frac{35}{64}\)For \(V(Y)\):- \(E(Y^2) = (-2)^2 \frac{1}{8} + (-1)^2 \frac{1}{4} + (1)^2 \frac{1}{2} + 2^2 \frac{1}{8} = \frac{4}{8} + \frac{1}{4} + \frac{1}{2} + \frac{4}{8} = \frac{7}{4}\)- \(V(Y) = \frac{7}{4} - 0^2 = \frac{7}{4}\)

08

Determine Marginal Distribution of X

The marginal distribution is obtained by summing probabilities across all values of \(y\) for each \(x\):- \(P(X = -1) = \frac{1}{8}\)- \(P(X = -0.5) = \frac{1}{4}\)- \(P(X = 0.5) = \frac{1}{2}\)- \(P(X = 1) = \frac{1}{8}\)

09

Determine Conditional Distribution of Y Given X=1

Evaluate \(P(Y | X=1)\) using \(P(Y=y_i, X=1)/P(X=1)\):- \(P(Y=2 | X=1) = \frac{\frac{1}{8}}{\frac{1}{8}} = 1\)Since \(X=1\) occurs only when \(Y=2\), \(P(Y=2 | X=1)=1\) and other probabilities are 0.

10

Determine Conditional Distribution of X Given Y=1

Calculate \(P(X | Y=1)\) using \(P(X=x_i, Y=1)/P(Y=1)\):- \(P(X=0.5 | Y=1) = \frac{\frac{1}{2}}{\frac{1}{2}} = 1\)Thus, \(P(X=0.5 | Y=1)=1\) and other probabilities are 0.

11

Calculate E(X | Y=1)

Since \(P(X=0.5 | Y=1)=1\), expectation:\(E(X | Y=1) = 0.5 \times 1 = 0.5\)

12

Assess Independence of X and Y

Check independence by determining whether the product of marginals equals the joint distribution for all pairs. For any given pair, if \(P(X=x_i, Y=y_i) eq P(X=x_i) \times P(Y=y_i)\), they are dependent.Example:- \(P(X = 0.5, Y = 1) = \frac{1}{2}\), but \(P(X = 0.5) \times P(Y = 1) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\).Since the products are not equal, \(X\) and \(Y\) are dependent.

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Key Concepts

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

Marginal Distribution

The concept of marginal distribution is essential in understanding the behavior of a single random variable within a joint probability distribution. To find the marginal distribution of a random variable, we sum over all possible values of the other variable. This isolates the random variable we're interested in.

For instance, consider a joint probability mass function (PMF) with variables \(X\) and \(Y\). To find the marginal distribution of \(X\), we sum the probabilities of all combinations of \(Y\) for each \(X\):

• \(P(X = -1) = \frac{1}{8}\)
• \(P(X = -0.5) = \frac{1}{4}\)
• \(P(X = 0.5) = \frac{1}{2}\)
• \(P(X = 1) = \frac{1}{8}\)

This tells us the standalone probability distribution of \(X\), which is useful in understanding how \(X\) behaves regardless of \(Y\).

Conditional Probability

Conditional probability helps us determine the likelihood of an event occurring given that another event has already occurred. It’s calculated using the formula:\[P(A | B) = \frac{P(A \cap B)}{P(B)}\]where \(P(A \cap B)\) is the probability of both events occurring, and \(P(B)\) is the probability of the given event.

For example, to find the probability of \(Y\) given \(X = 1\), we use: \[P(Y \mid X=1) = \frac{P(Y=2, X=1)}{P(X=1)}\] In this exercise:

• \(P(Y = 2 | X = 1) = 1\) because \(P(Y=2, X=1) = \frac{1}{8}\) and \(P(X=1) = \frac{1}{8}\).

Thus, given that \(X=1\), \(Y=2\) with certainty. Conditional probabilities provide insight into the dependent relationships between random variables.

Independence

In probability theory, two random variables are independent if the occurrence of one does not affect the probability of occurrence of the other. Mathematically, \(X\) and \(Y\) are independent if: \[P(X = x_i, Y = y_i) = P(X = x_i) \, \times \, P(Y = y_i)\] for all \(i\).

In this exercise, to check for independence:

• We note that for \(X = 0.5\) and \(Y = 1\), \(P(0.5, 1) = \frac{1}{2}\)
• However, \(P(0.5) \times P(1) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\)

The inequality indicates dependency since the joint and product of marginals do not match. Understanding independence is crucial as it simplifies the calculation of probabilities and expectations.

Expected Value

The expected value is a key concept in probability which summarizes the average outcome of a random variable by weighing each possible value by its probability. The formula for expected value of \(X\), \(E(X)\), is given by: \[E(X) = \sum x_i P(x_i)\]

In the exercise, to find \(E(X)\):

• \((-1)\times\frac{1}{8} + (-0.5)\times\frac{1}{4} + (0.5)\times\frac{1}{2} + 1\times\frac{1}{8} = \frac{1}{8}\)

Similarly, for \(E(Y)\), it is calculated using their respective probabilities and outcomes. Notably, expected values are fundamental in statistics as they predict average phenomena over time and trials, aiding in decision making and risk assessment.

Variance

Variance measures how much the values of a random variable differ from the expected value. It indicates the spread or dispersion of the possible values and is given by:\[V(X) = E(X^2) - (E(X))^2\]

Understanding variance helps in assessing the variability or consistency of a distribution. In practice:

• Compute \(E(X^2)\), the expected value of the squares of \(X\).
• Subtract the square of the expected value from it to get variance.

For \(X\) in the exercise,

• \(E(X^2) = \frac{9}{16}\)
• \(V(X) = \frac{9}{16} - (\frac{1}{8})^2 = \frac{35}{64}\)

High variance indicates more dispersion. Thus, variance is vital for understanding the reliability and predictability of a dataset.