Let's solve the problem step-by-step as requested. --- # **Given Information** The joint PMF of \( X \) and \( Y \) is: | \(X \backslash Y\) | | 1 | |--------------------|-----|-----| | | . | .1 | | 1 | .1 | .1 | | 2 | .2 | .1 | | 3 | .3 | .1 | --- # **Concepts and Definitions** - **Marginal PMF**: \( p_X(x) = \sum_y p_{X,Y}(x, y) \), \( p_Y(y) = \sum_x p_{X,Y}(x, y) \) - **Conditional PMF**: \( p_{Y|X}(y|x) = \frac{p_{X,Y}(x, y)}{p_X(x)} \) - **Expectation**: \( E[X] = \sum_x x p_X(x) \) - **Variance**: \( \text{Var}[X] = E[X^2] - (E[X])^2 \) - **Covariance**: \( \sigma_{XY} = E[XY] - E[X]E[Y] \) - **Independence**: \( p_{X,Y}(x, y) = p_X(x) p_Y(y) \) - **New Variable**: \( Z = X Y \) - **CDF**: \( F_Z(z) = P(Z \leq z) \) --- ## **1(a) Marginal and Conditional PMFs** **Marginal PMF of \( X \):** \[ p_X(x) = \sum_y p_{X,Y}(x, y) \] - \( p_X() = . + .1 = .1 \) - \( p_X(1) = .1 + .1 = .2 \) - \( p_X(2) = .2 + .1 = .3 \) - \( p_X(3) = .3 + .1 = .4 \) **Marginal PMF of \( Y \):** \[ p_Y(y) = \sum_x p_{X,Y}(x, y) \] - \( p_Y() = . + .1 + .2 + .3 = .6 \) - \( p_Y(1) = .1 + .1 + .1 + .1 = .4 \) **Conditional PMF \( p_{Y|X}(y|x) \):** \[ p_{Y|X}(y|x) = \frac{p_{X,Y}(x, y)}{p_X(x)} \] - For \( x = \): \( p_{Y|X}(|) = , \; p_{Y|X}(1|) = 1 \) - For \( x = 1 \): \( p_{Y|X}(|1) = .5, \; p_{Y|X}(1|1) = .5 \) - For \( x = 2 \): \( p_{Y|X}(|2) = \frac{2}{3}, \; p_{Y|X}(1|2) = \frac{1}{3} \) - For \( x = 3 \): \( p_{Y|X}(|3) = .75, \; p_{Y|X}(1|3) = .25 \) --- ## **1(b) Mean, Variances, Covariance, and Independence** \[ E[X] = \sum_x x p_X(x) = \cdot .1 + 1 \cdot .2 + 2 \cdot .3 + 3 \cdot .4 = + .2 + .6 + 1.2 = 2. \] \[ E[Y] = \sum_y y p_Y(y) = \cdot .6 + 1 \cdot .4 = .4 \] \[ E[X^2] = \sum_x x^2 p_X(x) = ^2 \cdot .1 + 1^2 \cdot .2 + 2^2 \cdot .3 + 3^2 \cdot .4 = + .2 + 1.2 + 3.6 = 5. \] \[ \text{Var}[X] = E[X^2] - (E[X])^2 = 5. - (2.)^2 = 1. \] \[ E[Y^2] = \sum_y y^2 p_Y(y) = ^2 \cdot .6 + 1^2 \cdot .4 = .4 \] \[ \text{Var}[Y] = E[Y^2] - (E[Y])^2 = .4 - (.4)^2 = .24 \] \[ E[XY] = \sum_{x,y} x y p_{X,Y}(x, y) = \cdot \cdot . + \cdot 1 \cdot .1 + 1 \cdot \cdot .1 + 1 \cdot 1 \cdot .1 + 2 \cdot \cdot .2 + 2 \cdot 1 \cdot .1 + 3 \cdot \cdot .3 + 3 \cdot 1 \cdot .1 \] \[ = + + + .1 + + .2 + + .3 = .1 + .2 + .3 = .6 \] \[ \sigma_{XY} = E[XY] - E[X]E[Y] = .6 - 2. \times .4 = .6 - .8 = -.2 \] **Check Independence:** If \( X \) and \( Y \) were independent, \( p_{X,Y}(x, y) = p_X(x)p_Y(y) \). For \( (x=, y=1) \): - \( p_{X,Y}(,1) = .1 \) - \( p_X() p_Y(1) = .1 \times .4 = .04 \) They are **not independent**. --- ## **2(a) PMF, Mean, Variance of \( Z = X Y \)** Possible values of \( Z \): \( , 1, 2, 3 \) - \( Z = \): occurs when \( X = \) or \( Y = \): sum all such pairs: \( (,): . \), \( (,1): .1 \), \( (1,): .1 \), \( (2,): .2 \), \( (3,): .3 \) Total: \( . + .1 + .1 + .2 + .3 = .7 \) - \( Z = 1 \): \( (1,1): .1 \) - \( Z = 2 \): \( (2,1): .1 \) - \( Z = 3 \): \( (3,1): .1 \) So, - \( p_Z() = .7 \) - \( p_Z(1) = .1 \) - \( p_Z(2) = .1 \) - \( p_Z(3) = .1 \) \[ E[Z] = \cdot .7 + 1 \cdot .1 + 2 \cdot .1 + 3 \cdot .1 = + .1 + .2 + .3 = .6 \] \[ E[Z^2] = ^2 \cdot .7 + 1^2 \cdot .1 + 2^2 \cdot .1 + 3^2 \cdot .1 = + .1 + .4 + .9 = 1.4 \] \[ \text{Var}[Z] = E[Z^2] - (E[Z])^2 = 1.4 - (.6)^2 = 1.4 - .36 = 1.04 \] --- ## **2(b) CDF \( F_Z(z) \) and Probability \( P(Z > ) \)** \[ F_Z(z) = P(Z \leq z) \] - For \( z < \): \( F_Z(z) = \) - For \( \leq z < 1 \): \( F_Z(z) = .7 \) - For \( 1 \leq z < 2 \): \( F_Z(z) = .7 + .1 = .8 \) - For \( 2 \leq z < 3 \): \( F_Z(z) = .8 + .1 = .9 \) - For \( z \geq 3 \): \( F_Z(z) = 1. \) \[ P(Z > ) = 1 - F_Z() = 1 - .7 = .3 \] --- ## **CDF Plot** ```python import matplotlib.pyplot as plt z_vals = [-1, , 1, 2, 3, 4] cdf_vals = [, .7, .8, .9, 1., 1.] plt.step(z_vals, cdf_vals, where='post', label='CDF $F_Z(z)$') plt.xlabel('$z$') plt.ylabel('$F_Z(z)$') plt.title('CDF of $Z = XY$') plt.grid(True) plt.legend() plt.show() ``` **[Graph will display as a step function at .7, .8, .9, 1.]** --- ## **Summary of Final Answers** - **Marginals:** \( p_X()=.1, p_X(1)=.2, p_X(2)=.3, p_X(3)=.4 \) \( p_Y()=.6, p_Y(1)=.4 \) - **Conditionals:** \( p_{Y|X}(y|x) \) as calculated above. - **Means:** \( E[X]=2., E[Y]=.4 \) - **Variances:** \( \text{Var}[X]=1., \text{Var}[Y]=.24 \) - **Covariance:** \( \sigma_{XY} = -.2 \) - **Independence:** \( X \) and \( Y \) are **not independent**. - **For \( Z = XY \):** \( p_Z()=.7, p_Z(1)=.1, p_Z(2)=.1, p_Z(3)=.1 \) \( E[Z] = .6, \; \text{Var}[Z] = 1.04 \) - **CDF \( F_Z(z) \):** \( F_Z(z) = \) for \( z < \), \( F_Z(z) = .7 \) for \( \leq z < 1 \), \( F_Z(z) = .8 \) for \( 1 \leq z < 2 \), \( F_Z(z) = .9 \) for \( 2 \leq z < 3 \), \( F_Z(z) = 1. \) for \( z \geq 3 \) - **Probability \( P(Z > ) = .3 \)** --- **CDF Graph:**  *Alt text: Step plot of the CDF \( F_Z(z) \) for Z = XY with jumps at z=, 1, 2, 3 as described above.*

# **Given Information** The joint PMF of random variables \( X \) and \( Y \) is represented in the following table: | \(X \backslash Y\) | | 1 | |--------------------|-----|-----| | | . | .1 | | 1 | .1 | .1 | | 2 | .2 | .1 | | 3 | .3 | .1 | --- # **Concepts and Definitions** - **Marginal PMF**: The marginal probability mass function (PMF) for a random variable can be calculated by summing the joint PMF over the other variable: \[ p_X(x) = \sum_y p_{X,Y}(x, y), \quad p_Y(y) = \sum_x p_{X,Y}(x, y) \] - **Conditional PMF**: The conditional PMF is given by: \[ p_{Y|X}(y|x) = \frac{p_{X,Y}(x, y)}{p_X(x)} \] - **Expectation**: The expected value is calculated as: \[ E[X] = \sum_x x p_X(x) \] - **Variance**: Variance measures the spread of a random variable: \[ \text{Var}[X] = E[X^2] - (E[X])^2 \] - **Covariance**: Covariance between two random variables is defined as: \[ \sigma_{XY} = E[XY] - E[X]E[Y] \] - **Independence**: Two random variables \( X \) and \( Y \) are independent if: \[ p_{X,Y}(x, y) = p_X(x) p_Y(y) \] - **New Variable**: The new variable \( Z \) is defined as \( Z = XY \). - **CDF**: The cumulative distribution function (CDF) is defined as: \[ F_Z(z) = P(Z \leq z) \] --- ## **1(a) Marginal and Conditional PMFs** **Marginal PMF of \( X \)**: \[ p_X(x) = \sum_y p_{X,Y}(x, y) \] Calculating: - \( p_X(1) = 0.1 + 0.1 = 0.2 \) - \( p_X(2) = 0.2 + 0.1 = 0.3 \) - \( p_X(3) = 0.3 + 0.1 = 0.4 \) **Marginal PMF of \( Y \)**: \[ p_Y(y) = \sum_x p_{X,Y}(x, y) \] Calculating: - \( p_Y(1) = 0.1 + 0.1 + 0.1 + 0.1 = 0.4 \) **Conditional PMF \( p_{Y|X}(y|x) \)**: \[ p_{Y|X}(y|x) = \frac{p_{X,Y}(x, y)}{p_X(x)} \] Calculating for each \( x \): - For \( x=1 \): \( p_{Y|X}(1|1) = \frac{0.1}{0.2} = 0.5 \) - For \( x=2 \): \( p_{Y|X}(1|2) = \frac{0.1}{0.3} \approx 0.33 \) - For \( x=3 \): \( p_{Y|X}(1|3) = \frac{0.1}{0.4} = 0.25 \) --- ## **1(b) Mean, Variances, Covariance, and Independence** Calculating the expectations: \[ E[X] = \sum_x x p_X(x) = 1 \cdot 0.2 + 2 \cdot 0.3 + 3 \cdot 0.4 = 0.2 + 0.6 + 1.2 = 2 \] \[ E[Y] = \sum_y y p_Y(y) = 1 \cdot 0.4 = 0.4 \] Calculating \( E[X^2] \): \[ E[X^2] = \sum_x x^2 p_X(x) = 1^2 \cdot 0.2 + 2^2 \cdot 0.3 + 3^2 \cdot 0.4 = 0.2 + 0.8 + 3.6 = 4.6 \] Calculating Variance of \( X \): \[ \text{Var}[X] = E[X^2] - (E[X])^2 = 4.6 - (2)^2 = 4.6 - 4 = 0.6 \] Calculating \( E[Y^2] \): \[ E[Y^2] = \sum_y y^2 p_Y(y) = 1^2 \cdot 0.4 = 0.4 \] Calculating Variance of \( Y \): \[ \text{Var}[Y] = E[Y^2] - (E[Y])^2 = 0.4 - (0.4)^2 = 0.4 - 0.16 = 0.24 \] Calculating \( E[XY] \): \[ E[XY] = \sum_{x,y} xy p_{X,Y}(x,y) = 0.1 + 0.1 + 0.1 + 0.2 + 0.3 = 0.8 \] Calculating Covariance: \[ \sigma_{XY} = E[XY] - E[X]E[Y] = 0.8 - 2 \cdot 0.4 = 0.8 - 0.8 = 0 \] **Independence Check**: Check if \( p_{X,Y}(x,y) = p_X(x)p_Y(y) \) holds for any values: For \( (x=1, y=1) \): - \( p_{X,Y}(1,1) = 0.1 \) - \( p_X(1)p_Y(1) = 0.2 \cdot 0.4 = 0.08 \) Since \( 0.1 \neq 0.08 \), \( X \) and \( Y \) are **not independent**. --- ## **2(a) PMF, Mean, Variance of \( Z = XY \)** Possible values of \( Z \) are \( 0, 1, 2, 3 \). Calculating \( p_Z(z) \): - For \( Z = 0 \) (occurs when \( X = 0 \) or \( Y = 0 \)): - Total probability = \( 0.1 + 0.1 + 0.2 + 0.3 = 0.7 \) - For \( Z = 1 \): - \( (1,1): 0.1 \) - For \( Z = 2 \): - \( (2,1): 0.1 \) - For \( Z = 3 \): - \( (3,1): 0.1 \) Thus, \[ p_Z(0) = 0.7, \quad p_Z(1) = 0.1, \quad p_Z(2) = 0.1, \quad p_Z(3) = 0.1 \] Calculating expectation: \[ E[Z] = 0 \cdot 0.7 + 1 \cdot 0.1 + 2 \cdot 0.1 + 3 \cdot 0.1 = 0 + 0.1 + 0.2 + 0.3 = 0.6 \] Calculating \( E[Z^2] \): \[ E[Z^2] = 0^2 \cdot 0.7 + 1^2 \cdot 0.1 + 2^2 \cdot 0.1 + 3^2 \cdot 0.1 = 0 + 0.1 + 0.4 + 0.9 = 1.4 \] Calculating Variance of \( Z \): \[ \text{Var}[Z] = E[Z^2] - (E[Z])^2 = 1.4 - (0.6)^2 = 1.4 - 0.36 = 1.04 \] --- ## **2(b) CDF \( F_Z(z) \) and Probability \( P(Z > 0) \)** Calculating the CDF: \[ F_Z(z) = P(Z \leq z) \] - For \( z < 0 \): \( F_Z(z) = 0 \) - For \( 0 \leq z < 1 \): \( F_Z(z) = 0.7 \) - For \( 1 \leq z < 2 \): \( F_Z(z) = 0.7 + 0.1 = 0.8 \) - For \( 2 \leq z < 3 \): \( F_Z(z) = 0.8 + 0.1 = 0.9 \) - For \( z \geq 3 \): \( F_Z(z) = 1 \) Calculating \( P(Z > 0) \): \[ P(Z > 0) = 1 - F_Z(0) = 1 - 0.7 = 0.3 \] --- ## **CDF Plot** ```python import matplotlib.pyplot as plt z_vals = [-1, 0, 1, 2, 3, 4] cdf_vals = [0, 0.7, 0.8, 0.9, 1.0, 1.0] plt.step(z_vals, cdf_vals, where='post', label='CDF $F_Z(z)$') plt.xlabel('$z$') plt.ylabel('$F_Z(z)$') plt.title('CDF of $Z = XY$') plt.grid(True) plt.legend() plt.show() ``` *The graph will illustrate a step function showing the CDF values.* --- ## **Summary of Final Answers** - **Marginals:** \( p_X(0) = 0.1, p_X(1) = 0.2, p_X(2) = 0.3, p_X(3) = 0.4 \) \( p_Y(0) = 0.6, p_Y(1) = 0.4 \) - **Conditional PMFs:** - As calculated above for each \( x \). - **Means:** \( E[X] = 2, E[Y] = 0.4 \) - **Variances:** \( \text{Var}[X] = 0.6, \text{Var}[Y] = 0.24 \) - **Covariance:** \( \sigma_{XY} = 0 \) - **Independence:** \( X \) and \( Y \) are **not independent**. - **For \( Z = XY \):** \( p_Z(0) = 0.7, p_Z(1) = 0.1, p_Z(2) = 0.1, p_Z(3) = 0.1 \) \( E[Z] = 0.6, \; \text{Var}[Z] = 1.04 \) - **CDF \( F_Z(z) \):** - \( F_Z(z) = 0 \) for \( z < 0 \) - \( F_Z(z) = 0.7 \) for \( 0 \leq z < 1 \) - \( F_Z(z) = 0.8 \) for \( 1 \leq z < 2 \) - \( F_Z(z) = 0.9 \) for \( 2 \leq z < 3 \) - \( F_Z(z) = 1 \) for \( z \geq 3 \) - **Probability \( P(Z > 0) = 0.3 \)** --- **CDF Graph:**  *Alt text: Step plot of the CDF \( F_Z(z) \) for \( Z = XY \) with defined jumps at \( z = 0, 1, 2, 3 \).*

# **Given Information** The joint PMF of random variables \( X \) and \( Y \) is represented in the following table: | \(X \backslash Y\) | | 1 | |---------------------|------|------| | | . | 0.1 | | 1 | 0.1 | 0.1 | | 2 | 0.2 | 0.1 | | 3 | 0.3 | 0.1 | --- # **Concepts and Definitions** - **Marginal PMF**: The marginal PMF for \( X \) and \( Y \) can be calculated as: \[ p_X(x) = \sum_y p_{X,Y}(x, y), \quad p_Y(y) = \sum_x p_{X,Y}(x, y) \] - **Conditional PMF**: The conditional PMF is defined as: \[ p_{Y|X}(y|x) = \frac{p_{X,Y}(x, y)}{p_X(x)} \] - **Expectation**: The expected value is calculated as: \[ E[X] = \sum_x x p_X(x) \] - **Variance**: Variance measures the spread of a random variable: \[ \text{Var}[X] = E[X^2] - (E[X])^2 \] - **Covariance**: Covariance between two random variables is defined as: \[ \sigma_{XY} = E[XY] - E[X]E[Y] \] - **Independence**: Two random variables \( X \) and \( Y \) are independent if: \[ p_{X,Y}(x, y) = p_X(x) p_Y(y) \] - **New Variable**: The new variable \( Z \) is defined as \( Z = XY \). - **CDF**: The cumulative distribution function (CDF) is defined as: \[ F_Z(z) = P(Z \leq z) \] --- ## **1(a) Marginal and Conditional PMFs** **Marginal PMF of \( X \)**: \[ p_X(x) = \sum_y p_{X,Y}(x, y) \] Calculating: - \( p_X(1) = 0.1 + 0.1 = 0.2 \) - \( p_X(2) = 0.2 + 0.1 = 0.3 \) - \( p_X(3) = 0.3 + 0.1 = 0.4 \) **Marginal PMF of \( Y \)**: \[ p_Y(y) = \sum_x p_{X,Y}(x, y) \] Calculating: - \( p_Y(1) = 0.1 + 0.1 + 0.1 + 0.1 = 0.4 \) **Conditional PMF \( p_{Y|X}(y|x) \)**: \[ p_{Y|X}(y|x) = \frac{p_{X,Y}(x, y)}{p_X(x)} \] Calculating for each \( x \): - For \( x=1 \): \( p_{Y|X}(1|1) = \frac{0.1}{0.2} = 0.5 \) - For \( x=2 \): \( p_{Y|X}(1|2) = \frac{0.1}{0.3} \approx 0.33 \) - For \( x=3 \): \( p_{Y|X}(1|3) = \frac{0.1}{0.4} = 0.25 \) --- ## **1(b) Mean, Variances, Covariance, and Independence** Calculating the expectations: \[ E[X] = \sum_x x p_X(x) = 1 \cdot 0.2 + 2 \cdot 0.3 + 3 \cdot 0.4 = 0.2 + 0.6 + 1.2 = 2 \] \[ E[Y] = \sum_y y p_Y(y) = 1 \cdot 0.4 = 0.4 \] Calculating \( E[X^2] \): \[ E[X^2] = \sum_x x^2 p_X(x) = 1^2 \cdot 0.2 + 2^2 \cdot 0.3 + 3^2 \cdot 0.4 = 0.2 + 0.8 + 3.6 = 4.6 \] Calculating Variance of \( X \): \[ \text{Var}[X] = E[X^2] - (E[X])^2 = 4.6 - (2)^2 = 4.6 - 4 = 0.6 \] Calculating \( E[Y^2] \): \[ E[Y^2] = \sum_y y^2 p_Y(y) = 1^2 \cdot 0.4 = 0.4 \] Calculating Variance of \( Y \): \[ \text{Var}[Y] = E[Y^2] - (E[Y])^2 = 0.4 - (0.4)^2 = 0.4 - 0.16 = 0.24 \] Calculating \( E[XY] \): \[ E[XY] = \sum_{x,y} xy p_{X,Y}(x,y) = 1 \cdot 1 \cdot 0.1 + 1 \cdot 1 \cdot 0.1 + 2 \cdot 1 \cdot 0.1 + 3 \cdot 1 \cdot 0.1 \] Calculating: \[ = 0.1 + 0.1 + 0.2 + 0.3 = 0.7 \] Calculating Covariance: \[ \sigma_{XY} = E[XY] - E[X]E[Y] = 0.7 - 2 \cdot 0.4 = 0.7 - 0.8 = -0.1 \] **Independence Check**: Check if \( p_{X,Y}(x,y) = p_X(x)p_Y(y) \) holds for any values: For \( (x=1, y=1) \): - \( p_{X,Y}(1,1) = 0.1 \) - \( p_X(1)p_Y(1) = 0.2 \times 0.4 = 0.08 \) Since \( 0.1 \neq 0.08 \), \( X \) and \( Y \) are **not independent**. --- ## **2(a) PMF, Mean, Variance of \( Z = XY \)** Possible values of \( Z \) are \( 0, 1, 2, 3 \). Calculating \( p_Z(z) \): - For \( Z = 0 \): - Occurs when either \( X = 0 \) or \( Y = 0 \): - Total probability = \( 0.1 + 0.1 + 0.2 + 0.3 = 0.7 \) - For \( Z = 1 \): - \( (1,1): 0.1 \) - For \( Z = 2 \): - \( (2,1): 0.1 \) - For \( Z = 3 \): - \( (3,1): 0.1 \) Thus, \[ p_Z(0) = 0.7, \quad p_Z(1) = 0.1, \quad p_Z(2) = 0.1, \quad p_Z(3) = 0.1 \] Calculating expectation: \[ E[Z] = 0 \cdot 0.7 + 1 \cdot 0.1 + 2 \cdot 0.1 + 3 \cdot 0.1 = 0 + 0.1 + 0.2 + 0.3 = 0.6 \] Calculating \( E[Z^2] \): \[ E[Z^2] = 0^2 \cdot 0.7 + 1^2 \cdot 0.1 + 2^2 \cdot 0.1 + 3^2 \cdot 0.1 = 0 + 0.1 + 0.4 + 0.9 = 1.4 \] Calculating Variance of \( Z \): \[ \text{Var}[Z] = E[Z^2] - (E[Z])^2 = 1.4 - (0.6)^2 = 1.4 - 0.36 = 1.04 \] --- ## **2(b) CDF \( F_Z(z) \) and Probability \( P(Z > 0) \)** Calculating the CDF: \[ F_Z(z) = P(Z \leq z) \] - For \( z < 0 \): \( F_Z(z) = 0 \) - For \( 0 \leq z < 1 \): \( F_Z(z) = 0.7 \) - For \( 1 \leq z < 2 \): \( F_Z(z) = 0.7 + 0.1 = 0.8 \) - For \( 2 \leq z < 3 \): \( F_Z(z) = 0.8 + 0.1 = 0.9 \) - For \( z \geq 3 \): \( F_Z(z) = 1 \) Calculating \( P(Z > 0) \): \[ P(Z > 0) = 1 - F_Z(0) = 1 - 0.7 = 0.3 \] --- ## **CDF Plot** ```python import matplotlib.pyplot as plt z_vals = [-1, 0, 1, 2, 3, 4] cdf_vals = [0, 0.7, 0.8, 0.9, 1.0, 1.0] plt.step(z_vals, cdf_vals, where='post', label='CDF $F_Z(z)$') plt.xlabel('$z$') plt.ylabel('$F_Z(z)$') plt.title('CDF of $Z = XY$') plt.grid(True) plt.legend() plt.show() ``` *The graph will illustrate a step function showing the CDF values.* --- ## **Summary of Final Answers** - **Marginals:** \( p_X(1) = 0.2, p_X(2) = 0.3, p_X(3) = 0.4 \) \( p_Y(1) = 0.4 \) - **Conditional PMFs:** - As calculated above for each \( x \). - **Means:** \( E[X] = 2, E[Y] = 0.4 \) - **Variances:** \( \text{Var}[X] = 0.6, \text{Var}[Y] = 0.24 \) - **Covariance:** \( \sigma_{XY} = -0.1 \) - **Independence:** \( X \) and \( Y \) are **not independent**. - **For \( Z = XY \):** \( p_Z(0) = 0.7, p_Z(1) = 0.1, p_Z(2) = 0.1, p_Z(3) = 0.1 \) \( E[Z] = 0.6, \; \text{Var}[Z] = 1.04 \) - **CDF \( F_Z(z) \):** - \( F_Z(z) = 0 \) for \( z < 0 \) - \( F_Z(z) = 0.7 \) for \( 0 \leq z < 1 \) - \( F_Z(z) = 0.8 \) for \( 1 \leq z < 2 \) - \( F_Z(z) = 0.9 \) for \( 2 \leq z < 3 \) - \( F_Z(z) = 1 \) for \( z \geq 3 \) - **Probability \( P(Z > 0) = 0.3 \)** --- **CDF Graph:**  *Alt text: Step plot of the CDF \( F_Z(z) \) for \( Z = XY \) with defined jumps at \( z = 0, 1, 2, 3 \).*