Problem 13 - Applied Statistics and Probability for Engineers

Question

Determine the value of \(c\) such that the function \(f(x, y)=c x y\) for \(01.8,1

Step-by-Step Solution

Answer

(a) \(\frac{4}{9}\); (b) \(\frac{25}{36}\); (c) \(\frac{21}{36}\); (d) \(0.345\); (e) \(2\); (f) \(0\); (g) \(\frac{2x}{9}\); (h) \(\frac{2y}{9}\); (i) \(2\); (j) \(\frac{4}{9}\); (k) \(\frac{2x}{9}\).

Step 1: Verify properties for a joint probability density function

The function \( f(x, y) = cxy \) must satisfy two conditions to be a joint probability density function: 1. Non-negativity: \( f(x, y) \geq 0 \) for all \(x, y\) in the domain.2. Normalization: \( \int_0^3 \int_0^3 f(x, y) \, dx \, dy = 1 \).Here, \(cxy\) is non-negative for \(0 < x < 3 \) and \(0 < y < 3 \). Now we compute the integral: \[ \int_0^3 \int_0^3 cxy \, dx \, dy = c \int_0^3 x \, dx \int_0^3 y \, dy. \]Computing the integrals individually: \[ \int_0^3 x \, dx = \left[ \frac{x^2}{2} \right]_0^3 = \frac{9}{2}, \quad \int_0^3 y \, dy = \left[ \frac{y^2}{2} \right]_0^3 = \frac{9}{2}. \]So, \[ \int_0^3 \int_0^3 cxy \, dx \, dy = c \times \frac{9}{2} \times \frac{9}{2} = \frac{81c}{4}. \]Setting this equal to 1 for normalization gives \( \frac{81c}{4} = 1 \), solving for \(c\) we find \( c = \frac{4}{81} \).

Step 2: Calculate (a) \( P(X<2, Y<3) \)

For the probability \( P(X<2, Y<3) \), we integrate the joint density function \( f(x, y) = \frac{4}{81}xy \) over the region \( 0 < x < 2 \) and \( 0 < y < 3 \):\[ P(X<2, Y<3) = \int_0^2 \int_0^3 \frac{4}{81}xy \, dy \, dx. \]First solve the integral with respect to \(y\):\[ \int_0^3 \frac{4}{81}xy \, dy = \frac{4x}{81} \left[ \frac{y^2}{2} \right]_0^3 = \frac{4x}{81} \cdot \frac{9}{2} = \frac{18x}{81} = \frac{2x}{9}. \]Now, integrate with respect to \(x\):\[ \int_0^2 \frac{2x}{9} \, dx = \frac{2}{9} \left[ \frac{x^2}{2} \right]_0^2 = \frac{2}{9} \cdot 2 = \frac{4}{9}. \]

Step 3: Calculate (b) \( P(X<2.5) \)

Since we need the marginal probability distribution of \(X\) to solve this, take:\[ f_X(x) = \int_0^3 \frac{4}{81}xy \, dy = \frac{2x}{9} \].Now compute \( P(X<2.5) \): \[ P(X<2.5) = \int_0^{2.5} \frac{2x}{9} \, dx = \frac{2}{9} \left[ \frac{x^2}{2} \right]_0^{2.5} = \frac{2}{9} \cdot \frac{6.25}{2} = \frac{6.25}{9} = \frac{25}{36}. \]

Step 4: Calculate (c) \( P(1<Y<2.5) \)

First, find the marginal probability distribution of \(Y\): \[ f_Y(y) = \int_0^3 \frac{4}{81}xy \, dx = \left[ \frac{4}{81} \cdot \frac{x^2}{2} \right]_0^3 y = \frac{6y}{27} = \frac{2y}{9}. \]Calculate \( P(1

Step 5: Calculate (d) \( P(X>1.8, 1<Y<2.5) \)

For this probability, integrate over the region \(1.8 < x < 3\) and \(1 < y < 2.5\):\[ P(X>1.8, 1

Step 6: Calculate (e) \( E(X) \)

The expected value \(E(X)\) is calculated using the marginal density \(f_X(x)\):\[ E(X) = \int_0^3 x f_X(x) \, dx = \int_0^3 x \frac{2x}{9} \, dx = \frac{2}{9} \int_0^3 x^2 \, dx. \]Compute the integral:\[ \int_0^3 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^3 = \frac{27}{3} = 9. \]Thus, \[ E(X) = \frac{2}{9} \times 9 = 2. \]

Step 7: Calculate (f) \( P(X<0, Y<4) \)

Since \(X\) and \(Y\) are strictly positive variables within \(0 < x < 3\) and \(0 < y < 3\), the probability of \(X < 0\) or \(Y > 3\) is zero. Therefore,\[ P(X<0, Y<4) = 0. \]

Step 8: Find (g) Marginal probability distribution of \(X\)

From Step 3:The marginal distribution of \(X\) was previously found as:\[ f_X(x) = \frac{2x}{9}. \]

Step 9: Find (h) Conditional probability distribution of \(Y\) given \(X=1.5\)

The conditional probability is given by:\[ f_{Y \mid X}(y \mid X=1.5) = \frac{f(1.5, y)}{f_X(1.5)}. \]Where,\[ f(1.5, y) = \frac{4}{81} \cdot 1.5 \cdot y = \frac{6y}{81} = \frac{2y}{27}, \]and from Step 3, \( f_X(1.5) = \frac{3}{9} = \frac{1}{3} \).Thus,\[ f_{Y \mid X}(y \mid X=1.5) = \frac{2y/27}{1/3} = \frac{2y}{9}. \]

Step 10: Calculate (i) \( E(Y \mid X=1.5) \)

Using the conditional distribution \( f_{Y \mid X}(y \mid X=1.5) = \frac{2y}{9} \), compute:\[ E(Y \mid X=1.5) = \int_0^3 y \cdot \frac{2y}{9} \, dy = \frac{2}{9} \int_0^3 y^2 \, dy. \]Compute the integral:\[ \int_0^3 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^3 = \frac{27}{3} = 9. \]Therefore,\[ E(Y \mid X=1.5) = \frac{2}{9} \times 9 = 2. \]

Step 11: Calculate (j) \( P(Y<2 \mid X=1.5) \)

From the conditional density \( f_{Y \mid X}(y \mid X=1.5) = \frac{2y}{9} \), find:\[ P(Y<2 \mid X=1.5) = \int_0^2 \frac{2y}{9} \, dy = \frac{2}{9} \left[ \frac{y^2}{2} \right]_0^2 = \frac{2}{9} \cdot 2 = \frac{4}{9}. \]

Step 12: Find (k) Conditional probability distribution of \(X\) given \(Y=2\)

By finding \( f_{Y}(2) \) as part of the joint and marginal integration:From Step 4, we know the marginal distribution of \( Y \):\[ f_{Y}(y) = \frac{2y}{9}. \]Then \( f_{Y}(2) = \frac{2 \times 2}{9} = \frac{4}{9}. \)Now, calculate:\[ f_{X \mid Y}(x \mid Y=2) = \frac{f(x, 2)}{f_Y(2)} = \frac{\frac{8x}{81}}{4/9} = \frac{72x/81}{4/9} = \frac{2x}{9}. \]This is the conditional probability distribution for \(X\) given \(Y=2\).

Key Concepts

Conditional Probability

Conditional probability allows us to understand how one event affects the probability of another event. In the context of random variables, it helps to calculate the probability of one variable given the value of another variable. For example, to find the conditional probability distribution of \(Y\) given \(X = 1.5\), we use the formula:\[f_{Y \mid X}(y \mid X=1.5) = \frac{f(1.5, y)}{f_X(1.5)}\]Here, you compute the probability density function at a specific condition (\(X=1.5\)) and divide it by the marginal probability of \(X\) at that condition. By using this formula, conditional probability translates the influence of \(X\) on \(Y\). Therefore, the computed density, \(f_{Y \mid X}(y \mid X=1.5)\), illustrates how likely different values of \(Y\) are when we know \(X = 1.5\). This method helps understand dependencies between random variables in a joint distribution.

Marginal Distribution

Marginal distribution answers the question, "What is the probability distribution of a single random variable, ignoring the presence of others?". This is achieved by integrating out the other variables from the joint distribution. For example, if you want to find the marginal distribution of \(X\), you integrate over all possible values of \(Y\):\[f_X(x) = \int_0^3 f(x, y) \, dy = \int_0^3 \frac{4}{81}xy \, dy = \frac{2x}{9}\]Similarly, the marginal distribution of \(Y\) can be found by integrating over all \(X\):\[f_Y(y) = \int_0^3 f(x, y) \, dx\]These marginal distributions are useful because they provide a complete probability model for each individual variable, disregarding the others. They are fundamental for understanding the behavior of a single variable when considered in isolation from the joint setting.

Expectation of Random Variables

The expectation of a random variable, often referred to as the expected value, is a measure of its average or central value. For example, to find \(E(X)\), you use the marginal distribution of \(X\):\[E(X) = \int_0^3 x f_X(x) \, dx = \int_0^3 x \frac{2x}{9} \, dx\]The integration process results in the average value of \(X\) over its range. Similarly, the conditional expectation \(E(Y \mid X=1.5)\) is the expected value of \(Y\) given a condition on \(X\):\[E(Y \mid X=1.5) = \int_0^3 y f_{Y \mid X}(y \mid X=1.5) \, dy\]These calculations offer insight into the expected behavior of random variables, either overall or conditional upon certain scenarios. By interpreting these values, one gains a deeper understanding of the random variables' tendency to assume particular values.

Integration in Probability

Integration is a fundamental tool in probability, especially when working with continuous random variables. It allows us to compute probabilities, expectations, and marginal distributions. In the example of determining the constant \(c\) to satisfy the joint probability density function, integration is used to ensure that the total probability across the domain equals 1:\[\int_0^3 \int_0^3 cxy \, dx \, dy = 1\]For probability calculations, integration over a specified region gives the probability of the random variable outcomes within that range. When calculating expectations or marginal distributions, integration helps sum up all possible contributions of values across their entire domain. Each integral evaluation in probability represents the area under the curve of the density function, and thus directly relates to the probability of events. Mastery of integration is crucial for effectively analyzing and solving problems in probability theory.

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Problem 12

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Problem 14

Other exercises in this chapter

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Problem 12

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Problem 14

Determine the value of \(c\) that makes the function \(f(x, y)=c(x+y)\) a joint probability density function over the range \(01)\) (d) \(P(X2 \mid X=1)\) (k) C

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Problem 15

Determine the value of \(c\) that makes the function \(f(x, y)=c(x+y)\) a joint probability density function over the range \(01)\) (d) \(P(X2 \mid X=1)\) (k) C

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Key Concepts

Conditional Probability

Conditional probability allows us to understand how one event affects the probability of another event. In the context of random variables, it helps to calculate the probability of one variable given the value of another variable. For example, to find the conditional probability distribution of \(Y\) given \(X = 1.5\), we use the formula:\[f_{Y \mid X}(y \mid X=1.5) = \frac{f(1.5, y)}{f_X(1.5)}\]Here, you compute the probability density function at a specific condition (\(X=1.5\)) and divide it by the marginal probability of \(X\) at that condition. By using this formula, conditional probability translates the influence of \(X\) on \(Y\). Therefore, the computed density, \(f_{Y \mid X}(y \mid X=1.5)\), illustrates how likely different values of \(Y\) are when we know \(X = 1.5\). This method helps understand dependencies between random variables in a joint distribution.

Marginal Distribution

Marginal distribution answers the question, "What is the probability distribution of a single random variable, ignoring the presence of others?". This is achieved by integrating out the other variables from the joint distribution. For example, if you want to find the marginal distribution of \(X\), you integrate over all possible values of \(Y\):\[f_X(x) = \int_0^3 f(x, y) \, dy = \int_0^3 \frac{4}{81}xy \, dy = \frac{2x}{9}\]Similarly, the marginal distribution of \(Y\) can be found by integrating over all \(X\):\[f_Y(y) = \int_0^3 f(x, y) \, dx\]These marginal distributions are useful because they provide a complete probability model for each individual variable, disregarding the others. They are fundamental for understanding the behavior of a single variable when considered in isolation from the joint setting.

Expectation of Random Variables

The expectation of a random variable, often referred to as the expected value, is a measure of its average or central value. For example, to find \(E(X)\), you use the marginal distribution of \(X\):\[E(X) = \int_0^3 x f_X(x) \, dx = \int_0^3 x \frac{2x}{9} \, dx\]The integration process results in the average value of \(X\) over its range. Similarly, the conditional expectation \(E(Y \mid X=1.5)\) is the expected value of \(Y\) given a condition on \(X\):\[E(Y \mid X=1.5) = \int_0^3 y f_{Y \mid X}(y \mid X=1.5) \, dy\]These calculations offer insight into the expected behavior of random variables, either overall or conditional upon certain scenarios. By interpreting these values, one gains a deeper understanding of the random variables' tendency to assume particular values.

Integration in Probability

Integration is a fundamental tool in probability, especially when working with continuous random variables. It allows us to compute probabilities, expectations, and marginal distributions. In the example of determining the constant \(c\) to satisfy the joint probability density function, integration is used to ensure that the total probability across the domain equals 1:\[\int_0^3 \int_0^3 cxy \, dx \, dy = 1\]For probability calculations, integration over a specified region gives the probability of the random variable outcomes within that range. When calculating expectations or marginal distributions, integration helps sum up all possible contributions of values across their entire domain. Each integral evaluation in probability represents the area under the curve of the density function, and thus directly relates to the probability of events. Mastery of integration is crucial for effectively analyzing and solving problems in probability theory.