# Exercise on Joint Probability Density Function### For the joint PDF of Exercise 30 in Chapter 4 (page 170):**(a)** Find the correlation coefficient of $ X $ and $ Y $.**(b)** Find the conditional expectation, $ E(Y \mid x) $.**(c)** Find the conditional variance, $ \text{Var}(Y \mid x) $.This exercise involves analyzing the joint probability density function (PDF) to determine key statistical measures such as the correlation coefficient, the conditional expectation, and the conditional variance. Each part requires a different approach to derive these statistical parameters based on the provided joint PDF.The correlation coefficient quantifies the degree to which two variables $ X $ and $ Y $ are linearly related. The conditional expectation $ E(Y \mid x) $ provides the expected value of $ Y $ given a specific value of $ x $. Lastly, the conditional variance $ \text{Var}(Y \mid x) $ describes the variability of $ Y $ given $ x $. ### Problem 30Suppose $ X $ and $ Y $ are continuous random variables with joint probability density function (pdf) $ f(x, y) = 60x^2y $ if $ 0 < x, 0 < y, x + y < 1 $, and zero otherwise.**(a) Find the marginal pdf of $ X $.**To find the marginal pdf of $ X $, we integrate the joint pdf over all possible values of $ Y $:\[ f_X(x) = \int_{0}^{1-x} 60x^2y \, dy \]**(b) Find the conditional pdf of $ Y $ given $ X = x $.**The conditional pdf $ f_{Y|X}(y|x) $ is given by:\[ f_{Y|X}(y|x) = \frac{f(x,y)}{f_X(x)} \]**(c) Find $ P(Y > .1 | X = .5) $.**The probability $ P(Y > .1 | X = .5) $ can be found by integrating the conditional pdf $ f_{Y|X}(y|x) $ over the desired range:\[ P(Y > .1 | X = .5) = \int_{0.1}^{0.5} f_{Y|X}(y|x=0.5) \, dy \]### Detailed Instructions:1. **Marginal pdf of $ X $:** - Determine the range for $ y $: $ 0 < y < 1 - x $ - Integrate $ 60x^2y $ with respect to $ y $ from $ 0 $ to $ 1 - x $2. **Conditional pdf of $ Y $ given $ X = x $:** - Compute $ f_X(x) $ from part (a) - Substitute into the formula $ f_{Y|X}(y|x) $3. **Probability Calculation:** - Use the conditional pdf obtained in part (b) - Integrate from $ 0.1 $ to $ 0.5 $By carefully following these steps, you will be able to solve for the marginal pdf, conditional pdf, and the specific conditional probability as required.

(a). Determine the correlation coefficient of X and Y: The probability density function is given…

Answered: For the joint pdf of Exercise 30 in…

For the joint pdf of Exercise 30 in Chapter 4 (page 170): (a) Find the correlation coefficient of X and Y. (b) Find the conditional expectation, E(Y|x). (c) Find the conditional variance, Var(Y|x).

MATLAB: An Introduction with Applications

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ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

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# Exercise on Joint Probability Density Function

### For the joint PDF of Exercise 30 in Chapter 4 (page 170):

**(a)** Find the correlation coefficient of $ X $ and $ Y $.

**(b)** Find the conditional expectation, $ E(Y \mid x) $.

**(c)** Find the conditional variance, $ \text{Var}(Y \mid x) $.

This exercise involves analyzing the joint probability density function (PDF) to determine key statistical measures such as the correlation coefficient, the conditional expectation, and the conditional variance. Each part requires a different approach to derive these statistical parameters based on the provided joint PDF.

The correlation coefficient quantifies the degree to which two variables $ X $ and $ Y $ are linearly related. The conditional expectation $ E(Y \mid x) $ provides the expected value of $ Y $ given a specific value of $ x $. Lastly, the conditional variance $ \text{Var}(Y \mid x) $ describes the variability of $ Y $ given $ x $.

### Problem 30

Suppose $ X $ and $ Y $ are continuous random variables with joint probability density function (pdf) $ f(x, y) = 60x^2y $ if $ 0 < x, 0 < y, x + y < 1 $, and zero otherwise.

**(a) Find the marginal pdf of $ X $.**

To find the marginal pdf of $ X $, we integrate the joint pdf over all possible values of $ Y $:

\[ f_X(x) = \int_{0}^{1-x} 60x^2y \, dy \]

**(b) Find the conditional pdf of $ Y $ given $ X = x $.**

The conditional pdf $ f_{Y|X}(y|x) $ is given by:

\[ f_{Y|X}(y|x) = \frac{f(x,y)}{f_X(x)} \]

**(c) Find $ P(Y > .1 | X = .5) $.**

The probability $ P(Y > .1 | X = .5) $ can be found by integrating the conditional pdf $ f_{Y|X}(y|x) $ over the desired range:

\[ P(Y > .1 | X = .5) = \int_{0.1}^{0.5} f_{Y|X}(y|x=0.5) \, dy \]

### Detailed Instructions:

1. **Marginal pdf of $ X $:**
- Determine the range for $ y $: $ 0 < y < 1 - x $
- Integrate $ 60x^2y $ with respect to $ y $ from $ 0 $ to $ 1 - x $

2. **Conditional pdf of $ Y $ given $ X = x $:**
- Compute $ f_X(x) $ from part (a)
- Substitute into the formula $ f_{Y|X}(y|x) $

3. **Probability Calculation:**
- Use the conditional pdf obtained in part (b)
- Integrate from $ 0.1 $ to $ 0.5 $

By carefully following these steps, you will be able to solve for the marginal pdf, conditional pdf, and the specific conditional probability as required.

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