Problem 149 Let $Y_{1}$ and $Y_{2}$ have... [FREE SOLUTION]

Chapter 5: Problem 149

Let $Y_{1}$ and $Y_{2}$ have a joint density function given by $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 3 y_{1}, & 0 \leq y_{2} \leq y_{1} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. Find the marginal density functions of $Y_{1}$ and $Y_{2}$ b. Find $P\left(Y_{1} \leq 3 / 4 | Y_{2} \leq 1 / 2\right)$ c. Find the conditional density function of $Y_{1}$ given $Y_{2}=y_{2}$ d. Find $P\left(Y_{1} \leq 3 / 4 | Y_{2}=1 / 2\right)$

Short Answer

Expert verified

a. Marginal densities: $f_{Y_{1}}(y_1) = 3y_1^2$, $f_{Y_{2}}(y_2) = \frac{3}{2}(1 - y_2^2)$; b. $\frac{25}{44}$; c. $f_{Y_1|Y_2}(y_1|y_2) = \frac{2y_1}{1-y_2^2}$; d. $\frac{5}{12}$.

Step by step solution

Find marginal density of Y1

To find the marginal density function of $Y_1$, $f_{Y_{1}}(y_1)$, integrate the joint density over $y_2$: \[f_{Y_{1}}(y_1) = \int_{0}^{y_1} 3y_1 \, dy_2\]Compute the integral: \[f_{Y_{1}}(y_1) = 3y_1 \times \left[y_2\right]_{0}^{y_1} = 3y_1^2\]Thus, the marginal density function of $Y_1$ is $f_{Y_{1}}(y_1) = 3y_1^2$ for $0 \leq y_1 \leq 1$.

Find marginal density of Y2

To find the marginal density function of $Y_2$, $f_{Y_{2}}(y_2)$, integrate the joint density over $y_1$: \[f_{Y_{2}}(y_2) = \int_{y_2}^{1} 3y_1 \, dy_1\]Compute the integral:\[f_{Y_{2}}(y_2) = 3\left[\frac{y_1^2}{2} - \frac{y_2^2}{2}\right]_{y_2}^{1} = 3\left(\frac{1}{2} - \frac{y_2^2}{2}\right)\]Thus, the marginal density function of $Y_2$ is $f_{Y_{2}}(y_2) = \frac{3}{2}(1 - y_2^2)$ for $0 \leq y_2 \leq 1$.

Calculate conditional probability

To find $P(Y_1 \leq \frac{3}{4} | Y_2 \leq \frac{1}{2})$, use the fact that this is a conditional probability:\[P(Y_1 \leq \frac{3}{4} | Y_2 \leq \frac{1}{2}) = \frac{P(Y_1 \leq \frac{3}{4}, Y_2 \leq \frac{1}{2})}{P(Y_2 \leq \frac{1}{2})}\]First, find $P(Y_1 \leq \frac{3}{4}, Y_2 \leq \frac{1}{2})$: Integrate the joint probability:\[\int_{0}^{1/2}\int_{y_2}^{3/4} 3y_1 \, dy_1 \, dy_2\]Integrate with respect to $y_1$:\[\int_{0}^{1/2} \left.\frac{3y_1^2}{2}\right|_{y_2}^{3/4} \, dy_2 = \int_{0}^{1/2} (\frac{27}{32} - \frac{3y_2^2}{2}) \, dy_2\]Integrate with respect to $y_2$:\[\frac{27y_2}{32} - \frac{y_2^3}{2} |_{0}^{1/2} = \frac{27}{64} - \frac{1}{16} = \frac{25}{64}\]Now, find $P(Y_2 \leq \frac{1}{2})$:\[\int_{0}^{1/2} \frac{3}{2} (1 - y_2^2) \, dy_2\]Integrate:\[\left.\frac{3}{2}(y_2 - \frac{y_2^3}{3})\right|_{0}^{1/2} = \left.\frac{3}{2}(y_2 - \frac{y_2^3}{3})\right|_{0}^{1/2} = \frac{3}{2}(\frac{1}{2} - \frac{1}{24}) = \frac{3}{2} \times \frac{11}{24} = \frac{11}{16}\]Thus, the probability is: \[\frac{25/64}{11/16} = \frac{25}{44}\].

Conditional density function of Y1 given Y2=y2

To find the conditional density function $f_{Y_1 | Y_2}(y_1 | y_2)$, use:\[f_{Y_1 | Y_2}(y_1 | y_2) = \frac{f(y_1, y_2)}{f_{Y_2}(y_2)}\]Using the previously found marginal densities, determine:\[f_{Y_1 | Y_2}(y_1 | y_2) = \frac{3y_1}{\frac{3}{2}(1 - y_2^2)} = \frac{2y_1}{1 - y_2^2}\]This is valid for $y_2 \leq y_1 \leq 1$.

Find probability of Y1 given specific Y2

To find $P(Y_1 \leq \frac{3}{4} | Y_2 = \frac{1}{2})$, integrate the conditional density function over $[\frac{1}{2}, \frac{3}{4}]$:\[\int_{1/2}^{3/4} \frac{2y_1}{1 - (1/2)^2} \, dy_1\]Since $(1 - \frac{1}{4}) = \frac{3}{4}$, it simplifies to:\[\int_{1/2}^{3/4} \frac{8}{3}y_1 \, dy_1\]Calculate the integral:\[\left.\frac{8}{3} \frac{y_1^2}{2}\right|_{1/2}^{3/4} = \frac{4}{3}(\frac{9}{16} - \frac{1}{4}) = \frac{4}{3}(\frac{5}{16}) = \frac{5}{12}\]So, $P(Y_1 \leq \frac{3}{4} | Y_2 = \frac{1}{2}) = \frac{5}{12}$.

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Key Concepts

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

Joint Density Function

When dealing with probability distributions, we often encounter situations involving two or more random variables. The **joint density function** describes the probability of two continuous random variables, say $ Y_1 $ and $ Y_2 $, occurring together. Essentially, it tells us how the probabilities are distributed over a two-dimensional domain. For example, in the exercise, we have a joint density function $ f(y_1, y_2) = 3y_1 $ valid for parameters $ 0 \leq y_2 \leq y_1 \leq 1 $. This function conveys the likelihood of $ Y_1 $ and $ Y_2 $ falling within a particular range simultaneously. Beyond the given range, the function is zero, meaning the probability is nonexistent elsewhere.
Understanding joint density functions is essential because it lays the foundation for finding other related functions like marginal densities and can illustrate how random variables are dependent or independent of each other.

Conditional Probability

**Conditional probability** refers to the likelihood of an event occurring, given that another event has already taken place. More formally, for random variables such as $ Y_1 $ and $ Y_2 $, the conditional probability $ P(Y_1 \leq a | Y_2 \leq b) $ calculates the probability of $ Y_1 $ being less than or equal to $ a $ given that $ Y_2 $ is less than or equal to $ b $.
In the solution, we computed $ P(Y_1 \leq 3/4 | Y_2 \leq 1/2) $ by dividing the joint probability of both conditions鈥擻( Y_1 \leq 3/4 \) and $ Y_2 \leq 1/2 $鈥攂y the probability of the conditioning event $ Y_2 \leq 1/2 $. Breaking down these probabilities through integration helps us gain insights into the dependency relationships between events.
Mastering conditional probability is crucial for comprehending more complex concepts in probability and statistics, such as Bayesian inference.

Conditional Density Function

The **conditional density function** describes the probability distribution of a random variable conditional on another variable's fixed value. For instance, to determine how $ Y_1 $ behaves when $ Y_2 $ has a specific value $ y_2 $, we use the conditional density formula: \[ f_{Y_1 | Y_2}(y_1 | y_2) = \frac{f(y_1, y_2)}{f_{Y_2}(y_2)} \] where $ f(y_1, y_2) $ is the joint density and $ f_{Y_2}(y_2) $ is the marginal density of $ Y_2 $.
For the exercise, the calculation results in $ f_{Y_1 | Y_2}(y_1 | y_2) = \frac{2y_1}{1 - y_2^2} $, illustrating how $ Y_1 $'s probabilities are adjusted given a specific $ Y_2 $ value. Understanding this function is vital in applications like regression analysis, where we explore how one variable influences another.

Integration in Probability

In probability, **integration** is a fundamental technique used to compute probabilities, especially for continuous random variables. Through integration, we can determine areas under density curves, essentially giving us the probabilities of a variable falling within certain ranges.
Marginal densities are found by integrating the joint density with respect to the unwanted variable. For example, to derive the marginal density of $ Y_1 $, we integrated $ 3y_1 $ over $ y_2 $. Similarly, to find the conditional probability $ P(Y_1 \leq 3/4 | Y_2 \leq 1/2) $, we performed a series of integrations over specified limits.
Incorporating integration in probability allows us to handle complex distributions and solve nuanced problems involving multiple and conditional probabilities.

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Short Answer

Step by step solution

Find marginal density of Y1

Find marginal density of Y2

Calculate conditional probability

Conditional density function of Y1 given Y2=y2

Find probability of Y1 given specific Y2

Key Concepts

Joint Density Function

Conditional Probability

Conditional Density Function

Integration in Probability

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