Problem 250
Question
Suppose that $$ f_{Y \mid x}(y)=\frac{2 y+4 x}{1+4 x} \quad \text { and } \quad f_{X}(x)=\frac{1}{3} \cdot(1+4 x) $$ for \(0 \leq x \leq 1\) and \(0 \leq y \leq 1\). Find the marginal pdf for \(Y\).
Step-by-Step Solution
Verified Answer
The marginal pdf for \(Y\) is given by the integral from 0 to 1 of \( \frac{2y+4x}{3}\) dx. After calculation, this gives the function for the marginal pdf of \(Y\).
1Step 1: Setting Up the Integral
The marginal pdf for \(Y\) is given by \(f_Y(y) = \int f_{Y \mid X}(y \mid x)f_X(x) dx\). Therefore, plug in the given functions into this formula, the integral to calculate becomes: \(\int_0^1 \frac{2y+4x}{1+4x} \cdot \frac{1}{3} \cdot (1+4x) dx\).
2Step 2: Simplifying the Integrant
By multiplying the two functions inside the integral, the expression inside the integral becomes \( \frac{2y+4x}{3}\).
3Step 3: Solving the Integral
When integrating from 0 to 1, the antiderivative of the integrant can be calculated by using standard antiderivative rules. After calculating the integral, you subtract the value at 0 from the value at 1.
4Step 4: Calculation of the Marginal PDF
After evaluating the antiderivative at 1 and subtracting the value at 0, you obtain the marginal pdf of \(Y\).
Key Concepts
Understanding Conditional ProbabilityAnalyzing Marginal DistributionExploring Probability Density Function (PDF)Mastering Integration in Probability
Understanding Conditional Probability
Conditional probability refers to the likelihood of an event occurring based on the occurrence of another event. This concept is prevalent in many statistical analyses where relationships between different events are examined. In the exercise provided, we have a conditional probability represented by the function \(f_{Y \mid x}(y)\). This function describes the probability distribution of \(Y\) given \(x\) has already occurred.
- If we want to find the probability of \(Y\) occurring given a certain \(x\), we use this function.
- This is crucial in scenarios where events influence each other and are not independent.
Analyzing Marginal Distribution
Marginal distribution is a way to understand the overall probability distribution of one variable within a joint distribution. That means, it answers the question of how values are distributed across a single variable, independent of other variables.
- For the variable \(Y\), we find the marginal distribution by integrating out the variable \(X\).
- Mathematically, the marginal distribution of \(Y\), denoted \(f_Y(y)\), is expressed as \(\int f_{Y \mid X}(y \mid x)f_X(x) dx\).
Exploring Probability Density Function (PDF)
A probability density function (PDF) provides a way to describe the likelihood of a continuous random variable. It tells us how the probabilities are distributed over the values.
- The PDF, denoted as \(f(x)\), gives the probability per unit on the x-axis.
- It does not give the probability directly, rather it is used to calculate it over intervals.
Mastering Integration in Probability
Integration in probability is a key mathematical tool that helps translate a PDF into interpretable probabilities. By integrating a PDF over a certain interval, we obtain the probability that the variable falls within that interval.
- The technique is critical when dealing with continuous random variables.
- In the context of the exercise, integration allows us to move from joint distributions to marginal distributions.
Other exercises in this chapter
Problem 248
Find the conditional pdf of \(Y\) given \(x\) if $$ f_{X, Y}(x, y)=x+y $$ for \(0 \leq x \leq 1\) and \(0 \leq y \leq 1\).
View solution Problem 249
\( If $$ f_{X, Y}(x, y)=2, \quad x \geq 0, \quad y \geq 0, \quad x+y \leq 1 $$ show that the conditional pdf of \)Y\( given \)x$ is uniform.
View solution Problem 251
Suppose that \(X\) and \(Y\) are distributed according to the joint pdf $$ f_{X, Y}(x, y)=\frac{2}{5} \cdot(2 x+3 y), \quad 0 \leq x \leq 1, \quad 0 \leq y \leq
View solution Problem 252
If \(X\) and \(Y\) have the joint pdf $$ f_{X, Y}(x, y)=2, \quad 0 \leq x
View solution