Problem 187
Question
Let \(X\) and \(Y\) be two independent random variables, whose marginal pdfs are given below. Find the pdf of \(X+Y\). (Hint: Consider two cases, \(0 \leq w<1\) and \(1 \leq w \leq 2 .)\) $$ f_{X}(x)=1,0 \leq x \leq 1, \text { and } f_{Y}(y)=1,0 \leq y \leq 1 $$
Step-by-Step Solution
Verified Answer
The pdf of \(W=X+Y\) is \(f_{W}(w) = w\) for \(0 \leq w<1\), \(f_{W}(w) = 2-w\) for \(1 \leq w \leq 2\), and \(f_{W}(w) = 0\) otherwise.
1Step 1: Set up the equations
First, suppose \(W=X+Y\). The probability density function (pdf) of \(W\) can be found by integrating the joint pdf of \(X\) and \(Y\) over the region for which the sum of \(X\) and \(Y\) equals \(w\). Since \(X\) and \(Y\) are independent, the joint pdf of \(X\) and \(Y\), \(f_{X,Y}(x,y)\), is simply the product of the marginal pdfs, \(f_{X}(x) * f_{Y}(y)\). Thus, for each case of \(w\), the pdf of \(W\) can be calculated as follows: For the case \(0 \leq w<1\), the boundary values of \(x\) that make \(x + y = w\) are 0 and \(w\). So the integral to compute is \( \int_{0}^{w} f_{X,Y}(x,w-x) dx = \int_{0}^{w} dx \). For the case \(1 \leq w \leq 2\), the boundary values of \(x\) that make \(x + y = w\) are \(w-1\) and 1. So the integral to compute is \( \int_{w-1}^{1} f_{X,Y}(x,w-x) dx = \int_{w-1}^{1} dx \).
2Step 2: Compute the integrals
Now, it's time to compute these integrals: For the case \(0 \leq w<1\), \( \int_{0}^{w} dx = [x]_{0}^{w} = w - 0 = w \). For the case \(1 \leq w \leq 2\), \( \int_{w-1}^{1} dx = [x]_{w-1}^{1} = 1 - (w-1) = 2 - w \).
3Step 3: Formulate the pdf of \(W\)
Now that the integrals have been calculated, we can formulate the pdf of \(W\), \(f_{W}(w)\): It is \(w\) if \(0 \leq w<1\), and \(2-w\) if \(1 \leq w \leq 2\). The pdf is 0 otherwise. This completes the derivation.
Key Concepts
Independent Random VariablesJoint PDF CalculationIntegration in Probability
Independent Random Variables
Understanding independent random variables is crucial when dealing with probabilities. Independent random variables are those whose outcomes do not affect each other. For example, flipping a coin and rolling a dice are independent events because the result of one does not influence the result of the other. In probability theory, this concept is vital as it simplifies many calculations.
In the given exercise, the random variables 'X' and 'Y' are independent, and their marginal probability density functions (pdfs) are given. Since they do not influence each other, we can express the joint probability density function simply as the product of the two individual pdfs. When considering the sum of these random variables, this property of independence is particularly handy. It allows us to compute the probability density function of their sum, designated as 'W=X+Y', using certain techniques which include integration.
In the given exercise, the random variables 'X' and 'Y' are independent, and their marginal probability density functions (pdfs) are given. Since they do not influence each other, we can express the joint probability density function simply as the product of the two individual pdfs. When considering the sum of these random variables, this property of independence is particularly handy. It allows us to compute the probability density function of their sum, designated as 'W=X+Y', using certain techniques which include integration.
Joint PDF Calculation
To find the probability density function (pdf) of the sum of two independent random variables, one must understand the concept of a joint pdf. A joint pdf quantifies the likelihood that two (or more) random variables fall within specific ranges of values. When the variables are independent, as in our current example, the joint pdf is the multiplicative result of their individual pdfs.
This is the foundation for calculating the pdf of 'X+Y'. The exercise outlines a step-by-step approach to determine the pdf for the sum 'W' by considering two separate cases for 'w'. By integrating the joint pdf over specific boundaries for each case, we find the values that ultimately define 'W's' pdf. This process hinges on recognizing the bounds of integration, which relate directly to the value ranges for which 'X+Y' equates to a particular 'w'. These calculations highlight the significance of correctly setting up and interpreting the joint pdf in probability problems.
This is the foundation for calculating the pdf of 'X+Y'. The exercise outlines a step-by-step approach to determine the pdf for the sum 'W' by considering two separate cases for 'w'. By integrating the joint pdf over specific boundaries for each case, we find the values that ultimately define 'W's' pdf. This process hinges on recognizing the bounds of integration, which relate directly to the value ranges for which 'X+Y' equates to a particular 'w'. These calculations highlight the significance of correctly setting up and interpreting the joint pdf in probability problems.
Integration in Probability
Integration is a fundamental tool in probability theory, particularly when we want to combine or manipulate the pdf of continuous random variables. In our exercise, integration allows us to sum the probabilities across a range of values to find the pdf of 'W', the sum of 'X' and 'Y'.
The method involves setting the appropriate limits of integration according to the value of 'w' we are investigating. For the scenario where '0 \( \leq \) w<1', we integrated from 0 to 'w', and for '1 \( \leq \) w \leq 2', we integrated from 'w-1' to 1. These steps are essentially adding up all the infinitely small probabilities over the range of 'X' values that satisfy 'X+Y=w' for each piece of the defined interval.
By successfully evaluating these integrals, we arrived at the desired pdf piecewise function for 'W'. This showcases how integration is used in a probability context to aggregate and manipulate distributions, thus forming a key component in the analysis of random variables and in solving many types of probabilistic problems.
The method involves setting the appropriate limits of integration according to the value of 'w' we are investigating. For the scenario where '0 \( \leq \) w<1', we integrated from 0 to 'w', and for '1 \( \leq \) w \leq 2', we integrated from 'w-1' to 1. These steps are essentially adding up all the infinitely small probabilities over the range of 'X' values that satisfy 'X+Y=w' for each piece of the defined interval.
By successfully evaluating these integrals, we arrived at the desired pdf piecewise function for 'W'. This showcases how integration is used in a probability context to aggregate and manipulate distributions, thus forming a key component in the analysis of random variables and in solving many types of probabilistic problems.
Other exercises in this chapter
Problem 185
Let \(X\) and \(Y\) be two independent random variables. Given the marginal pdfs shown below, find the pdf of \(X+Y\). In each case, check to see if \(X+Y\) bel
View solution Problem 186
Suppose \(f_{X}(x)=x e^{-x}, x \geq 0\), and \(f_{Y}(y)=e^{-y}\), \(y \geq 0\), where \(X\) and \(Y\) are independent. Find the pdf of \(X+Y\).
View solution Problem 188
If a random variable \(V\) is independent of two independent random variables \(X\) and \(Y\), prove that \(V\) is independent of \(X+Y\).
View solution Problem 189
Let \(Y\) be a continuous nonnegative random variable. Show that \(W=Y^{2}\) has pdf \(f_{W}(w)=\frac{1}{2 \sqrt{w}} f_{Y}(\sqrt{w})\). (Hint: First find \(F_{W
View solution