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Repeat Problem 6.56 when \(X\) and \(Y\) are independent exponential random variables, each with parameter \(\lambda=1.\)

Short Answer

Expert verified
The probability density function for the sum of two independent exponential random variables X and Y, both with parameter λ=1, is: \[f_Z(z) = \left\{ \begin{array}{ll} ze^{-z} & \text{if } z \geq 0\\ 0 & \text{otherwise} \end{array}\right.\]

Step by step solution

01

Recall the properties of exponential random variables

Recall that the probability density function (pdf) of an exponential random variable X with parameter λ is given by: \[f_X(x) = \lambda e^{-\lambda x}\] for \(x \geq 0\), and \(f_X(x) = 0\) otherwise. In our case, since both X and Y have the same parameter λ=1, their pdfs are: \[f_X(x) = e^{-x}\] for \(x \geq 0\) \[f_Y(y) = e^{-y}\] for \(y \geq 0\)
02

Apply the convolution formula

To find the probability density function (pdf) of the sum Z = X + Y, we use the convolution formula: \[f_Z(z) = \int_{-\infty}^{\infty} f_X(x) f_Y(z - x) dx\] Now plug in the pdfs for X and Y: \[f_Z(z) = \int_{-\infty}^{\infty} e^{-x} e^{-(z - x)} dx\]
03

Calculate the integral

Now, calculate the integral by considering the definition domain of X and Y: \[f_Z(z) = \int_{0}^{z} e^{-x} e^{-(z - x)} dx\] To simplify the integrand, notice that: \[e^{-x} e^{-(z - x)} = e^{-x} e^{-z} e^{x} = e^{-z}\] Now, f_Z(z) can be rewritten as: \[f_Z(z) = e^{-z} \int_{0}^{z} dx\] Then, integrate with respect to x: \[f_Z(z) = e^{-z} [x]_0^z = ze^{-z}\] for \(z \geq 0\), and \(f_Z(z) = 0\) otherwise.
04

Present the final result

The probability density function for the sum of two independent exponential random variables X and Y, both with parameter λ=1, is: \[f_Z(z) = \left\{ \begin{array}{ll} ze^{-z} & \text{if } z \geq 0\\ 0 & \text{otherwise} \end{array}\right.\]

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Key Concepts

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

Probability Density Function
Understanding the Probability Density Function (PDF) is essential in the study of continuous random variables. The PDF, represented as \( f_X(x) \), tells us the likelihood of a random variable X taking on a value at a particular point. Think of it like a snapshot of the distribution of probabilities across a range of possible outcomes.

For an exponential random variable with parameter \( \lambda \), the PDF is defined as \( f_X(x) = \lambda e^{-\lambda x} \) for \( x \geq 0 \), and zero otherwise. This specific type of PDF is characteristic for exponential distributions and indicates that the variable has a constant rate of decay, \( \lambda \). In practical terms, this rate can represent the time between occurrences of an event in a Poisson process.
Convolution Formula
The Convolution Formula is a powerful mathematical tool used to find the probability density function of the sum of two independent random variables. Imagine you have two separate systems or processes, each with their own distributions of probability. Now, if you want to know how these two independent variables interact additively, you would use the convolution formula.

It's defined as \( f_{Z}(z) = \int_{-\infty}^{\infty} f_{X}(x) f_{Y}(z - x) dx \) for continuous variables X and Y. The beauty of this formula is that it allows us to combine two independent processes into a new one, represented by the random variable \( Z = X + Y \). It's like blending two ingredients together to get a new flavor; only, in this case, we're blending probabilities.
Integral Calculus
The area of mathematics known as Integral Calculus is essential when dealing with continuous random variables and their associated probabilities. In the context of probability and statistics, it is used to calculate areas under the PDF curves, which translate to probabilities.

To draw a parallel, if PDF is the recipe, integral calculus is the method by which you measure the ingredients. It's what we use to sum up all the infinitesimally small probabilities across a continuous interval. For example, the integral calculation \( \int_0^{z} dx \) represents the summing process over a range from 0 to z. In probabilistic terms, this gives us the accumulated likelihood of a random variable falling within that range.
Independent Random Variables
The concept of Independent Random Variables is crucial in probability theory when analyzing the joint behavior of multiple random elements. Two random variables X and Y are considered independent if the occurrence of one does not influence the likelihood of occurrence of the other.

This independence is what makes tools like the convolution formula applicable. When dealing with independent exponential random variables, for instance, it means that the interval of one event does not affect the interval of the other, making calculations of their combined behavior, such as the sum, predictable and based solely on their individual PDFs. Independence is essentially the assumption that the random variables play fair and don't interfere with each other's outcomes.

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Most popular questions from this chapter

The random vector \((X, Y)\) is said to be uniformly distributed over a region \(R\) in the plane if, for some constant \(c,\) its joint density is $$f(x, y)=\left\\{\begin{array}{ll}c & \text { if }(x, y) \in R \\\0 & \text { otherwise }\end{array}\right.$$ (a) Show that \(1 / c=\) area of region \(R\) Suppose that \((X, Y)\) is uniformly distributed over the square centered at (0,0) and with sides of length 2. (b) Show that \(X\) and \(Y\) are independent, with each being distributed uniformly over (-1,1). (c) What is the probability that \((X, Y)\) lies in the circle of radius 1 centered at the origin? That is, find \(P\left\\{X^{2}+Y^{2} \leq 1\right\\}.\)

The joint density function of \(X\) and \(Y\) is given by $$f(x, y)=x e^{-x(y+1)} \quad x>0, y>0$$ (a) Find the conditional density of \(X,\) given \(Y=y,\) and that of \(Y,\) given \(X=x\). (b) Find the density function of \(Z=X Y\).

A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a time, until the defective ones are identified. Denote by \(N_{1}\) the number of tests made until the first defective is identified and by \(N_{2}\) the number of additional tests until the second defective is identified. Find the joint probability mass function of \(N_{1}\) and \(N_{2}.\)

The number of people who enter a drugstore in a given hour is a Poisson random variable with parameter \(\lambda=10 .\) Compute the conditional probability that at most 3 men entered the drugstore, given that 10 women entered in that hour. What assumptions have you made?

Jill's bowling scores are approximately normally distributed with mean 170 and standard deviation \(20,\) while Jack's scores are approximately normally distributed with mean 160 and standard deviation \(15 .\) If Jack and Jill each bowl one game, then assuming that their scores are independent random variables, approximate the probability that (a) Jack's score is higher; (b) the total of their scores is above \(350 .\)

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