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Let \(X\) be a nonnegative random variable. We say that \(X\) is memoryless if $$ P(X>s+t \mid X>t)=P(X>s) \text { for all } s, t \geq 0 $$ Show that a random variable with pdf \(f_{X}(x)=(1 / \lambda) e^{-x / \lambda}\), \(x>0\), is memoryless.

Short Answer

Expert verified
Yes, a random variable with the given probability density function \(f_{X}(x)=\frac{1}{\lambda} e^{-x/\lambda}, x>0\) is memoryless because it satisfies the property \(P(X>s+t | X>t)=P(X>s)\) for all \(s, t \geq 0\).

Step by step solution

01

Understanding the definition of a memoryless random variable

A random variable \(X\) is said to be memoryless if the probability \(P(X>s+t | X>t)=P(X>s)\) holds for all \(s, t \geq 0\). This means, the probability that \(X\) exceeds \(s+t\), given that it already exceeded \(t\), is equal to the probability that it exceeds \(s\).
02

Enumerate the given pdf for the random variable \(X\)

Here, the given pdf of \(X\) is \(f_{X}(x)=\frac{1}{\lambda} e^{-x/\lambda}, x>0\) . This is the exponential distribution with parameter \(1/\lambda\).
03

Check the memoryless property

Let's calculate \(P(X>s+t | X>t)\) by definition of conditional probability:\(P(X>s+t | X>t)=\frac{P(X>s+t \cap X>t)}{P(X>t)}=\frac{P(X>s+t)}{P(X>t)}\), because \(X>s+t \Rightarrow X>t\).Now, \(P(X>s+t)=\int_{s+t}^{\infty} f_{X}(x) dx =\int_{s+t}^{\infty} \frac{1}{\lambda} e^{-x/\lambda} dx = e^{-\frac{s+t}{\lambda}}\).And \(P(X>t)=\int_{t}^{\infty} f_{X}(x) dx =\int_{t}^{\infty} \frac{1}{\lambda} e^{-x/\lambda} dx = e^{-\frac{t}{\lambda}}\).Therefore, \(P(X>s+t | X>t)=\frac{e^{-\frac{s+t}{\lambda}}}{e^{-\frac{t}{\lambda}}} = e^{-\frac{s}{\lambda}}\)but \(P(X>s)=\int_{s}^{\infty} f_{X}(x) dx =\int_{s}^{\infty} \frac{1}{\lambda} e^{-x/\lambda} dx = e^{-\frac{s}{\lambda}}\),so \(P(X>s+t | X>t) = P(X>s)\), proving that \(X\) is memoryless.

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Key Concepts

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

Conditional Probability
Understanding conditional probability is essential when delving into topics like the memoryless property of random variables. At its core, conditional probability is the probability of an event occurring given that another event has already taken place. In mathematical terms, the conditional probability of an event A given that event B has occurred is denoted as P(A|B). This can be calculated using the formula:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
provided that P(B) is not equal to zero. The idea is that the occurrence of event B changes the sample space, creating a new 'universe' where our probabilities are recalculated with B assumed.

For example, if we want to find out the probability of rolling a six on a die given that we've rolled an even number, we're only considering the outcomes in the sample space where the result is even (2, 4, or 6), thus altering the probability from \(\frac{1}{6}\) to \(\frac{1}{3}\). This concept is pivotal in understanding the memoryless property of random variables, as it deals with the probability of an event relative to the occurrence of another.
Probability Density Function (pdf)
A probability density function (pdf) is a function that describes the likelihood of a continuous random variable assuming a value at any point in the variable's range. If you have a pdf f(x) for a random variable X, you can find the probability that X falls within a particular interval by integrating the pdf over that interval.

For example, to calculate the probability that X is between a and b, the integral would look like \[ P(a \leq X \leq b) = \int_{a}^{b} f(x) dx \.\] The pdf is so named because, unlike discrete random variables which have a probability mass function (pmf), continuous random variables have probabilities spread out over a continuous range, and thus density - rather than mass - is the appropriate concept. The area under the pdf curve over the entire range is always equal to 1, representing the certainty that the random variable will take on a value within its range. When dealing with pdfs, we often handle integrals and operate within the realm of calculus to solve probability problems.
Exponential Distribution
The exponential distribution is a continuous probability distribution that is often used to model the time until an event occurs, such as the time between arrivals in a queue. The key feature of an exponential distribution is its memoryless property, which states that the probability of the event occurring in the next s units of time does not depend on how much time has already elapsed.

The probability density function of an exponential distribution is defined by: \[ f_{X}(x)= \begin{cases} \lambda e^{-\lambda x}, & x \geq 0,\ 0, & x < 0 \end{cases} \] with \(\backslash lambda\) being the rate parameter. The mean and variance of the exponential distribution are both \(\backslash frac{1}{\backslash lambda}\).

Notably, the memoryless property of the exponential distribution makes it unique among continuous distributions, and thus it has applications in areas such as reliability engineering and queuing theory, where the lack of memory is a desired characteristic of the model. It is this property that our above exercise explores when attempting to prove that a random variable with a given pdf is, indeed, memoryless.

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Most popular questions from this chapter

A typical day's production of a certain electronic component is twelve. The probability that one of these components needs rework is \(0.11\). Each component needing rework costs \(\$ 100\). What is the average daily cost for defective components?

Suppose that random variables \(X\) and \(Y\) vary in accordance with the joint pdf, \(f_{X, Y}(x, y)=c(x+y), 0

Calculate \(E(Y)\) for the following pdfs: (a) \(f_{Y}(y)=3(1-y)^{2}, 0 \leq y \leq 1\) (b) \(f_{Y}(y)=4 y e^{-2 y}, y \geq 0\) (c) \(f_{Y}(y)= \begin{cases}\frac{3}{4}, & 0 \leq y \leq 1 \\ \frac{1}{4}, & 2 \leq y \leq 3 \\ 0, & \text { elsewhere }\end{cases}\) (d) \(f_{Y}(y)=\sin y, \quad 0 \leq y \leq \frac{\pi}{2}\)

An urn contains one white chip and one black chip. A chip is drawn at random. If it is white, the "game" is over; if it is black, that chip and another black one are put into the urn. Then another chip is drawn at random from the "new" urn and the same rules for ending or continuing the game are followed (i.e., if the chip is white, the game is over; if the chip is black, it is placed back in the urn, together with another chip of the same color). The drawings continue until a white chip is selected. Show that the expected number of drawings necessary to get a white chip is not finite.

Suppose that two fair dice are tossed one time. Let \(X\) denote the number of 2 's that appear, and \(Y\) the number of 3 's. Write the matrix giving the joint probability density function for \(X\) and \(Y\). Suppose a third random variable, \(Z\), is defined, where \(Z=X+Y\). Use \(p_{X, Y}(x, y)\) to find \(p_{Z}(z)\).

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