/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 95 A continuous random variable \(X... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A continuous random variable \(X\) has the following probability distribution: $$ f(x)=4 x e^{-2 x}, \quad x>0 $$ (a) Find the moment-generating function for \(X\). (b) Find the mean and variance of \(X\).

Short Answer

Expert verified
(a) MGF is \( \frac{4}{(2-t)^2} \) for \( t < 2 \); (b) Mean is 1, variance is 0.5.

Step by step solution

01

Identify the Moment-Generating Function (MGF) Formula

The moment-generating function (MGF) for a continuous random variable is defined as \( M_X(t) = E[e^{tX}] = \int_{-fty}^{fty} e^{tx}f(x)dx \). For our given density function \( f(x) = 4x e^{-2x} \), we integrate over the range \( x>0 \). Thus, the expression becomes \( M_X(t) = \int_0^{fty} e^{tx} \cdot 4x e^{-2x} dx \).
02

Simplify and Solve the Integral for MGF

The integral becomes \( \int_0^{fty} 4x e^{(t-2)x} dx \). This is a standard form \( \int x e^{ax} dx \), which evaluates to: \( \left[ \frac{x e^{ax}}{a} - \frac{e^{ax}}{a^2} \right]_0^{fty} \). Calculating with limits, we finally get the simplified form of the MGF, which evaluates to \( M_X(t) = \frac{4}{(2-t)^2} \) for \( t < 2 \).
03

Find the Mean Using the MGF

The mean \( \mu \) of the distribution can be found using the MGF by taking the first derivative at \( t = 0 \). This is given by \( \mu = M_X'(0) \). Therefore, we have \( M_X'(t) = \frac{8}{(2-t)^3} \). Evaluating this at \( t = 0 \), \( \mu = \frac{8}{8} = 1 \).
04

Find the Variance Using the MGF

To find the variance \( \sigma^2 \), we first take the second derivative of the MGF: \( M_X''(t) = \frac{24}{(2-t)^4} \). Then, evaluate \( M_X''(0) \), which gives \( \frac{24}{16} = 1.5 \). We've \( \text{Var}(X) = M_X''(0) - [M_X'(0)]^2 = 1.5 - 1^2 = 0.5 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Moment-Generating Function
The Moment-Generating Function (MGF) is a powerful tool in probability and statistics used to derive moments of a probability distribution. For a continuous random variable like our example, the MGF is defined as \( M_X(t) = E[e^{tX}] \). This translates to evaluating the integral of \( e^{tx} \) multiplied by the probability density function \( f(x) \).
In our case, we start with the given density function \( f(x) = 4x e^{-2x} \) for \( x > 0 \). The MGF is computed by solving the integral
  • \( M_X(t) = \int_0^{\infty} e^{tx} \, 4x e^{-2x} \, dx \)
  • This integral, after applying integration techniques like substitution or by recognizing the pattern, simplifies to \( M_X(t) = \frac{4}{(2-t)^2} \)
It's crucial to note that this result holds only for \( t < 2 \), ensuring the convergence of the integral.
Thus, MGFs are extremely useful in finding other properties like the mean and variance efficiently.
Mean and Variance
Once we have the MGF, determining the mean and variance becomes straightforward. These are fundamental characteristics of the distribution of a random variable, highlighting its central tendency and spread, respectively.
To find the mean \( \mu \), we take the first derivative of the MGF with respect to \( t \) and evaluate it at \( t = 0 \). For the given MGF \( M_X(t) = \frac{4}{(2-t)^2} \), the first derivative is
  • \( M_X'(t) = \frac{8}{(2-t)^3} \)
  • When \( t = 0 \), \( M_X'(0) = \frac{8}{8} = 1 \), indicating a mean of 1
Variance \( \sigma^2 \) is derived from the second derivative. It indicates how much values vary from the mean.
We first calculate
  • \( M_X''(t) = \frac{24}{(2-t)^4} \)
  • Evaluate it at zero, \( M_X''(0) = \frac{24}{16} = 1.5 \)
Variance is then given by the formula: \( \text{Var}(X) = M_X''(0) - [M_X'(0)^2] = 1.5 - 1^2 = 0.5 \)
Mean and variance, obtained from derivatives of the MGF, provide a clear, concise measure of the distribution characteristics.
Continuous Random Variables
Continuous random variables are variables that can take an infinite number of values within a given range. These are contrasted with discrete random variables, which can only take particular values. Understanding continuous random variables is essential in probability theory, as they describe phenomena that are not restricted to distinct outcomes.
For continuous random variables, a probability density function (PDF), as opposed to a probability mass function in discrete cases, describes the likelihood of different outcomes. Such a function must satisfy
  • It integrates to 1 over all possible values
  • Function values are always non-negative\( f(x) \geq 0 \)
The given function \( f(x) = 4x e^{-2x} \) outlines a specific example, where the range is \( x > 0 \). You integrate this function across its range to calculate probabilities, such as in deriving the MGF.
Continuous random variables play a vital role across numerous applications, from natural sciences to engineering, giving a more precise model for non-discrete phenomena.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Based on the number of voids, a ferrite slab is classified as either high, medium, or low. Historically, \(5 \%\) of the slabs are classified as high, \(85 \%\) as medium, and \(10 \%\) as low. A sample of 20 slabs is selected for testing. Let \(X, Y,\) and \(Z\) denote the number of slabs that are independently classified as high, medium, and low, respectively. (a) What are the name and the values of the parameters of the joint probability distribution of \(X, Y,\) and \(Z ?\) (b) What is the range of the joint probability distribution of \(X\), \(Y,\) and \(Z ?\) (c) What are the name and the values of the parameters of the marginal probability distribution of \(X ?\) (d) Determine \(E(X)\) and \(V(X)\). Determine the following: (e) \(P(X=1, Y=17, Z=3)\) (f) \(P(X \leq 1, Y=17, Z=3)\) (g) \(P(X \leq 1)\) (h) \(E(Y)\) (i) \(P(X=2, Z=3 \mid Y=17)\) (j) \(P(X=2 \mid Y=17)\) (k) \(E(X \mid Y=17)\)

Determine the covariance and correlation for the joint probability density function \(f_{X Y}(x, y)=e^{-x-y}\) over the range \(0

The blade and the bearings are important parts of a lathe. The lathe can operate only when both of them work properly. The lifetime of the blade is exponentially distributed with the mean three years; the lifetime of the bearings is also exponentially distributed with the mean four years. Assume that each lifetime is independent. (a) What is the probability that the lathe will operate for at least five years? (b) The lifetime of the lathe exceeds what time with \(95 \%\) probability?

Suppose that the random variables \(X, Y,\) and \(Z\) have the joint probability density function \(f_{X Y Z}(x, y, z)=c\) over the cylinder \(x^{2}+y^{2}<4\) and \(0

An aircraft is flying at a constant altitude with velocity magnitude \(r_{1}\) (relative to the air) and angle \(\theta_{1}\) (in a twodimensional coordinate system). The magnitude and direction of the wind are \(r_{2}\) and \(\theta_{2},\) respectively. Suppose that the wind angle is uniformly distributed between 10 and 20 degrees and all other parameters are constant. Determine the probability density function of the magnitude of the resultant vector \(r=\left[r_{1}^{2}+r_{2}^{2}+r_{1} r_{2}\left(\cos \theta_{1}-\cos \theta_{2}\right)\right]^{0.5}\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.