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A continuous random variable \(X\) has the density function $$f(x)=\left\\{\begin{array}{l}e^{-x} \\\0\end{array}\right.,$$ $$x>0$$ Find the expected value of \(g(X)=e^{2 X / 3}\).

Short Answer

Expert verified
The expected value of the function \( g(X) = e^{2X/3} \) is \( +\infty \).

Step by step solution

01

Identify the Integral to Solve

We are required to find the expected value of a function \( g(X) = e^{2X/3} \) where \( X \) is a random variable with a given density function \( f(x) = e^{-x} \). The calculation of an expected value involves solving an integral of the form \( E[g(X)] = \int g(x)f(x) dx \). Therefore, we substitute \( g(x) \) and \( f(x) \) into this formula to find the integral that we need to solve: \( E[g(X)] = \int e^{2x/3} \cdot e^{-x} dx \) over \( x > 0 \).
02

Simplify the Integral to Solve

The product \( e^{2x/3} \cdot e^{-x} \) simplifies to \( e^{x/3} \). Therefore, the integral becomes \( E[g(X)] = \int_{0}^{+\infty} e^{x/3} dx \).
03

Perform the Integration using Substitution

To solve the integral, we perform a substitution. Let \( u = x/3 \), and hence \( du = dx/3 \). So the integral is converted to \( 3 \int_{0}^{+\infty} e^udu \), and can be further solved. The integral of \( e^u \) is \( e^u + C \), where \( C \) is the constant of integration. Therefore, the integral is \( 3 [e^u]_{0}^{+\infty} = 3 (e^{+\infty} - e^0) \).
04

Simplify the Expression for the Answer

The expression \( 3 (e^{+\infty} - e^0) \) simplifies to \( 3 (+\infty - 1) \). As we cannot subtract a finite number from infinity, the result is \( +\infty \).
05

Interpret the Result

The expected value of \( g(X) = e^{2X/3} \) is \( +\infty \). This indicates that the average or expected value computed from the function \( g(X) \) applied to the random variable \( X \), with its given density function, is infinitely large.

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Key Concepts

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

Continuous Random Variable
A continuous random variable is a variable that can take any value within a given range. Unlike discrete random variables, which have distinct values, continuous variables are like variables that can move smoothly across values. Imagine a thermometer measuring temperature. It doesn’t just jump from one degree to another but smoothly transitions, capturing every possible temperature. This continuity makes continuous random variables vital for modern statistics and probability. In our exercise, the variable \(X\) is continuous since it covers all positive values \(>0\). This characteristic influences how we compute expectations because we work with integrations instead of summations.
Density Function
A density function is a mathematical description of the probability distribution of a continuous random variable. This function shows us how dense or compressed the data can be at different points. In simpler terms, it tells us how likely it is for a variable to be at a particular value.
  • The function must be non-negative.
  • The total area under the curve of the density function must be equal to 1, which symbolizes total probability.
In our exercise, the density function \(f(x)=e^{-x}\) is provided for the random variable \(X\). Here, \(e^{-x}\) is the density function for an exponential distribution, emphasizing that lower values of \(X\) are more probable than higher values. As \(x\) increases, \(e^{-x}\) decreases, indicating a rapid drop-off in probability.
Integration
Integration is a mathematical tool used to sum up infinitesimally small data slices over a continuous range. When dealing with continuous random variables, integration becomes essential because it allows us to consider all potential values the variable might assume. It is analogous to finding the area under a curve on a graph. In our problem, determining the expected value of \(g(X)\) requires integrating the function \(e^{x/3}\) over the range \(0\) to \(+\infty\). The integration process involves simplifying the function, often using methods like substitution. This is similar to how you might change variables in an algebraic equation to make it easier to solve. Mastery of integration techniques is crucial for students to handle expected value calculations for continuous random variables effectively.
Exponential Function
An exponential function is a mathematical expression where a constant base is raised to a variable exponent. It grows rapidly and is vital in many scientific fields, including probability. The function \(f(x)=e^{-x}\) is a typical exponential function; its rapid decrease makes it apt for modeling decay processes, like radioactive decay or discharging capacitors.
  • Exponential functions have unique properties: they never go negative, and their derivatives and integrals are proportionate to the maximum value of the function.
  • In probability, exponential functions often express waiting times or life spans of systems.
In this exercise, the exponential function \(g(X)=e^{2X/3}\) modifies the behavior of the density function \(e^{-x}\), creating a new composite exponential that we ultimately integrate. The interplay between \(e^{2X/3}\) and \(e^{-x}\) showcases how different exponential functions can come together to model real-world probability scenarios.

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Most popular questions from this chapter

As we shall illustrate in Chapter \(12,\) statistical methods associated with linear and nonlinear models are very important. In fact, exponential functions are often used in a wide variety of scientific and engineering problems. Consider a model that is fit to a set of data involving measured values \(k \mid\) and \(k_{2}\) and a certain response \(Y\) to the measurements. The model postulated is $$\mathbf{Y}=e^{b_{0}}+b_{1} k_{1}+b_{2} k_{2}$$ where \(Y\) denotes estimated value of \(Y . k_{1}\) and \(k_{2}\) are fixed values and \(b_{0}, b_{1},\) and \(b_{2}\) are estimates of constants and hence are random variables. Assume that these random variables are independent and use the approximate formula. For the variance of a non-linear function of more than one variable. Give an expression for \(\operatorname{Var}(Y)\). Assume that the means of \(b o, b_{1},\) and \(b_{2}\) are known and are \(B_{0}, \beta_{1},\) and \(\beta_{2}\) and assume that the variances of \(b o, b_{1}\) and \(b_{2}\) are known and are \(\sigma_{0}^{2}, \sigma_{1}^{2}\), and \(\sigma_{2}^{2}\).

The hospital period, in days, for patients following treatment for a certain typo of kidney disorder is a random variable \(Y=X+4,\) where \(X\) has the density function $$f(x)=\left\\{\begin{array}{l}\frac{32}{(x-4)^{3}} \\\0\end{array}\right.,$$ $$x>0$$ Find the average number of days that a person is hospitalized following treatment for this disorder.

In a gambling game a woman is paid \(\$ 3\) if she draws a jack or a queen and \(\$ 5\) if she draws a king or an ace from an ordinary deck of 52 playing cards. If she draws any other card, she loses. How much should she pay to play if the game is fair?

The probability distribution of \(X,\) the number of imperfections per 10 meters of a synthetic fabric in continuous rolls of uniform width, is given in Exercise 3.13 on page 89 as $$\begin{array}{cccccc}x & \mathrm{I} & 0 & 1 & 2 & 3 & 4 \\\\\hline f(x) & 0.41 & 0.37 & 0.16 & 0.05 & 0.01\end{array}$$ Find the average number of imperfections per 10 meters of this fabric.

In Exercise 3.13 on page \(89,\) the distribution of the number of imperfections per 10 meters of synthetic fabric is given by $$\begin{array}{l|ccccc}x & 0 & 1 & 2 & 3 & 4 \\\\\hline f(x) & 0.41 & 0.37 & 0.16 & 0.05 & 0.01\end{array}$$ (a) Plot the probability function. (b) Find the expected number of imperfections. \(E(X)=\mu\) (c) Find \(E\left(X^{2}\right)\).

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