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Let \(X\) be a random variable that takes on nonnegative values and has distribution function \(F(x) .\) Show that $$E(X)=\int_{0}^{\infty}(1-F(x)) d x$$ Hint: Integrate by parts. Illustrate this result by calculating \(E(X)\) by this method if \(X\) has an exponential distribution \(F(x)=1-e^{-\lambda x}\) for \(x \geq 0,\) and \(F(x)=0\) otherwise.

Short Answer

Expert verified
For an exponential distribution with rate \( \lambda \), the expectation \( E(X) = \frac{1}{\lambda} \).

Step by step solution

01

Understanding Expectation Definition

The expectation of a continuous random variable \( X \) with probability distribution function \( F(x) \) is given by \( E(X) = \int_{0}^{\infty} x \cdot f(x) \, dx \), where \( f(x) \) is the probability density function of \( X \).
02

Express the Expectation Using Integration by Parts

Apply the integration by parts formula to \( E(X) = \int_{0}^{\infty} x f(x) \, dx \). When using integration by parts, let \( u = x \) and \( dv = f(x) \, dx \), hence \( du = dx \) and \( v = F(x) \). The formula gives\[ E(X) = \left. xF(x) \right|_{0}^{\infty} - \int_{0}^{\infty} F(x) \, dx. \]
03

Evaluate Boundary Conditions

As \( x \to \infty \), \( F(x) \to 1 \), so \( xF(x) \to \infty \cdot 1 = \infty\), hence the boundary term is zero because the distribution approaches 1 smoothly. Also, \( xF(x) \bigg|_{x=0} = 0 \cdot F(0) = 0 \).
04

Simplify and Rearrange

Substitute the boundary terms into the integration by parts formula to get:\[ E(X) = 0 - \int_{0}^{\infty} F(x) \, dx = -\int_{0}^{\infty} F(x) \, dx. \]Since \( \int_{0}^{\infty} F(x) \, dx = \int_{0}^{\infty} (1 - F(x)) \, dx \), we obtain \( E(X) = \int_{0}^{\infty} (1 - F(x)) \, dx. \)
05

Apply the Formula for the Exponential Distribution

For the exponential distribution, \( F(x) = 1 - e^{-\lambda x} \). Hence,\[ E(X) = \int_{0}^{\infty} (1 - (1-e^{-\lambda x})) \, dx = \int_{0}^{\infty} e^{-\lambda x} \, dx. \]
06

Compute the Integral for Exponential Distribution

The integral becomes:\[ E(X) = \int_{0}^{\infty} e^{-\lambda x} \, dx = \left[-\frac{1}{\lambda} e^{-\lambda x} \right]_{0}^{\infty} = 0 - \left(-\frac{1}{\lambda}\right) = \frac{1}{\lambda}. \]
07

Conclusion

The expectation for \( X \), if it follows an exponential distribution with parameter \( \lambda \), is \( \frac{1}{\lambda} \). This result aligns with the properties of the exponential distribution.

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Key Concepts

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

Integration by Parts
Integration by parts is a valuable technique used in calculus to simplify integrals. It is based on the product rule for differentiation. The formula is given by \[\int u \, dv = uv - \int v \, du\]. This technique is particularly useful when dealing with products of functions that are not simple to integrate directly.
In the context of expectation of random variables, we use integration by parts to transform the expression into a more manageable form. Here, if we set \(u = x\) and \(dv = f(x) \, dx\), we can find that \(du = dx\) and \(v = F(x)\), where \(F(x)\) is the cumulative distribution function and \(f(x)\) is the probability density function.
  • This method effectively allows us to express the expectation \(E(X)\) in a form involving \(F(x)\), making calculations considerably easier, especially at boundaries where this method often simplifies to zero.
  • Integration by parts is a fundamental tool not only in probability but also in solving various types of integrals encountered in mathematics and physics.
Exponential Distribution
The exponential distribution is a commonly used continuous probability distribution. It is often used to model the time between events in a process where events occur continuously and independently at a constant average rate.
This distribution is defined by its rate parameter \( \lambda \), where the probability density function is given by \[f(x) = \lambda e^{-\lambda x}\] for \(x \geq 0\) and \(f(x) = 0\) otherwise.
The cumulative distribution function (CDF) for the exponential distribution is \[F(x) = 1 - e^{-\lambda x}\]. In other words, it represents the probability that the random variable \(X\) is less than or equal to \(x\).
  • For the exponential distribution, the expected value \(E(X)\) is \(\frac{1}{\lambda}\), which represents the average time or space until a certain event happens.
  • This property makes the exponential distribution extremely useful in fields like reliability analysis, queuing theory, and various other stochastic modeling domains.
Continuous Probability Distributions
Continuous probability distributions are used to model situations where outcomes can take any value within a certain range. Unlike discrete distributions where outcomes are distinct and separate, continuous distributions account for an infinite number of possible outcomes.
Characterized by their probability density functions (PDFs), these distributions are essential for calculating probabilities of random variables falling within particular intervals. The area under the PDF curve within an interval gives the probability of the variable being in that interval.
Some key points:
  • The total area under the entire curve of a PDF equals 1, which corresponds to the certainty that the event of any possible outcome will happen.
  • The cumulative distribution function (CDF) is crucial as it provides the probability that a random variable is less than or equal to a certain value.
  • Continuous distributions include normal, exponential, and uniform distributions, each serving different models and applications.
Continuous distributions are foundational in statistical analysis, supporting theories and applications ranging from quality control to financial modeling.

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Most popular questions from this chapter

A die is rolled twice. Let \(X\) denote the sum of the two numbers that turn up, and \(Y\) the difference of the numbers (specifically, the number on the first roll minus the number on the second). Show that \(E(X Y)=E(X) E(Y) .\) Are \(X\) and \(Y\) independent?

For a sequence of Bernoulli trials, let \(X_{1}\) be the number of trials until the first success. For \(j \geq 2,\) let \(X_{j}\) be the number of trials after the \((j-1)\) st success until the \(j\) th success. It can be shown that \(X_{1}, X_{2}, \ldots\) is an independent trials process. (a) What is the common distribution, expected value, and variance for \(X_{j} ?\) (b) Let \(T_{n}=X_{1}+X_{2}+\cdots+X_{n} .\) Then \(T_{n}\) is the time until the \(n\) th success. Find \(E\left(T_{n}\right)\) and \(V\left(T_{n}\right)\) (c) Use the results of (b) to find the expected value and variance for the number of tosses of a coin until the \(n\) th occurrence of a head.

A deck of ESP cards consists of 20 cards each of two types: say ten stars, ten circles (normally there are five types). The deck is shuffled and the cards turned up one at a time. You, the alleged percipient, are to name the symbol on each card before it is turned up. Suppose that you are really just guessing at the cards. If you do not get to see each card after you have made your guess, then it is easy to calculate the expected number of correct guesses, namely ten. If, on the other hand, you are guessing with information, that is, if you see each card after your guess, then, of course, you might expect to get a higher score. This is indeed the case, but calculating the correct expectation is no longer easy. But it is easy to do a computer simulation of this guessing with information, so we can get a good idea of the expectation by simulation. (This is similar to the way that skilled blackjack players make blackjack into a favorable game by observing the cards that have already been played. See Exercise 29.) (a) First, do a simulation of guessing without information, repeating the experiment at least 1000 times. Estimate the expected number of correct answers and compare your result with the theoretical expectation. (b) What is the best strategy for guessing with information? (c) Do a simulation of guessing with information, using the strategy in (b). Repeat the experiment at least 1000 times, and estimate the expectation in this case. (d) Let \(S\) be the number of stars and \(C\) the number of circles in the deck. Let \(h(S, C)\) be the expected winnings using the optimal guessing strategy in (b). Show that \(h(S, C)\) satisfies the recursion relation $$h(S, C)=\frac{S}{S+C} h(S-1, C)+\frac{C}{S+C} h(S, C-1)+\frac{\max (S, C)}{S+C}$$ and \(h(0,0)=h(-1,0)=h(0,-1)=0 .\) Using this relation, write a program to compute \(h(S, C)\) and find \(h(10,10)\). Compare the computed value of \(h(10,10)\) with the result of your simulation in (c). For more about this exercise and Exercise 26 see Diaconis and Graham. \(^{11}\)

You have 80 dollars and play the following game. An urn contains two white balls and two black balls. You draw the balls out one at a time without replacement until all the balls are gone. On each draw, you bet half of your present fortune that you will draw a white ball. What is your expected final fortune?

A card is drawn at random from a deck of playing cards. If it is red, the player wins 1 dollar; if it is black, the player loses 2 dollars. Find the expected value of the game.

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