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Chapter 9: Problem 543

Find the expected value of the random variable \(\mathrm{X}\) if \(\mathrm{X}\) is distributed with probability density function \(f(x)=\lambda e^{-\lambda x} \quad\) for \(0

Short Answer

Expert verified
The expected value, E[X], of the random variable X with the probability density function \(f(x) = \lambda e^{-\lambda x}\) for \(0 < x < \infty\) is \(\frac{1}{\lambda}\).

Step by step solution

01

Write down the given pdf and the definition of the expected value

We are given the pdf: \(f(x) = \lambda e^{-\lambda x}\) for \(0 < x < \infty\) And the definition of the expected value: E[X] = \(\int_0^\infty xf(x)dx\)
02

Substitute the given pdf into the expected value integral

Now, we will substitute the given pdf into the expected value integral: E[X] = \(\int_0^\infty x(\lambda e^{-\lambda x})dx\)
03

Perform Integration By Parts

To solve this integral, we can use the integration by parts technique. Let's choose our parts: \(u = x\) ; \(dv = \lambda e^{-\lambda x}dx\) Now, differentiate u: \(du = dx\) Integrate dv: \(v = - e^{-\lambda x}\) Now we can use the integration by parts formula: \(\int u \, dv = uv - \int v \, du\) In our case: E[X] = \([-x e^{-\lambda x}]_0^\infty - \int_0^\infty -e^{-\lambda x} dx\)
04

Evaluate the boundaries and solve the remaining integral

Now we need to evaluate the boundaries for the term \([-x e^{-\lambda x}]_0^\infty\) and solve the remaining integral: E[X] = \(-[\lim_{x\to\infty} x e^{-\lambda x} - 0] + \int_0^\infty e^{-\lambda x} dx\) (By limit properties, we know that \(x e^{-\lambda x}\) approaches zero as \(x\) approaches infinity, since \(e^{-\lambda x}\) decreases faster than \(x\) increases.) E[X] = \(0 + \int_0^\infty e^{-\lambda x} dx\) Now, we can evaluate the last integral: E[X] = \([-\frac{1}{\lambda}e^{-\lambda x}]_0^\infty\) E[X] = \(-\frac{1}{\lambda}[\lim_{x\to\infty} e^{-\lambda x} - e^{-0}]\)
05

Evaluate the limit and simplify the result

Evaluating the limit and simplifying the expression: E[X] = \(- \frac{1}{\lambda}[0 - 1]\) E[X] = \(\frac{1}{\lambda}\) Thus, the expected value of the random variable X following the given distribution is \(\frac{1}{\lambda}\).

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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 crucial when working with continuous random variables. A pdf, like the one given in our exercise \(f(x) = \textbackslash lambda e^{-\textbackslash lambda x}\) for \(0 < x < \textbackslash infty\), describes how the probability is distributed over the values of the random variable. It's important to remember that the area under the pdf over an interval gives the probability that the random variable falls within that interval.

For a valid pdf, there are two key conditions: the function must be non-negative (\(f(x) \geq 0\)) for all \(x\), and the total area under the curve must equal 1, which ensures that the probability of all possible outcomes sums to 100%.

In the context of finding the expected value, the pdf plays a pivotal role as it serves as the 'weight' for each value of the random variable in the calculation.
Integration by Parts
Integration by parts is a technique derived from the product rule for differentiation and is used to simplify the integration of products of functions. The formula is \(\textbackslash int u \textbackslash, dv = uv - \textbackslash int v \textbackslash, du\). To apply this method, we choose one part to differentiate (\(u\)) and another to integrate (\(dv\)).

Application in the Expected Value Calculation

For our exercise, \(u = x\) and \(dv = \textbackslash lambda e^{-\textbackslash lambda x}dx\). Differentiating \(u\) and integrating \(dv\) leads us to \(du = dx\) and \(v = - e^{-\textbackslash lambda x}\), which we then use in the integration by parts formula. Carefully selecting \(u\) and \(dv\) is a strategic move to simplify the calculation, especially when dealing with exponential functions common in probability problems.

This technique often requires evaluating the definite integral at the bounds, which may involve limits if the interval is infinite, as in our exercise.
Random Variable
A random variable is a numerical description of the outcome of a statistical experiment. It assigns a real number to each outcome in the experiment's sample space. Random variables can be discrete (having separate distinct values) or continuous (having values over a continuum of possibilities).

Our problem discusses a continuous random variable, \(X\), which can take any value within an interval, in this case, from zero to infinity. This continuity impacts how we calculate probabilities and related measures, like the expected value. Instead of summing probabilities as with discrete random variables, we use integration to find the expected value for continuous ones as it involves adding up an infinite number of probabilities over a range.

The expected value, sometimes called the mean, provides us with a measure of the center or 'average' outcome we'd expect from the random variable's distribution and is a fundamental concept in the fields of statistics and probability theory.

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Most popular questions from this chapter

Flapjack Computers is interested in developing a new tape drive for a proposed new computer. Flapjack does not have research personnel available to develop the new drive itself and so is going to subcontract the development to an independent research firm. Flapjack has set a fee of \(\$ 250,000\) for developing the new tape drive and has asked for bids from various research firms. The bid is to be awarded not on the basis of price (set at \(\$ 250,000\) ) but on the basis of both the technical plan shown in the bid and the firm's reputation. Dyna Research Institute is considering submitting a proposal (i.e., a bid) to Flapjack to develop the new tape drive. Dyna Research Management estimated that it would cost about \(\$ 50,000\) to prepare a proposal; further they estimated that the chances were about \(50-50\) that they would be awarded the contract. There was a major concern among Dyna Research engineers concerning exactly how they would develop the tape drive if awarded the contract. There were three alternative approaches that could be tried. One involved the use of certain electronic components. The engineers estimated that it would cost only \(\$ 50,000\) to develop a prototype of the tape drive using the electronic approach, but that there was only a 50 percent chance that the prototype would be satisfactory. A second approach involved the use of certain magnetic apparatus. The cost of developing a prototype using this approach would cost \(\$ 80,000\) with 70 percent chance of success. Finally, there was a mechanical approach with cost of \(\$ 120,000\), but the engineers were certain of success. Dyna Research could have sufficient time to try only two approaches. Thus, if either the magnetic or the electronic approach tried and failed, the second attempt would have to use the mechanical approach in order to guarantee a successful prototype. The management of Dyna Research was uncertain how to take all this information into account in making the immediate decision-whether to spend \(\$ 50,000\) to develop a proposal for Flapjack. Can you help?

Show, by altering the joint density of \(\mathrm{X}\) and \(\mathrm{Y}\) in the previous problem, that it is not always possible to construct a unique joint distribution from a pair of given marginal distributions.

Assume you have two populations \(\mathrm{N}\left(\mu_{1}, \sigma^{2}\right)\) and \(\mathrm{N}\left(\mu_{2}, \sigma^{2}\right)\). The distributions have the same, but unknown, variance \(\sigma^{2}\). Derive a method for determining a confidence interval for $$ \mu_{1}-\mu_{2} $$

A research worker wishes to estimate the mean of a population using a sample large enough that the probability will be \(.95\) that the sample mean will not differ from the population mean by more than 25 percent of the standard deviation. How large a sample should he take?

Let \(X\) be the random variable defined as the number of dots observed on the upturned face of a fair die after a single toss. Find the expected value of \(\mathrm{X}\).
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