/*! 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 34 Suppose that \(Y\) is a continuo... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that \(Y\) is a continuous random variable with density \(f(y)\) that is positive only if \(y \geq 0 .\) If \(F(y)\) is the distribution function, show that $$E(Y)=\int_{0}^{\infty} y f(y) d y=\int_{0}^{\infty}[1-F(y)] d y$$ $$\left[\text {Hint: If } y>0, y=\int_{0}^{y} d t, \text { and } E(Y)=\int_{0}^{\infty} y f(y) d y=\int_{0}^{\infty}\left\\{\int_{0}^{y} d t\right\\} f(y) d y .\text { Exchange }\right.$$ the order of integration to obtain the desired result.]

Short Answer

Expert verified
The expected value \(E(Y)\) can be expressed as \(\int_{0}^{\infty} [1-F(y)] dy\) by changing the order of integration.

Step by step solution

01

Understand the Problem

We are given a continuous random variable. We want to show that the expected value \(E(Y)\) can be expressed using the cumulative distribution function \(F(y)\). The hint suggests using integration by parts and exchanging the order of integration.
02

Express the Expected Value

The expected value of a continuous random variable \(Y\) with probability density function \(f(y)\) is given by \(E(Y) = \int_{0}^{\infty} y f(y) \, dy\). This is our starting point to derive the alternative expression.
03

Apply the Hint

According to the hint, when \(y > 0, y\) can be rewritten as \(\int_{0}^{y} \, dt\). Substituting this into the expression for \(E(Y)\), we get: \[E(Y) = \int_{0}^{\infty} \left(\int_{0}^{y} \, dt\right) f(y) \, dy\].
04

Interchange the Order of Integration

We can change the order of integration to simplify the double integral. This gives us: \[E(Y) = \int_{0}^{\infty} \int_{t}^{\infty} f(y) \, dy \, dt\].
05

Use the Cumulative Distribution Function

Notice that \(\int_{t}^{\infty} f(y) \, dy\) represents the probability that \(Y > t\), which is \(1 - F(t)\). Substituting this into our expression gives: \[E(Y) = \int_{0}^{\infty} (1 - F(t)) \, dt\].
06

Conclusion

We have derived that \(E(Y)\) can be expressed as \(\int_{0}^{\infty} [1-F(y)] \, dy\), confirming the desired result of the problem.

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Key Concepts

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

Cumulative Distribution Function
In probability theory, the cumulative distribution function (CDF), represented as \( F(y) \), is a powerful way to describe the distribution of a continuous random variable. CDF gives you the probability that the random variable \( Y \) takes a value less than or equal to \( y \). In mathematical terms, it's expressed as:\[ F(y) = P(Y \leq y) = \int_{0}^{y} f(t) \, dt \]where \( f(y) \) is the probability density function (PDF). The CDF increases monotonically from 0 to 1 as \( y \) goes from negative infinity to positive infinity. For \( y \geq 0 \), as is the case in our current problem, the CDF is often related to its complementary, \( 1 - F(y) \), which gives the probability that the variable takes a value greater than \( y \). This aspect is particularly useful when working with expected values as it simplifies the expressions and calculations. We use this property in the formula for the expected value derivation, as seen in the exercise.
Integration by Parts
Integration by parts is a key technique for integrating products of functions. It's especially handy when direct integration is difficult or impossible. The rule is derived from the product rule of differentiation and is expressed as:\[ \int u \, dv = uv - \int v \, du \]where \( u \) and \( dv \) are differentiable functions of \( x \). To apply this method effectively, we choose parts of the integrand to differentiate and integrate wisely. This technique is integral in transitioning certain complex forms into more manageable expressions. In the given problem, while the integration by parts is not directly applied between parts, understanding this concept helps in structuring the integration, such as rewriting \( y \) in terms of another integral, \( \int_{0}^{y} dt \), which allows for easier manipulation and substitution of variables in the integration process.
Order of Integration
A double integral involves integrating a function of two variables over a region in the plane. Often, it's set up as a nested integral, and the order of integration (which variable we integrate first) can significantly simplify the integration process. In technical terms, changing the order of integration refers to switching from \( \int_a^b \int_c^d \cdots \, dy \, dx \) to \( \int_c^d \int_a^b \cdots \, dx \, dy \), and vice versa.In our current problem, exchanging the order of integration transforms the step from integrating \( y f(y) \) first to managing \( dt \) initially. This strategic switch results in an inside-out exploration of the problem, giving way to gaining insights from an alternative angle on the infinite plane. Altering this order converts the initial expression, making it easier to observe the cumulative properties \( 1 - F(t) \), leading to a streamlined expression for the expected value. Applying this technique requires careful adjustment but opens avenues to unravel complex integrals effectively.

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Most popular questions from this chapter

Use the applet Comparison of Beta Density Functions to compare beta -density functions with \((\alpha=0.5, \beta=0.7),(\alpha=0.7, \beta=0.7),\) and \((\alpha=0.9, \beta=0.7)\). a. What is the general shape of these densities? b. What do you observe as the value of \(\alpha\) gets larger?

The temperature \(Y\) at which a thermostatically controlled switch turns on has probability density function given by $$f(y)=\left\\{\begin{array}{ll} 1 / 2, & 59 \leq y \leq 61 \\ 0, & \text { elsewhere } \end{array}\right.$$ Find \(E(Y)\) and \(V(Y)\)

The length of time \(Y\) necessary to complete a key operation in the construction of houses has an exponential distribution with mean 10 hours. The formula \(C=100+40 Y+3 Y^{2}\) relates the cost \(C\) of completing this operation to the square of the time to completion. Find the mean and variance of \(C\).

Suppose that the waiting time for the first customer to enter a retail shop after 9:00 A.M. is a random variable \(Y\) with an exponential density function given by $$f(y)=\left\\{\begin{array}{ll} \left(\frac{1}{\theta}\right) e^{-y / \theta}, & y>0 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. Find the moment-generating function for \(Y\). b. Use the answer from part (a) to find \(E(Y)\) and \(V(Y)\)

Errors in measuring the time of arrival of a wave front from an acoustic source sometimes have an approximate beta distribution. Suppose that these errors, measured in microseconds, have approximately a beta distribution with \(\alpha=1\) and \(\beta=2\). a. What is the probability that the measurement error in a randomly selected instance is less than .5 \(\mu \mathrm{s} ?\) b. Give the mean and standard deviation of the measurement errors.

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