/*! 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 6 In the text we noted that $$ ... [FREE SOLUTION] | 91Ó°ÊÓ

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In the text we noted that $$ E\left[\sum_{i=1}^{=} X_{i}\right]=\sum_{i=1}^{\infty} E\left[X_{i}\right] $$ when the \(X_{i}\) are all nonnegative random variables. Since an integral is a limit of sums, one might expect that $$ E\left[\int_{0}^{x} X(t) d t\right]=\int_{0}^{x} E[X(t)] d t $$ whenever \(X(t), 0 \leq t<\infty\), are all nonnegative random variables; and this result is indeed true. Use it to give another proof of the result that, for a nonnegative random variable \(X\), $$ E[X)=\int_{0}^{\pi} P\\{X>t\\} d t $$ Hint: Define, for each nonnegative \(t\), the random variable \(X(t)\) by $$ X(t)= \begin{cases}1 & \text { if } t

Short Answer

Expert verified
To prove that for a nonnegative random variable \(X\), $$ E[X]=\int_{0}^{\infty} P\{X>t\} dt $$ we define a random variable \(X(t)\) as $$ X(t)= \begin{cases}1 & \text { if } tt\}\) and also compute the integral of \(X(t)\) to be simply \(X\). Using the given relation, \(E\left[\int_{0}^{\infty} X(t) dt\right] = \int_{0}^{\infty} E[X(t)] dt\), we substitute these results to obtain the desired proof: $$ E[X] = \int_{0}^{\infty} P\{X>t\} dt $$

Step by step solution

01

Find E[X(t)]

We need to find the expected value of the random variable \(X(t)\). Using the definition of \(X(t)\), $$ E[X(t)] = 1 \cdot P\{t < X\} + 0 \cdot P\{t \geq X\} = P\{X>t\} $$ Now, we have the relation between \(E[X(t)]\) and \(P\{X>t\}\).
02

Relate Σ_{0}^{∞} X(t) dt to X

We need to relate the integral of \(X(t)\) to \(X\). Let's compute the integral: $$ \int_{0}^{\infty} X(t) dt = \int_{0}^{X} 1 dt + \int_{X}^{\infty} 0 dt = [t]_0^X + [0]_{X}^{\infty} = X - 0 + 0 = X $$ Now, we have the relation between the integral of \(X(t)\) and \(X\).
03

Use the given relation in the problem

Now let's apply the given relation in the problem: $$ E\left[\int_{0}^{\infty} X(t) dt\right] = \int_{0}^{\infty} E[X(t)] dt $$ We have found in Steps 1 and 2, that \(\int_{0}^{\infty} X(t) dt = X\) and \(E[X(t)] = P\{X>t\}\). Substitute these into the relation: $$ E[X] = \int_{0}^{\infty} P\{X>t\} dt $$ This completes the proof of the result.

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Key Concepts

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

Nonnegative Random Variables
A nonnegative random variable is a type of random variable that can only take values that are zero or positive. It's like having a scale that never goes below zero. In many probabilistic models, nonnegative random variables are common since they often represent quantities that can't be negative, like the amount of time until something happens or the number of customers entering a store.

Understanding nonnegative random variables is important because they simplify mathematical operations and expectations. Working with nonnegative variables means we avoid negative values, which can complicate computations or interpretations. They are often involved in integration and summation problems where the total value must be at least zero.

In probability, when dealing with sums of nonnegative random variables, these properties help ensure that results are logically consistent, such as with sums or expectations. The article examines how the expectation of the sum of these variables can be broken down into the sum of their individual expectations.
Proof Techniques in Probability
Proof techniques in probability often involve demonstrating that certain properties hold for random variables or probability distributions. One common method is exploiting relationships between sums, integrals, and expectations.

In probabilistic proofs, integration can often be seen as the limit of sums. Thus, by understanding each piece of a problem, like considering an integral as a continuous version of a sum, one can prove or derive important formulas or relationships.

For nonnegative random variables, a significant proof technique involves using these integral relationships. As shown in the exercise, by defining an appropriate function or variable—in the case of the article, using the function \(X(t)\)—one can relate integrals to expected values in probability. This method is powerful in flipping between discrete and continuous interpretations of a problem.

Another valuable technique involves substituting known relationships to uncover new insights. For instance, if we know how one function behaves, applying this to integrals or expectations can help confirm broader relationships, such as calculating the expected sum of infinite variables.
Expected Value
Expected value is a fundamental concept in probability and statistics, often thought of as the "average" or "mean" outcome of a random variable over many trials. For a nonnegative random variable, the expected value helps to understand the long-term average outcome when the event is repeated.

The expected value can be calculated by integrating over all possible outcomes weighted by their probabilities. In simpler terms, it's like gathering all possible values a random variable can take, multiplying each by its chance of occurring, and then summing these products.
  • For discrete random variables, we sum all possible outcomes: \(E[X] = \sum x_i P(X = x_i)\).
  • For continuous variables, we use integration: \(E[X] = \int x f(x) \, dx\).
In the context of nonnegative random variables, the expected value shines when calculating or predicting accumulated values such as total time, quantity, or cost. The exercise cleverly shows this using a specific example with \(X(t)\) and shows that by integral substitution, you can convert a complex looking integral into an understandable expected value. This highlights how the expected value serves as a bridge between the tangible realities that random variables represent and their abstract mathematical foundations.

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Most popular questions from this chapter

For Example 2 j show that the variance of the number of coupons needed to amass a full set is equal to $$ \sum_{i=1}^{N-1} \frac{i N}{(N-i)^{2}} $$ When \(N\) is large, this can be shown to be approximately equal (in the sense that their ratio approaches 1 as \(N \rightarrow \infty\) ) to \(N^{2}\left(\pi^{2} / 6\right)\).

Let \(X\) be the value of the first die and \(Y\) the sum of the values when two dice are rolled. Compute the joint moment generating function of \(X\) and \(Y\).

Consider a population consisting of individuals able to produce offspring of the same kind. Suppose that each individual will, by the end of its lifetime, have produced \(j\) new offspring with probability \(P_{j}, j \geq 0\), independently of the number produced by any other individual. The number of individuals initially present, denoted by \(X_{0}\), is called the size of the zeroth generation. All offspring of the zeroth generation constitute the first generation, and their number is denoted by \(X_{1} .\) In general, let \(X_{n}\) denote the size of the \(n\)th generation. Let \(\mu=\sum_{j=0}^{x} j P_{j}\) and \(\sigma^{2}=\sum_{j=0}^{x}(j-\mu)^{2} P_{j}\) denote, respectively, the mean and the variance of the number of offspring produced by a single individual. Suppose that \(X_{0}=1\) - that is, initially there is a single individual in the population. (a) Show that $$ E\left[X_{n}\right]=\mu E\left[X_{n-1}\right] $$ (b) Use part (a) to conclude that $$ E\left[X_{n}\right]=\mu^{n} $$ (c) Show that $$ \operatorname{Var}\left(X_{n}\right)=\sigma^{2} \mu^{n-1}+\mu^{2} \operatorname{Var}\left(X_{n-1}\right) $$ (d) Use part (c) to conclude that $$ \operatorname{Var}\left(X_{n}\right)= \begin{cases}\sigma^{2} \mu^{n-1}\left(\frac{\mu^{n}-1}{\mu-1}\right) & \text { if } \mu \neq 1 \\ n \sigma^{2} & \text { if } \mu=1\end{cases} $$ The case described above is known as a branching process, and an important question for a population that evolves along such lines is the probability that the population will eventually die out. Let \(\pi\) denote this probability when the population starts with a single individual. That is, $$ \pi=P\left\\{\text { population eventually dies out } \mid X_{0}=1\right. \text { ) } $$ (e) Argue that \(\pi\) satisfies $$ \pi=\sum_{j=0}^{\alpha} P_{j} \pi^{j} $$ HINT: Condition on the number of offspring of the initial member of the population.

Show \(\operatorname{Cov}(X, E[Y \mid X])=\operatorname{Cov}(X, Y)\)

Two envelopes, each containing a check, are placed in front of you. You are to choose one of the envelopes, open it, and see the amount of the check. At this point you can either accept that amount or you can exchange it for the check in the unopened envelope. What should you do? Is it possible to devise a strategy that does better than just accepting the first envelope? Let \(A\) and \(B, A

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