/*! 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 5 Let the pmf of \(Y_{n}\) be \(p_... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 5

Let the pmf of \(Y_{n}\) be \(p_{n}(y)=1, y=n\), zero elsewhere. Show that \(Y_{n}\) does not have a limiting distribution. (In this case, the probability has "escaped" to infinity.)

Short Answer

Expert verified
The given series of random variables \(Y_{n}\) does not have a limiting distribution. This is because the limit of the cumulative distribution function as \(n\) tends to infinity does not exist, indicating that the probability has 'escaped' to infinity.

Step by step solution

01

Understand the problem statement

The series of random variables \(Y_{n}\) given has the pmf \(p_{n}(y)=1\) for \(y=n\) and zero elsewhere. This means that for each \(n\), the random variable \(Y_{n}\) takes the value \(n\) with probability 1. The question is whether there is a limiting probability distribution as \(n\) goes to infinity.
02

Understand limiting distribution

A series of random variables \(X_{n}\) has a limiting distribution if the probabilities \(P(X_{n} \leq x)\) converge to a limit \(F(x)\) for each \(x\) as \(n\) goes to infinity, where \(F(x)\) is a cumulative distribution function.
03

Apply the definition to the given series

First, compute the cumulative distribution function \(F_{n}(x)\) of \(Y_{n}\). It is the probability that \(Y_{n}\) is less than or equal to \(x\), or 1 if \(x\geq n\) and 0 otherwise. Now, consider \(n\) tending to infinity, the limit \(lim_{n \to \infty} F_{n}(x)\) does not exist because It either becomes 0 or 1 based on the value of \(x\). Hence, \(Y_{n}\) does not have a limiting distribution.

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Most popular questions from this chapter

Let \(\bar{X}_{n}\) denote the mean of a random sample of size \(n\) from a Poisson distribution with parameter \(\mu=1\) (a) Show that the mgf of \(Y_{n}=\sqrt{n}\left(\bar{X}_{n}-\mu\right) / \sigma=\sqrt{n}\left(\bar{X}_{n}-1\right)\) is given by \(\exp \left[-t \sqrt{n}+n\left(e^{t / \sqrt{n}}-1\right)\right]\) (b) Investigate the limiting distribution of \(Y_{n}\) as \(n \rightarrow \infty\). Hint: Replace, by its MacLaurin's series, the expression \(e^{t / \sqrt{n}}\), which is in the exponent of the mgf of \(Y_{n}\).

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