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Chapter 5: 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
Therefore, it's clear that \(Y_{n}\) does not have a limiting distribution because the probability doesn't get concentrated at any particular outcome but rather continues to move towards higher outcomes as \(n\) increases, showing an indication that the probability has 'escaped' to infinity.

Step by step solution

01

Understanding the Probability Mass Function (pmf)

The pmf of \(Y_{n}\) is given by \(p_{n}(y)=1, y=n\), zero elsewhere. This means that for each \(n\), \(Y_{n}\) gives the probability 1 to outcome \(n\) and 0 to all other outcomes.
02

Determine the distribution of \(Y_{n}\)

Here, it's important to note that the distribution of each \(Y_{n}\) is different from all other \(Y_{n}\) for \(n > 1\), no two are the same. This is because \(Y_{n}\) only allocates the entire probability to the outcome \(n\). Hence each \(Y_{n}\) specifies a different unique outcome to which all the probability is given.
03

Consider the limiting distribution

Now, think of the limiting distribution. Assume there is a limiting distribution, say \(Y\). This would imply that the probability gets concentrated on some outcomes as \(n\) increases infinitely. But, for each \(n\), \(Y_{n}\) gives the probability only to \(n\) and nowhere else. This means the probability will keep moving to higher values as \(n\) increases, it won't get concentrated at any point. Hence the assumption of a limiting distribution is invalid.

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Most popular questions from this chapter

Let \(Y_{1}\) denote the minimum of a random sample of size \(n\) from a distribution that has pdf \(f(x)=e^{-(x-\theta)}, \theta

Let \(\bar{X}\) denote the mean of a random sample of size 100 from a distribution that is \(\chi^{2}(50)\). Compute an approximate value of \(P(49<\bar{X}<51)\).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample of size \(n\) from a distribution that is \(N\left(\mu, \sigma^{2}\right)\), where \(\sigma^{2}>0 .\) Show that the sum \(Z_{n}=\sum_{1}^{n} X_{i}\) does not have a limiting distribution.

Let \(p=0.95\) be the probability that a man, in a certain age group, lives at least 5 years. (a) If we are to observe 60 such men and if we assume independence, find the probability that at least 56 of them live 5 or more years. (b) Find an approximation to the result of part (a) by using the Poisson distribution. Hint: Redefine \(p\) to be \(0.05\) and \(1-p=0.95\).

If \(Y\) is \(b\left(100, \frac{1}{2}\right)\), approximate the value of \(P(Y=50)\).
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