Question

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

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

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.

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.

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.