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Chapter 5: Problem 15

Let the random variable \(Z_{n}\) have a Poisson distribution with parameter \(\mu=n .\) Show that the limiting distribution of the random variable \(Y_{n}=\left(Z_{n}-n\right) / \sqrt{n}\) is normal with mean zero and variance \(1 .\)

Short Answer

Expert verified
The limiting distribution of the random variable \(Y_{n}=\left(Z_{n}-n\right) / \sqrt{n}\) is a normal distribution with a mean of 0 and a variance of 1.

Step by step solution

01

Compute the Moment Generating Function (MGF) for \(Z_{n}\).

The moment generating function (MGF) of a Poisson random variable with parameter \(\mu\) is given by \(M(t) = e^{\mu(e^{t}-1)}\). Since the given parameter for \(Z_{n}\) is \(n\), its MGF would be \(M_{Z_{n}}(t) = e^{n(e^{t}-1)}\)
02

Compute the MGF for \(Y_{n}\).

The moment generating function of a transformed random variable can be obtained by plugging the transformation into the MGF of the original random variable. For \(Y_{n}=(Z_{n}-n)/\sqrt{n}\), plug this into the MGF of \(Z_{n}\) to get \(M_{Y_{n}}(t) = e^{n(e^{t/\sqrt{n}}-1)-t}\)
03

Find the limit of the MGF for \(Y_{n}\) as \(n\) approaches infinity.

To find the limiting distribution as \(n\) approaches infinity, we have to find the limit as \(n\) goes to infinity of \(M_{Y_{n}}(t)\). This yields \(\lim_{n \to \infty} e^{n(e^{t/\sqrt{n}}-1)-t}\). Apply L'Hopital's rule to this limit. After simplification, it can be shown that this limit equals \(e^{t^{2}/2}\), the MGF of a normal distribution with mean zero and variance 1.
04

Final Conclusion about the limiting distribution of \(Y_{n}\).

Given that the moment generating function of the limiting distribution is the same as that of a normal distribution with mean zero and variance 1, it can be concluded that the limiting distribution of \(Y_{n}\) is indeed normal with mean zero and variance 1.

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Most popular questions from this chapter

Let \(Y\) be \(b(n, 0.55)\). Find the smallest value of \(n\) so that (approximately) \(\operatorname{Pr}\left(Y / n>\frac{1}{2}\right) \geq 0.95\)

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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 \(1-p=0.95\).

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