Department of Statistics, Yale University



Department of Statistics, Yale University

STAT242b Theory of Statistics

Suggested Solutions to Homework 2

Compiled by Marco Pistagnesi

Problem 2.9

Note that [pic] , thus we have [pic]. Also note that for the Poisson we have [pic] and [pic]. Hence we can apply the CLT to state:

[pic]

Problem 6.1-3

Let us recall the relevant definitions;

[pic] ; [pic] ; [pic].

So for 6.1 we get;

[pic],

[pic], [pic].

For 6.2, in order to find the moments of X(n) we need to figure out its distribution. By independence, we get:

[pic].

Hence, the density will be: [pic]. It follows:

[pic]

[pic], from which:

[pic]

[pic]

[pic].

For 6.3, similarly we get:

[pic],

[pic] , [pic].

Problem 7.4

For this problem I will propose a fairly detailed solution, as the vast majority of you approached it in a slightly imprecise way.

Recall that, be definition, [pic].

We know that [pic] are iid, and thus we also know that the [pic] are also iid (note that the indicator is also a random variable). The RHS of the formula above is a sample average of indicators, and thus the CLT applies to it. In order to apply this theorem we first must find the mean and variance of each of the identical random variables [pic]. We simplify those calculations by computing the expectation value of I to every power:

[pic]. Hence:

[pic] and [pic][1].

Now the CLT allows as to make the following one statement and no others:

[pic] (1)

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[1] Note that we could have reached the same conclusion by arguing that the indicator takes only 2 values with success probability given by F(x), hence is distributed as Bernoulli(F(x)).

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