Let X be a continuous rv with density f. and let X1, X2 lie two independent rvs,
ID: 2944940 • Letter: L
Question
Let X be a continuous rv with density f. and let X1, X2 lie two independent rvs, both distributed as is X. It is then not usually the case that the rv 2X is distributed as is X1 + X2. However, the Cauchy density whose standardized form is given by f(x) = 1/pi.1/1 + x2 possesses this property: X1 + X2 has the same distribution as the rv 2X. It is illustrated and compared to the standard normal distribution in Figure 1.1. Based on the property described earlier, argue without any explicit calculation that the variance of the Cauchy distribution is necessarily infinite. Give an inductive argument for the rather unintuitive feature that for the Cauchy distribution the arithmetic mean from a sample of n = 2k independent realizations of X has exactly the same distribution as each contributing summand itself.Explanation / Answer
a. Let 2 denote the variance of X. According to the general rules, thevariance of the rv 2X is 42, whereas the variance of the rv X1 +X2 is equalto 22. Therefore, if both of these rvs are to follow the same — Cauchy —distribution, then 2 must be either infinite or zero. However, in the lattercase the probability mass would be concentrated at a single point, and so Xwould be a constant, not a Cauchy rv. Therefore, the first case applies, and2 = .
b. Consider first the case of n = 2. If, as stated, X1 + X2 is distributedas is 2X, then it is also true that X is distributed as is (X1 +X2)/2, i.e., thearithmetic mean of a sample of two. Thus, the property holds for n = 2k = 2,i.e. for k = 1.Next, assume that the sum of 2k independent Cauchy rvs is distributedas is 2kX. Consider then the sum of 2k+1 independent Cauchy rvs: it may bethought of as 2k independent Cauchy rvs plus another 2k independent Cauchyrvs. By assumption, both components are individually distributed as is 2kX,and so their sum has the same distribution as 2k(X1+X2). But we saw earlierthat X1 + X2 is distributed as is 2X, and so the sum of 2k+1 independentCauchy rvs has the same distribution as 2k(2X) = 2k+1X.By induction, then, the sum of 2k independent Cauchy rvs is distributedas is 2kX. Dividing by the scalar n = 2k yields the assertion: the arithmeticmean of n = 2k independent Cauchy rvs is distributed as is any of the contributingsummands individually. This property seems rather counterintuitiveand appears to many as deeply disturbing: it defies our nearly universal beliefthat the precision of arithmetic means necessarily increases with increasingsample size. What seems especially disturbing is that the Cauchy distribution(see Figure 1.1 looks fairly “well-behaved”: it does not have particularlybizarre or exoctic shape and is not easily dismissed as a mere mathematicalcuriosity. The property actually holds for any n, not just for n = 2k, but thisfact requires slightly more elaborate arguments.
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