Re: Probability space

C6L1V@xxxxxxx wrote:

bbmcmung@xxxxxxxxx wrote:


If a random variable X is in L^n(P), prove that (x^n)P(|X|>x) tends to
0 as x tends to infinity.



Since int_0^infinity |x|^n dP(|x|) is finite, we know that if we are
given any e>0, there exists L > 0 such that int_M^infinity |x|^n
dP(|x|) < e for all M >= L. Thus, e > int_M^infinity |x|^n dP(|x|) >=
M^n int_M^infinity dP(|x|)  =  M^n P(|X| >= M)  >=   M^n P(|X| > M),
for all M >= L.

R.G. Vickson


Here is a variation on this proof.

For simplicity of typing, assume X is nonnegative. For nonnegative x, let I_x be the indicator variable for the event {X >= x}. Since X^n I_x decreases to 0 as x grows arbitrarily large and E[X^n I_0] is finite, E[X^n I_x] vainishes as x grows arbitrarily large by monotone convergence.
X^n I_x >= x^n I_x >= 0 for all x, whence E[x^n I_x] = x^n EI_x vanishes as well.


I wonder if the following approach could also be used to come up with an (albeit less elegant) proof.

Pringsheim's theorem[*] states that if a is a summable, nonincreasing sequence of nonnegative real numbers, then n a_n vanishes as n grows arbitrarily large. We use this to show that if f: R+ -> R+ is nonincreasing and the improper Riemann integral int(0..infty, f) converges, then x f(x) vanishes as x grows large. Since
EX = int(x=0..infty, P{X > x}) for X nonnegative, this gives us the result for n = 1.

Can we use this to come up with some indcutive proof? E[X^n] = n int(x=0..infty, x^(n-1) P{X > x}). We do have the problem that the integrand is not monotone. Perhaps a generalization of Pringsheim: if a is nonnegative and nonincreasing and (n^k a_n) is summable, then n^(k+1) a_n vanishes as n grows large? Of course, the result we are trying to show implies this generalization.

[*] E.g., Goldberg, Methods of Real Analysis (an introductory undergraduate text), p.88 in the second edition.

--
Stephen J. Herschkorn                        sjherschko@xxxxxxxxxxxx
Math Tutor on the Internet and in Central New Jersey and Manhattan

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