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Moment generating function is finite given limsup property

Convergence properties of a moment generating function for a random variable without a finite upper bound.Finding the Moment Generating Function given f(x)Express moment generating function as sum of two othersProbability: Deriving The Moment Generating Function Given the Definition of a Continuous Random Variablelinear transformation of factorial moment generating functionmixture distribution moment generating functionIdentifying random variables from moment generating functions and using characteristic functionsMoment generating function within another functionmoment-generating function is well definedExpectation property of moment generating function

0

$begingroup$

For a random variable $X$ define the moment-generating function $M_X(t) = mathbb{E}[e^{tX}]$. If it is known that

$$limsup_{xtoinfty}frac{logmathbb{P}(X>x)}{x} =- c < 0,$$

how can it be shown that $M_X(t) < infty$ for all $t in [0, c)$? (source, Ex. 2)

probability probability-theory random-variables limsup-and-liminf moment-generating-functions

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asked Oct 30 '18 at 22:37

jIIjII

1,22021327

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0

$begingroup$

For a random variable $X$ define the moment-generating function $M_X(t) = mathbb{E}[e^{tX}]$. If it is known that

$$limsup_{xtoinfty}frac{logmathbb{P}(X>x)}{x} =- c < 0,$$

how can it be shown that $M_X(t) < infty$ for all $t in [0, c)$? (source, Ex. 2)

probability probability-theory random-variables limsup-and-liminf moment-generating-functions

share|cite|improve this question

asked Oct 30 '18 at 22:37

jIIjII

1,22021327

$endgroup$

add a comment | 

$begingroup$

For a random variable $X$ define the moment-generating function $M_X(t) = mathbb{E}[e^{tX}]$. If it is known that

$$limsup_{xtoinfty}frac{logmathbb{P}(X>x)}{x} =- c < 0,$$

how can it be shown that $M_X(t) < infty$ for all $t in [0, c)$? (source, Ex. 2)

probability probability-theory random-variables limsup-and-liminf moment-generating-functions

share|cite|improve this question

asked Oct 30 '18 at 22:37

jIIjII

1,22021327

$endgroup$

share|cite|improve this question

asked Oct 30 '18 at 22:37

jIIjII

1,22021327

share|cite|improve this question

asked Oct 30 '18 at 22:37

jIIjII

1,22021327

asked Oct 30 '18 at 22:37

jIIjII

1,22021327

asked Oct 30 '18 at 22:37

jIIjII

1,22021327

1 Answer
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$Ee^{tX} =int_0^{infty} P{e^{tX }>x}, dx$. It is enough to show that $int_M^{infty} P{e^{tX }>x}, dx <infty$ for some $M$. Let $t<c$ and $epsilon <c-t$. By hypothesis there exists $x_0$ such that $log P{X>x} <x( -c+epsilon)$ for $x > x_0$. Hence $P{X>x} <e^{x(-c+epsilon)}$ and $int_M^{infty} P{e^{tX }>x}, dx =int_M^{infty} P{X >(log , x) /t}dx<int_M^{infty} x^{frac {-c+epsilon} t} dx<infty$ because $(-c+epsilon) /t <-1$.

share|cite|improve this answer

answered Oct 31 '18 at 0:10

Kavi Rama MurthyKavi Rama Murthy

73.3k53170

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1

$begingroup$

$Ee^{tX} =int_0^{infty} P{e^{tX }>x}, dx$. It is enough to show that $int_M^{infty} P{e^{tX }>x}, dx <infty$ for some $M$. Let $t<c$ and $epsilon <c-t$. By hypothesis there exists $x_0$ such that $log P{X>x} <x( -c+epsilon)$ for $x > x_0$. Hence $P{X>x} <e^{x(-c+epsilon)}$ and $int_M^{infty} P{e^{tX }>x}, dx =int_M^{infty} P{X >(log , x) /t}dx<int_M^{infty} x^{frac {-c+epsilon} t} dx<infty$ because $(-c+epsilon) /t <-1$.

share|cite|improve this answer

answered Oct 31 '18 at 0:10

Kavi Rama MurthyKavi Rama Murthy

73.3k53170

$endgroup$

add a comment | 

1

$begingroup$

$Ee^{tX} =int_0^{infty} P{e^{tX }>x}, dx$. It is enough to show that $int_M^{infty} P{e^{tX }>x}, dx <infty$ for some $M$. Let $t<c$ and $epsilon <c-t$. By hypothesis there exists $x_0$ such that $log P{X>x} <x( -c+epsilon)$ for $x > x_0$. Hence $P{X>x} <e^{x(-c+epsilon)}$ and $int_M^{infty} P{e^{tX }>x}, dx =int_M^{infty} P{X >(log , x) /t}dx<int_M^{infty} x^{frac {-c+epsilon} t} dx<infty$ because $(-c+epsilon) /t <-1$.

share|cite|improve this answer

answered Oct 31 '18 at 0:10

Kavi Rama MurthyKavi Rama Murthy

73.3k53170

$endgroup$

add a comment | 

$begingroup$

$Ee^{tX} =int_0^{infty} P{e^{tX }>x}, dx$. It is enough to show that $int_M^{infty} P{e^{tX }>x}, dx <infty$ for some $M$. Let $t<c$ and $epsilon <c-t$. By hypothesis there exists $x_0$ such that $log P{X>x} <x( -c+epsilon)$ for $x > x_0$. Hence $P{X>x} <e^{x(-c+epsilon)}$ and $int_M^{infty} P{e^{tX }>x}, dx =int_M^{infty} P{X >(log , x) /t}dx<int_M^{infty} x^{frac {-c+epsilon} t} dx<infty$ because $(-c+epsilon) /t <-1$.

share|cite|improve this answer

answered Oct 31 '18 at 0:10

Kavi Rama MurthyKavi Rama Murthy

73.3k53170

$endgroup$

share|cite|improve this answer

answered Oct 31 '18 at 0:10

Kavi Rama MurthyKavi Rama Murthy

73.3k53170

answered Oct 31 '18 at 0:10

Kavi Rama MurthyKavi Rama Murthy

73.3k53170

answered Oct 31 '18 at 0:10

Kavi Rama MurthyKavi Rama Murthy

73.3k53170