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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 $M_X(t) = mathbbE[e^tX]$. If it is known that

$$limsup_xtoinftyfraclogmathbbP(X>x)x =- c < 0,$$

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 $M_X(t) = mathbbE[e^tX]$. If it is known that

$$limsup_xtoinftyfraclogmathbbP(X>x)x =- c < 0,$$

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 $M_X(t) = mathbbE[e^tX]$. If it is known that

$$limsup_xtoinftyfraclogmathbbP(X>x)x =- c < 0,$$

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

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answered Oct 31 '18 at 0:10

Kavi Rama MurthyKavi Rama Murthy

71.7k53170

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1

$begingroup$

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

share|cite|improve this answer

answered Oct 31 '18 at 0:10

Kavi Rama MurthyKavi Rama Murthy

71.7k53170

$endgroup$

add a comment | 

1

$begingroup$

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

share|cite|improve this answer

answered Oct 31 '18 at 0:10

Kavi Rama MurthyKavi Rama Murthy

71.7k53170

$endgroup$

add a comment | 

$begingroup$

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

share|cite|improve this answer

answered Oct 31 '18 at 0:10

Kavi Rama MurthyKavi Rama Murthy

71.7k53170

$endgroup$

share|cite|improve this answer

answered Oct 31 '18 at 0:10

Kavi Rama MurthyKavi Rama Murthy

71.7k53170

answered Oct 31 '18 at 0:10

Kavi Rama MurthyKavi Rama Murthy

71.7k53170

answered Oct 31 '18 at 0:10

Kavi Rama MurthyKavi Rama Murthy

71.7k53170