When is moment generating function finite on an interval?
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I'm working on the following exercise:
Let $X$ be a random variable on $(Omega, mathcal A, mathbf P)$ and let $$Lambda(t) := logleft(mathbf Eleft[e^{tX}right]right) quad textrm{for all } t in mathbb R.$$ Show that $D := {t in mathbb R : Lambda(t) < infty}$ is a nonempty interval.
Clearly $Lambda(0) = 0$ regardless of $X$, but I don't think it's generally true that $D$ is a nonempty interval. A counterexample could be $Omega = left[ -frac 1 2, frac 1 2 right]$, $mathbf P = mathrm{Leb}$, and $X : Omega to mathbb R$ is defined by
$$
X(omega) = begin{cases}
frac 1 omega & textrm{if } omega neq 0, \
0 &textrm{if } omega = 0.
end{cases}
$$
Then for $t > 0$,
$$
mathbf Eleft[e^{tX}right] geq int_0^{1/2} e^{t/omega} , domega = int_{-1/2}^0 e^{-t/omega} , domega = infty,
$$
so $Lambda(t) < infty$ only for $t=0$. Unless we're defining ${0} = [0,0]$ as a nonempty interval with empty interior, this statement seems false in generally. Poking around online, I found lots of results with $Lambda(t)$ being finite on $(-delta, delta)$ for some $delta > 0$, but very little on when this hypothesis is satisfied.
What are the conditions on $X$ for there being an interval around the origin (or an interval containing the origin, such as $[0, delta]$) such that $Lambda(t) < infty$?
probability integration probability-theory moment-generating-functions
asked Dec 22 '18 at 18:54
D FordD Ford
677313
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add a comment |
$begingroup$
I'm working on the following exercise:
Let $X$ be a random variable on $(Omega, mathcal A, mathbf P)$ and let $$Lambda(t) := logleft(mathbf Eleft[e^{tX}right]right) quad textrm{for all } t in mathbb R.$$ Show that $D := {t in mathbb R : Lambda(t) < infty}$ is a nonempty interval.
Clearly $Lambda(0) = 0$ regardless of $X$, but I don't think it's generally true that $D$ is a nonempty interval. A counterexample could be $Omega = left[ -frac 1 2, frac 1 2 right]$, $mathbf P = mathrm{Leb}$, and $X : Omega to mathbb R$ is defined by
$$
X(omega) = begin{cases}
frac 1 omega & textrm{if } omega neq 0, \
0 &textrm{if } omega = 0.
end{cases}
$$
Then for $t > 0$,
$$
mathbf Eleft[e^{tX}right] geq int_0^{1/2} e^{t/omega} , domega = int_{-1/2}^0 e^{-t/omega} , domega = infty,
$$
so $Lambda(t) < infty$ only for $t=0$. Unless we're defining ${0} = [0,0]$ as a nonempty interval with empty interior, this statement seems false in generally. Poking around online, I found lots of results with $Lambda(t)$ being finite on $(-delta, delta)$ for some $delta > 0$, but very little on when this hypothesis is satisfied.
What are the conditions on $X$ for there being an interval around the origin (or an interval containing the origin, such as $[0, delta]$) such that $Lambda(t) < infty$?
probability integration probability-theory moment-generating-functions
asked Dec 22 '18 at 18:54
D FordD Ford
677313
$endgroup$
1
$begingroup$
For nonnegative random variables (or |X| in your case), the property you are looking for is often referred to as being sub-exponential. You can think of it is precisely the condition needed! But there are other equivalent forms, in terms of moments, tail probabilities, etc. See for example, Prop. 2.7.1 in this book
$endgroup$
– passerby51
Dec 22 '18 at 19:06
$begingroup$
You are interpreting the term 'interval' wrongly. ${0}$ is also an interval.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 0:07
add a comment |
$begingroup$
I'm working on the following exercise:
Let $X$ be a random variable on $(Omega, mathcal A, mathbf P)$ and let $$Lambda(t) := logleft(mathbf Eleft[e^{tX}right]right) quad textrm{for all } t in mathbb R.$$ Show that $D := {t in mathbb R : Lambda(t) < infty}$ is a nonempty interval.
Clearly $Lambda(0) = 0$ regardless of $X$, but I don't think it's generally true that $D$ is a nonempty interval. A counterexample could be $Omega = left[ -frac 1 2, frac 1 2 right]$, $mathbf P = mathrm{Leb}$, and $X : Omega to mathbb R$ is defined by
$$
X(omega) = begin{cases}
frac 1 omega & textrm{if } omega neq 0, \
0 &textrm{if } omega = 0.
end{cases}
$$
Then for $t > 0$,
$$
mathbf Eleft[e^{tX}right] geq int_0^{1/2} e^{t/omega} , domega = int_{-1/2}^0 e^{-t/omega} , domega = infty,
$$
so $Lambda(t) < infty$ only for $t=0$. Unless we're defining ${0} = [0,0]$ as a nonempty interval with empty interior, this statement seems false in generally. Poking around online, I found lots of results with $Lambda(t)$ being finite on $(-delta, delta)$ for some $delta > 0$, but very little on when this hypothesis is satisfied.
What are the conditions on $X$ for there being an interval around the origin (or an interval containing the origin, such as $[0, delta]$) such that $Lambda(t) < infty$?
probability integration probability-theory moment-generating-functions
asked Dec 22 '18 at 18:54
D FordD Ford
677313
$endgroup$
I'm working on the following exercise:
Let $X$ be a random variable on $(Omega, mathcal A, mathbf P)$ and let $$Lambda(t) := logleft(mathbf Eleft[e^{tX}right]right) quad textrm{for all } t in mathbb R.$$ Show that $D := {t in mathbb R : Lambda(t) < infty}$ is a nonempty interval.
Clearly $Lambda(0) = 0$ regardless of $X$, but I don't think it's generally true that $D$ is a nonempty interval. A counterexample could be $Omega = left[ -frac 1 2, frac 1 2 right]$, $mathbf P = mathrm{Leb}$, and $X : Omega to mathbb R$ is defined by
$$
X(omega) = begin{cases}
frac 1 omega & textrm{if } omega neq 0, \
0 &textrm{if } omega = 0.
end{cases}
$$
Then for $t > 0$,
$$
mathbf Eleft[e^{tX}right] geq int_0^{1/2} e^{t/omega} , domega = int_{-1/2}^0 e^{-t/omega} , domega = infty,
$$
so $Lambda(t) < infty$ only for $t=0$. Unless we're defining ${0} = [0,0]$ as a nonempty interval with empty interior, this statement seems false in generally. Poking around online, I found lots of results with $Lambda(t)$ being finite on $(-delta, delta)$ for some $delta > 0$, but very little on when this hypothesis is satisfied.
What are the conditions on $X$ for there being an interval around the origin (or an interval containing the origin, such as $[0, delta]$) such that $Lambda(t) < infty$?
probability integration probability-theory moment-generating-functions
probability integration probability-theory moment-generating-functions
asked Dec 22 '18 at 18:54
D FordD Ford
677313
asked Dec 22 '18 at 18:54
D FordD Ford
677313
asked Dec 22 '18 at 18:54
D FordD Ford
677313
asked Dec 22 '18 at 18:54
D FordD Ford
677313
asked Dec 22 '18 at 18:54
D FordD Ford
677313
677313
1
$begingroup$
For nonnegative random variables (or |X| in your case), the property you are looking for is often referred to as being sub-exponential. You can think of it is precisely the condition needed! But there are other equivalent forms, in terms of moments, tail probabilities, etc. See for example, Prop. 2.7.1 in this book
$endgroup$
– passerby51
Dec 22 '18 at 19:06
$begingroup$
You are interpreting the term 'interval' wrongly. ${0}$ is also an interval.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 0:07
add a comment |
1
$begingroup$
For nonnegative random variables (or |X| in your case), the property you are looking for is often referred to as being sub-exponential. You can think of it is precisely the condition needed! But there are other equivalent forms, in terms of moments, tail probabilities, etc. See for example, Prop. 2.7.1 in this book
$endgroup$
– passerby51
Dec 22 '18 at 19:06
$begingroup$
You are interpreting the term 'interval' wrongly. ${0}$ is also an interval.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 0:07
1
1
$begingroup$
For nonnegative random variables (or |X| in your case), the property you are looking for is often referred to as being sub-exponential. You can think of it is precisely the condition needed! But there are other equivalent forms, in terms of moments, tail probabilities, etc. See for example, Prop. 2.7.1 in this book
$endgroup$
– passerby51
Dec 22 '18 at 19:06
$begingroup$
For nonnegative random variables (or |X| in your case), the property you are looking for is often referred to as being sub-exponential. You can think of it is precisely the condition needed! But there are other equivalent forms, in terms of moments, tail probabilities, etc. See for example, Prop. 2.7.1 in this book
$endgroup$
– passerby51
Dec 22 '18 at 19:06
$begingroup$
You are interpreting the term 'interval' wrongly. ${0}$ is also an interval.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 0:07
$begingroup$
You are interpreting the term 'interval' wrongly. ${0}$ is also an interval.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 0:07
add a comment |
1 Answer
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See my comment to see that your example is not a counterexample. What is asserted is simply that $a<b<c$, $Lambda (a) <infty,Lambda (b) <infty $ imply $Lambda (c) <infty$ which follows from the fact that $e^{cX} <e^{aX}+e^{bX}$. No necessary and sufficient conditions for finiteness of $Lambda$ in a neighbourhood of $0$ are available. If $X$ is a bounded random variable then $Lambda (t)$ is finite in for all $t$. If $P{X>s} leq a e^{-cs}$ for $s$ sufficently large, for some $a,c >0$ then $Lambda (t)<infty$ in a neighbothood of $0$.
answered Dec 23 '18 at 0:13
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
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add a comment |
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See my comment to see that your example is not a counterexample. What is asserted is simply that $a<b<c$, $Lambda (a) <infty,Lambda (b) <infty $ imply $Lambda (c) <infty$ which follows from the fact that $e^{cX} <e^{aX}+e^{bX}$. No necessary and sufficient conditions for finiteness of $Lambda$ in a neighbourhood of $0$ are available. If $X$ is a bounded random variable then $Lambda (t)$ is finite in for all $t$. If $P{X>s} leq a e^{-cs}$ for $s$ sufficently large, for some $a,c >0$ then $Lambda (t)<infty$ in a neighbothood of $0$.
answered Dec 23 '18 at 0:13
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
add a comment |
$begingroup$
See my comment to see that your example is not a counterexample. What is asserted is simply that $a<b<c$, $Lambda (a) <infty,Lambda (b) <infty $ imply $Lambda (c) <infty$ which follows from the fact that $e^{cX} <e^{aX}+e^{bX}$. No necessary and sufficient conditions for finiteness of $Lambda$ in a neighbourhood of $0$ are available. If $X$ is a bounded random variable then $Lambda (t)$ is finite in for all $t$. If $P{X>s} leq a e^{-cs}$ for $s$ sufficently large, for some $a,c >0$ then $Lambda (t)<infty$ in a neighbothood of $0$.
answered Dec 23 '18 at 0:13
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
add a comment |
$begingroup$
See my comment to see that your example is not a counterexample. What is asserted is simply that $a<b<c$, $Lambda (a) <infty,Lambda (b) <infty $ imply $Lambda (c) <infty$ which follows from the fact that $e^{cX} <e^{aX}+e^{bX}$. No necessary and sufficient conditions for finiteness of $Lambda$ in a neighbourhood of $0$ are available. If $X$ is a bounded random variable then $Lambda (t)$ is finite in for all $t$. If $P{X>s} leq a e^{-cs}$ for $s$ sufficently large, for some $a,c >0$ then $Lambda (t)<infty$ in a neighbothood of $0$.
answered Dec 23 '18 at 0:13
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
See my comment to see that your example is not a counterexample. What is asserted is simply that $a<b<c$, $Lambda (a) <infty,Lambda (b) <infty $ imply $Lambda (c) <infty$ which follows from the fact that $e^{cX} <e^{aX}+e^{bX}$. No necessary and sufficient conditions for finiteness of $Lambda$ in a neighbourhood of $0$ are available. If $X$ is a bounded random variable then $Lambda (t)$ is finite in for all $t$. If $P{X>s} leq a e^{-cs}$ for $s$ sufficently large, for some $a,c >0$ then $Lambda (t)<infty$ in a neighbothood of $0$.
answered Dec 23 '18 at 0:13
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
edited Dec 23 '18 at 0:45
answered Dec 23 '18 at 0:13
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
answered Dec 23 '18 at 0:13
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
answered Dec 23 '18 at 0:13
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
71.2k53170
add a comment |
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$begingroup$
For nonnegative random variables (or |X| in your case), the property you are looking for is often referred to as being sub-exponential. You can think of it is precisely the condition needed! But there are other equivalent forms, in terms of moments, tail probabilities, etc. See for example, Prop. 2.7.1 in this book
$endgroup$
– passerby51
Dec 22 '18 at 19:06
$begingroup$
You are interpreting the term 'interval' wrongly. ${0}$ is also an interval.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 0:07