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Necessary and sufficient condition for $P(xigeq b)=1$
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I'm working on the following question and I don't see how the claim of the problem is even possible
Let $xi$ be a non-negative random variable. Prove that for $b>0$, $$lim_tto +inftyfrac1tlog mathbb Ee^-txi = -b$$ iff $P(xigeq b) = 1$.
Thing is, I don't see how this statement even makes sense, because there is only one possible value of that limit if it's defined, but there can potentially be many possible values of $b$ such that $P(xigeq b) = 1$
Source: Fall 2015
probability probability-theory
edited Jul 24 at 4:13
Math1000
18.4k31444
asked Jul 24 at 3:42
iYOA
60049
 |Â
show 6 more comments
up vote
1
down vote
favorite
I'm working on the following question and I don't see how the claim of the problem is even possible
Let $xi$ be a non-negative random variable. Prove that for $b>0$, $$lim_tto +inftyfrac1tlog mathbb Ee^-txi = -b$$ iff $P(xigeq b) = 1$.
Thing is, I don't see how this statement even makes sense, because there is only one possible value of that limit if it's defined, but there can potentially be many possible values of $b$ such that $P(xigeq b) = 1$
Source: Fall 2015
probability probability-theory
edited Jul 24 at 4:13
Math1000
18.4k31444
asked Jul 24 at 3:42
iYOA
60049
2
You might simplify by calling the random variable $X$ and/or by putting the random variable somewhere in the expectation.
– Michael
Jul 24 at 4:01
@Michael I changed the $x$ to $xi$ since it is clearly a typo. Also, according to the source, there was a missing $log$ in the expression, so I added that too.
– Math1000
Jul 24 at 4:03
Can you do any part of this? For example if you assume $X geq b$ then $e^-tX leq e^-tb$ (assuming $t>0$).
– Michael
Jul 24 at 4:17
@Michael I think it is safe to assume $t>0$ since we are considering the limit as $tto+infty$ here :)
– Math1000
Jul 24 at 4:18
1
Looks like $X=1/2$ (with prob 1) is a counter-example in one direction. Try $b=1/2$ and $b=1/4$.
– Michael
Jul 24 at 4:38
 |Â
show 6 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm working on the following question and I don't see how the claim of the problem is even possible
Let $xi$ be a non-negative random variable. Prove that for $b>0$, $$lim_tto +inftyfrac1tlog mathbb Ee^-txi = -b$$ iff $P(xigeq b) = 1$.
Thing is, I don't see how this statement even makes sense, because there is only one possible value of that limit if it's defined, but there can potentially be many possible values of $b$ such that $P(xigeq b) = 1$
Source: Fall 2015
probability probability-theory
edited Jul 24 at 4:13
Math1000
18.4k31444
asked Jul 24 at 3:42
iYOA
60049
I'm working on the following question and I don't see how the claim of the problem is even possible
Let $xi$ be a non-negative random variable. Prove that for $b>0$, $$lim_tto +inftyfrac1tlog mathbb Ee^-txi = -b$$ iff $P(xigeq b) = 1$.
Thing is, I don't see how this statement even makes sense, because there is only one possible value of that limit if it's defined, but there can potentially be many possible values of $b$ such that $P(xigeq b) = 1$
Source: Fall 2015
probability probability-theory
edited Jul 24 at 4:13
Math1000
18.4k31444
asked Jul 24 at 3:42
iYOA
60049
edited Jul 24 at 4:13
Math1000
18.4k31444
edited Jul 24 at 4:13
Math1000
18.4k31444
edited Jul 24 at 4:13
Math1000
18.4k31444
18.4k31444
asked Jul 24 at 3:42
iYOA
60049
asked Jul 24 at 3:42
iYOA
60049
asked Jul 24 at 3:42
iYOA
60049
60049
2
You might simplify by calling the random variable $X$ and/or by putting the random variable somewhere in the expectation.
– Michael
Jul 24 at 4:01
@Michael I changed the $x$ to $xi$ since it is clearly a typo. Also, according to the source, there was a missing $log$ in the expression, so I added that too.
– Math1000
Jul 24 at 4:03
Can you do any part of this? For example if you assume $X geq b$ then $e^-tX leq e^-tb$ (assuming $t>0$).
– Michael
Jul 24 at 4:17
@Michael I think it is safe to assume $t>0$ since we are considering the limit as $tto+infty$ here :)
– Math1000
Jul 24 at 4:18
1
Looks like $X=1/2$ (with prob 1) is a counter-example in one direction. Try $b=1/2$ and $b=1/4$.
– Michael
Jul 24 at 4:38
 |Â
show 6 more comments
2
You might simplify by calling the random variable $X$ and/or by putting the random variable somewhere in the expectation.
– Michael
Jul 24 at 4:01
@Michael I changed the $x$ to $xi$ since it is clearly a typo. Also, according to the source, there was a missing $log$ in the expression, so I added that too.
– Math1000
Jul 24 at 4:03
Can you do any part of this? For example if you assume $X geq b$ then $e^-tX leq e^-tb$ (assuming $t>0$).
– Michael
Jul 24 at 4:17
@Michael I think it is safe to assume $t>0$ since we are considering the limit as $tto+infty$ here :)
– Math1000
Jul 24 at 4:18
1
Looks like $X=1/2$ (with prob 1) is a counter-example in one direction. Try $b=1/2$ and $b=1/4$.
– Michael
Jul 24 at 4:38
2
2
You might simplify by calling the random variable $X$ and/or by putting the random variable somewhere in the expectation.
– Michael
Jul 24 at 4:01
You might simplify by calling the random variable $X$ and/or by putting the random variable somewhere in the expectation.
– Michael
Jul 24 at 4:01
@Michael I changed the $x$ to $xi$ since it is clearly a typo. Also, according to the source, there was a missing $log$ in the expression, so I added that too.
– Math1000
Jul 24 at 4:03
@Michael I changed the $x$ to $xi$ since it is clearly a typo. Also, according to the source, there was a missing $log$ in the expression, so I added that too.
– Math1000
Jul 24 at 4:03
Can you do any part of this? For example if you assume $X geq b$ then $e^-tX leq e^-tb$ (assuming $t>0$).
– Michael
Jul 24 at 4:17
Can you do any part of this? For example if you assume $X geq b$ then $e^-tX leq e^-tb$ (assuming $t>0$).
– Michael
Jul 24 at 4:17
@Michael I think it is safe to assume $t>0$ since we are considering the limit as $tto+infty$ here :)
– Math1000
Jul 24 at 4:18
@Michael I think it is safe to assume $t>0$ since we are considering the limit as $tto+infty$ here :)
– Math1000
Jul 24 at 4:18
1
1
Looks like $X=1/2$ (with prob 1) is a counter-example in one direction. Try $b=1/2$ and $b=1/4$.
– Michael
Jul 24 at 4:38
Looks like $X=1/2$ (with prob 1) is a counter-example in one direction. Try $b=1/2$ and $b=1/4$.
– Michael
Jul 24 at 4:38
 |Â
show 6 more comments
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Here is a fix: Let $X$ be any random variable such that $E[e^-tX]$ is finite for all $t >0$. Prove:
1) If $b in mathbbR$ and $P[X<b]=0$ then
$$ limsup_trightarrowinfty frac1tlog(E[e^-tX]) leq -b $$
2) If $b in mathbbR$ and $P[X<b]>0$ then
$$ liminf_trightarrowinfty frac1tlog(E[e^-tX]) > -b $$
3) Conclude that $lim_trightarrowinfty frac1tlog(E[e^-tX])$ exists (possibly being $infty$) and is equal to $-b^*$, where
$$ b^* = inf b in mathbbR : P[X<b]>0 $$
4) Conclude that $lim_trightarrowinfty frac1tlog(E[e^-tX]) leq -b$ if and only if $P[Xgeq b] =1$.
answered Jul 24 at 15:11
Michael
12.1k11325
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Here is a fix: Let $X$ be any random variable such that $E[e^-tX]$ is finite for all $t >0$. Prove:
1) If $b in mathbbR$ and $P[X<b]=0$ then
$$ limsup_trightarrowinfty frac1tlog(E[e^-tX]) leq -b $$
2) If $b in mathbbR$ and $P[X<b]>0$ then
$$ liminf_trightarrowinfty frac1tlog(E[e^-tX]) > -b $$
3) Conclude that $lim_trightarrowinfty frac1tlog(E[e^-tX])$ exists (possibly being $infty$) and is equal to $-b^*$, where
$$ b^* = inf b in mathbbR : P[X<b]>0 $$
4) Conclude that $lim_trightarrowinfty frac1tlog(E[e^-tX]) leq -b$ if and only if $P[Xgeq b] =1$.
answered Jul 24 at 15:11
Michael
12.1k11325
add a comment |Â
up vote
2
down vote
accepted
Here is a fix: Let $X$ be any random variable such that $E[e^-tX]$ is finite for all $t >0$. Prove:
1) If $b in mathbbR$ and $P[X<b]=0$ then
$$ limsup_trightarrowinfty frac1tlog(E[e^-tX]) leq -b $$
2) If $b in mathbbR$ and $P[X<b]>0$ then
$$ liminf_trightarrowinfty frac1tlog(E[e^-tX]) > -b $$
3) Conclude that $lim_trightarrowinfty frac1tlog(E[e^-tX])$ exists (possibly being $infty$) and is equal to $-b^*$, where
$$ b^* = inf b in mathbbR : P[X<b]>0 $$
4) Conclude that $lim_trightarrowinfty frac1tlog(E[e^-tX]) leq -b$ if and only if $P[Xgeq b] =1$.
answered Jul 24 at 15:11
Michael
12.1k11325
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Here is a fix: Let $X$ be any random variable such that $E[e^-tX]$ is finite for all $t >0$. Prove:
1) If $b in mathbbR$ and $P[X<b]=0$ then
$$ limsup_trightarrowinfty frac1tlog(E[e^-tX]) leq -b $$
2) If $b in mathbbR$ and $P[X<b]>0$ then
$$ liminf_trightarrowinfty frac1tlog(E[e^-tX]) > -b $$
3) Conclude that $lim_trightarrowinfty frac1tlog(E[e^-tX])$ exists (possibly being $infty$) and is equal to $-b^*$, where
$$ b^* = inf b in mathbbR : P[X<b]>0 $$
4) Conclude that $lim_trightarrowinfty frac1tlog(E[e^-tX]) leq -b$ if and only if $P[Xgeq b] =1$.
answered Jul 24 at 15:11
Michael
12.1k11325
Here is a fix: Let $X$ be any random variable such that $E[e^-tX]$ is finite for all $t >0$. Prove:
1) If $b in mathbbR$ and $P[X<b]=0$ then
$$ limsup_trightarrowinfty frac1tlog(E[e^-tX]) leq -b $$
2) If $b in mathbbR$ and $P[X<b]>0$ then
$$ liminf_trightarrowinfty frac1tlog(E[e^-tX]) > -b $$
3) Conclude that $lim_trightarrowinfty frac1tlog(E[e^-tX])$ exists (possibly being $infty$) and is equal to $-b^*$, where
$$ b^* = inf b in mathbbR : P[X<b]>0 $$
4) Conclude that $lim_trightarrowinfty frac1tlog(E[e^-tX]) leq -b$ if and only if $P[Xgeq b] =1$.
answered Jul 24 at 15:11
Michael
12.1k11325
edited Jul 24 at 15:26
answered Jul 24 at 15:11
Michael
12.1k11325
answered Jul 24 at 15:11
Michael
12.1k11325
answered Jul 24 at 15:11
Michael
12.1k11325
12.1k11325
add a comment |Â
add a comment |Â
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2
You might simplify by calling the random variable $X$ and/or by putting the random variable somewhere in the expectation.
– Michael
Jul 24 at 4:01
@Michael I changed the $x$ to $xi$ since it is clearly a typo. Also, according to the source, there was a missing $log$ in the expression, so I added that too.
– Math1000
Jul 24 at 4:03
Can you do any part of this? For example if you assume $X geq b$ then $e^-tX leq e^-tb$ (assuming $t>0$).
– Michael
Jul 24 at 4:17
@Michael I think it is safe to assume $t>0$ since we are considering the limit as $tto+infty$ here :)
– Math1000
Jul 24 at 4:18
1
Looks like $X=1/2$ (with prob 1) is a counter-example in one direction. Try $b=1/2$ and $b=1/4$.
– Michael
Jul 24 at 4:38