Mixing
One particular probabilistic inequality
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I've been studying for my exam in Probability theory and I found this exercise:
prove that for every $lambda$ greater than 0 this inequality holds:
$$int_0^+inftylog(x+1)lambda e^-lambda xdxgeq frac1elogleft(fraclambda +1lambdaright)$$
The left side is obviously equal to $mathbbE[log(X+1)]$, where $Xsim textExp(lambda)$, but I don't see what right side might be. Can anyone provide me some help?
probability inequality expectation
edited Jul 17 at 17:21
Shashi
6,1261523
asked Jul 17 at 14:39
Martin
354210
 |Â
show 2 more comments
up vote
1
down vote
favorite
I've been studying for my exam in Probability theory and I found this exercise:
prove that for every $lambda$ greater than 0 this inequality holds:
$$int_0^+inftylog(x+1)lambda e^-lambda xdxgeq frac1elogleft(fraclambda +1lambdaright)$$
The left side is obviously equal to $mathbbE[log(X+1)]$, where $Xsim textExp(lambda)$, but I don't see what right side might be. Can anyone provide me some help?
probability inequality expectation
edited Jul 17 at 17:21
Shashi
6,1261523
asked Jul 17 at 14:39
Martin
354210
Are you looking for some special meaning for this quantity ?
– ippiki-ookami
Jul 17 at 14:45
That is correct
– Martin
Jul 17 at 14:48
Is that the end of the exercise or are there follow up questions to that one ?
– ippiki-ookami
Jul 17 at 14:49
The whole exercise is to prove the inequality
– Martin
Jul 17 at 14:50
2
Please choose a more expressive title. The current title says basically nothing.
– M. Winter
Jul 17 at 16:37
 |Â
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I've been studying for my exam in Probability theory and I found this exercise:
prove that for every $lambda$ greater than 0 this inequality holds:
$$int_0^+inftylog(x+1)lambda e^-lambda xdxgeq frac1elogleft(fraclambda +1lambdaright)$$
The left side is obviously equal to $mathbbE[log(X+1)]$, where $Xsim textExp(lambda)$, but I don't see what right side might be. Can anyone provide me some help?
probability inequality expectation
edited Jul 17 at 17:21
Shashi
6,1261523
asked Jul 17 at 14:39
Martin
354210
I've been studying for my exam in Probability theory and I found this exercise:
prove that for every $lambda$ greater than 0 this inequality holds:
$$int_0^+inftylog(x+1)lambda e^-lambda xdxgeq frac1elogleft(fraclambda +1lambdaright)$$
The left side is obviously equal to $mathbbE[log(X+1)]$, where $Xsim textExp(lambda)$, but I don't see what right side might be. Can anyone provide me some help?
probability inequality expectation
edited Jul 17 at 17:21
Shashi
6,1261523
asked Jul 17 at 14:39
Martin
354210
edited Jul 17 at 17:21
Shashi
6,1261523
edited Jul 17 at 17:21
Shashi
6,1261523
edited Jul 17 at 17:21
Shashi
6,1261523
6,1261523
asked Jul 17 at 14:39
Martin
354210
asked Jul 17 at 14:39
Martin
354210
asked Jul 17 at 14:39
Martin
354210
354210
Are you looking for some special meaning for this quantity ?
– ippiki-ookami
Jul 17 at 14:45
That is correct
– Martin
Jul 17 at 14:48
Is that the end of the exercise or are there follow up questions to that one ?
– ippiki-ookami
Jul 17 at 14:49
The whole exercise is to prove the inequality
– Martin
Jul 17 at 14:50
2
Please choose a more expressive title. The current title says basically nothing.
– M. Winter
Jul 17 at 16:37
 |Â
show 2 more comments
Are you looking for some special meaning for this quantity ?
– ippiki-ookami
Jul 17 at 14:45
That is correct
– Martin
Jul 17 at 14:48
Is that the end of the exercise or are there follow up questions to that one ?
– ippiki-ookami
Jul 17 at 14:49
The whole exercise is to prove the inequality
– Martin
Jul 17 at 14:50
2
Please choose a more expressive title. The current title says basically nothing.
– M. Winter
Jul 17 at 16:37
Are you looking for some special meaning for this quantity ?
– ippiki-ookami
Jul 17 at 14:45
Are you looking for some special meaning for this quantity ?
– ippiki-ookami
Jul 17 at 14:45
That is correct
– Martin
Jul 17 at 14:48
That is correct
– Martin
Jul 17 at 14:48
Is that the end of the exercise or are there follow up questions to that one ?
– ippiki-ookami
Jul 17 at 14:49
Is that the end of the exercise or are there follow up questions to that one ?
– ippiki-ookami
Jul 17 at 14:49
The whole exercise is to prove the inequality
– Martin
Jul 17 at 14:50
The whole exercise is to prove the inequality
– Martin
Jul 17 at 14:50
2
2
Please choose a more expressive title. The current title says basically nothing.
– M. Winter
Jul 17 at 16:37
Please choose a more expressive title. The current title says basically nothing.
– M. Winter
Jul 17 at 16:37
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
$newcommandEmathbb EnewcommandPMmathbb P$Let $Xsim textExp(lambda)$ then your problem is equivalent with proving the following:
beginalign
E[log(X+1)]geq PM(X>E[X]) logE[X+1]
endalign
We know that
beginalign
PM(log(X+1)geq logE[X+1] ) leq fracE[log(X+1)]logE[X+1]
endalign
By Markov's inequality, hence:
beginalign
E[log(X+1)]geq PM(log(X+1)geq logE[X+1] )logE[X+1]
endalign
But we also know: $$PM(log(X+1)geq logE[X+1])=PM(X+1>E[X+1]) =PM(X>E[X])$$ (btw this manipulations is exactly what lead me to this answer) and this proves the inequality.
The only thing that we did is noticing that $e^-1=PM(X>E[X])$ and Markov's inequality.
answered Jul 17 at 17:10
Shashi
6,1261523
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$newcommandEmathbb EnewcommandPMmathbb P$Let $Xsim textExp(lambda)$ then your problem is equivalent with proving the following:
beginalign
E[log(X+1)]geq PM(X>E[X]) logE[X+1]
endalign
We know that
beginalign
PM(log(X+1)geq logE[X+1] ) leq fracE[log(X+1)]logE[X+1]
endalign
By Markov's inequality, hence:
beginalign
E[log(X+1)]geq PM(log(X+1)geq logE[X+1] )logE[X+1]
endalign
But we also know: $$PM(log(X+1)geq logE[X+1])=PM(X+1>E[X+1]) =PM(X>E[X])$$ (btw this manipulations is exactly what lead me to this answer) and this proves the inequality.
The only thing that we did is noticing that $e^-1=PM(X>E[X])$ and Markov's inequality.
answered Jul 17 at 17:10
Shashi
6,1261523
add a comment |Â
up vote
3
down vote
accepted
$newcommandEmathbb EnewcommandPMmathbb P$Let $Xsim textExp(lambda)$ then your problem is equivalent with proving the following:
beginalign
E[log(X+1)]geq PM(X>E[X]) logE[X+1]
endalign
We know that
beginalign
PM(log(X+1)geq logE[X+1] ) leq fracE[log(X+1)]logE[X+1]
endalign
By Markov's inequality, hence:
beginalign
E[log(X+1)]geq PM(log(X+1)geq logE[X+1] )logE[X+1]
endalign
But we also know: $$PM(log(X+1)geq logE[X+1])=PM(X+1>E[X+1]) =PM(X>E[X])$$ (btw this manipulations is exactly what lead me to this answer) and this proves the inequality.
The only thing that we did is noticing that $e^-1=PM(X>E[X])$ and Markov's inequality.
answered Jul 17 at 17:10
Shashi
6,1261523
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$newcommandEmathbb EnewcommandPMmathbb P$Let $Xsim textExp(lambda)$ then your problem is equivalent with proving the following:
beginalign
E[log(X+1)]geq PM(X>E[X]) logE[X+1]
endalign
We know that
beginalign
PM(log(X+1)geq logE[X+1] ) leq fracE[log(X+1)]logE[X+1]
endalign
By Markov's inequality, hence:
beginalign
E[log(X+1)]geq PM(log(X+1)geq logE[X+1] )logE[X+1]
endalign
But we also know: $$PM(log(X+1)geq logE[X+1])=PM(X+1>E[X+1]) =PM(X>E[X])$$ (btw this manipulations is exactly what lead me to this answer) and this proves the inequality.
The only thing that we did is noticing that $e^-1=PM(X>E[X])$ and Markov's inequality.
answered Jul 17 at 17:10
Shashi
6,1261523
$newcommandEmathbb EnewcommandPMmathbb P$Let $Xsim textExp(lambda)$ then your problem is equivalent with proving the following:
beginalign
E[log(X+1)]geq PM(X>E[X]) logE[X+1]
endalign
We know that
beginalign
PM(log(X+1)geq logE[X+1] ) leq fracE[log(X+1)]logE[X+1]
endalign
By Markov's inequality, hence:
beginalign
E[log(X+1)]geq PM(log(X+1)geq logE[X+1] )logE[X+1]
endalign
But we also know: $$PM(log(X+1)geq logE[X+1])=PM(X+1>E[X+1]) =PM(X>E[X])$$ (btw this manipulations is exactly what lead me to this answer) and this proves the inequality.
The only thing that we did is noticing that $e^-1=PM(X>E[X])$ and Markov's inequality.
answered Jul 17 at 17:10
Shashi
6,1261523
edited Jul 17 at 17:20
answered Jul 17 at 17:10
Shashi
6,1261523
answered Jul 17 at 17:10
Shashi
6,1261523
answered Jul 17 at 17:10
Shashi
6,1261523
6,1261523
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2854563%2fone-particular-probabilistic-inequality%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Are you looking for some special meaning for this quantity ?
– ippiki-ookami
Jul 17 at 14:45
That is correct
– Martin
Jul 17 at 14:48
Is that the end of the exercise or are there follow up questions to that one ?
– ippiki-ookami
Jul 17 at 14:49
The whole exercise is to prove the inequality
– Martin
Jul 17 at 14:50
2
Please choose a more expressive title. The current title says basically nothing.
– M. Winter
Jul 17 at 16:37