Expectation of Reciprocal of Sum of i.i.ds












-3












$begingroup$


$X_i$s are real continous i.i.ds. I am trying to prove the conjecture



$$
E[frac{1}{sum_{i=1}^nX_i}] = E[frac{1}{nX_i}]
$$



I can't seem to be getting anywhere. Prove or disprove.






probability expected-value







share|cite|improve this question











$endgroup$












  • $begingroup$
    You can't seem to be getting anywhere, because this is false.
    $endgroup$
    – zhoraster
    Jan 17 at 20:04












  • $begingroup$
    @adityadua write that down as an answer and I'll accept it.
    $endgroup$
    – user2268997
    Jan 17 at 20:16
















-3












$begingroup$


$X_i$s are real continous i.i.ds. I am trying to prove the conjecture



$$
E[frac{1}{sum_{i=1}^nX_i}] = E[frac{1}{nX_i}]
$$



I can't seem to be getting anywhere. Prove or disprove.






probability expected-value







share|cite|improve this question











$endgroup$












  • $begingroup$
    You can't seem to be getting anywhere, because this is false.
    $endgroup$
    – zhoraster
    Jan 17 at 20:04












  • $begingroup$
    @adityadua write that down as an answer and I'll accept it.
    $endgroup$
    – user2268997
    Jan 17 at 20:16














-3












-3








-3





$begingroup$


$X_i$s are real continous i.i.ds. I am trying to prove the conjecture



$$
E[frac{1}{sum_{i=1}^nX_i}] = E[frac{1}{nX_i}]
$$



I can't seem to be getting anywhere. Prove or disprove.






probability expected-value







share|cite|improve this question











$endgroup$




$X_i$s are real continous i.i.ds. I am trying to prove the conjecture



$$
E[frac{1}{sum_{i=1}^nX_i}] = E[frac{1}{nX_i}]
$$



I can't seem to be getting anywhere. Prove or disprove.






probability expected-value




probability expected-value






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 20:16







user2268997

















asked Jan 17 at 19:53









user2268997user2268997

973




973












  • $begingroup$
    You can't seem to be getting anywhere, because this is false.
    $endgroup$
    – zhoraster
    Jan 17 at 20:04












  • $begingroup$
    @adityadua write that down as an answer and I'll accept it.
    $endgroup$
    – user2268997
    Jan 17 at 20:16


















  • $begingroup$
    You can't seem to be getting anywhere, because this is false.
    $endgroup$
    – zhoraster
    Jan 17 at 20:04












  • $begingroup$
    @adityadua write that down as an answer and I'll accept it.
    $endgroup$
    – user2268997
    Jan 17 at 20:16
















$begingroup$
You can't seem to be getting anywhere, because this is false.
$endgroup$
– zhoraster
Jan 17 at 20:04






$begingroup$
You can't seem to be getting anywhere, because this is false.
$endgroup$
– zhoraster
Jan 17 at 20:04














$begingroup$
@adityadua write that down as an answer and I'll accept it.
$endgroup$
– user2268997
Jan 17 at 20:16




$begingroup$
@adityadua write that down as an answer and I'll accept it.
$endgroup$
– user2268997
Jan 17 at 20:16










1 Answer
1






active

oldest

votes


















2












$begingroup$

Arithmetic mean is greater than or equal to the harmonic mean.



$${1 over n}sum_{i=1}^n X_i geq {n over sum_{i=1}^n {1 over X_i}}$$
$$sum_{i=1}^n {1 over X_i} geq {n^2 over sum_{i=1}^n X_i}$$
$$n mathbb{E} left[ {1 over X_1} right] geq n^2 mathbb{E}left[ {1 over sum_{i=1}^n X_i} right] $$
$$mathbb{E}left[ {1 over sum_{i=1}^n X_i} right] leq {1 over n}mathbb{E} left[ {1 over X_1} right]$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is not the RHS...
    $endgroup$
    – user2268997
    Jan 18 at 1:33










  • $begingroup$
    @user2268997 Better?
    $endgroup$
    – Aditya Dua
    Jan 18 at 4:18










  • $begingroup$
    Sure. All you need now is to (formally) rule-out the equal in the less-than-or-equal.
    $endgroup$
    – user2268997
    Jan 18 at 17:53










  • $begingroup$
    There is at least one degenrate case ($X_i = const$) where the equality holds. Not sure how to prove that it is the only case.
    $endgroup$
    – Aditya Dua
    Jan 18 at 19:58










  • $begingroup$
    I know. An alternative way would be to plug in some distribution for Xi and show that the equality does not hold for it.
    $endgroup$
    – user2268997
    Jan 19 at 0:12












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Arithmetic mean is greater than or equal to the harmonic mean.



$${1 over n}sum_{i=1}^n X_i geq {n over sum_{i=1}^n {1 over X_i}}$$
$$sum_{i=1}^n {1 over X_i} geq {n^2 over sum_{i=1}^n X_i}$$
$$n mathbb{E} left[ {1 over X_1} right] geq n^2 mathbb{E}left[ {1 over sum_{i=1}^n X_i} right] $$
$$mathbb{E}left[ {1 over sum_{i=1}^n X_i} right] leq {1 over n}mathbb{E} left[ {1 over X_1} right]$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is not the RHS...
    $endgroup$
    – user2268997
    Jan 18 at 1:33










  • $begingroup$
    @user2268997 Better?
    $endgroup$
    – Aditya Dua
    Jan 18 at 4:18










  • $begingroup$
    Sure. All you need now is to (formally) rule-out the equal in the less-than-or-equal.
    $endgroup$
    – user2268997
    Jan 18 at 17:53










  • $begingroup$
    There is at least one degenrate case ($X_i = const$) where the equality holds. Not sure how to prove that it is the only case.
    $endgroup$
    – Aditya Dua
    Jan 18 at 19:58










  • $begingroup$
    I know. An alternative way would be to plug in some distribution for Xi and show that the equality does not hold for it.
    $endgroup$
    – user2268997
    Jan 19 at 0:12
















2












$begingroup$

Arithmetic mean is greater than or equal to the harmonic mean.



$${1 over n}sum_{i=1}^n X_i geq {n over sum_{i=1}^n {1 over X_i}}$$
$$sum_{i=1}^n {1 over X_i} geq {n^2 over sum_{i=1}^n X_i}$$
$$n mathbb{E} left[ {1 over X_1} right] geq n^2 mathbb{E}left[ {1 over sum_{i=1}^n X_i} right] $$
$$mathbb{E}left[ {1 over sum_{i=1}^n X_i} right] leq {1 over n}mathbb{E} left[ {1 over X_1} right]$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is not the RHS...
    $endgroup$
    – user2268997
    Jan 18 at 1:33










  • $begingroup$
    @user2268997 Better?
    $endgroup$
    – Aditya Dua
    Jan 18 at 4:18










  • $begingroup$
    Sure. All you need now is to (formally) rule-out the equal in the less-than-or-equal.
    $endgroup$
    – user2268997
    Jan 18 at 17:53










  • $begingroup$
    There is at least one degenrate case ($X_i = const$) where the equality holds. Not sure how to prove that it is the only case.
    $endgroup$
    – Aditya Dua
    Jan 18 at 19:58










  • $begingroup$
    I know. An alternative way would be to plug in some distribution for Xi and show that the equality does not hold for it.
    $endgroup$
    – user2268997
    Jan 19 at 0:12














2












2








2





$begingroup$

Arithmetic mean is greater than or equal to the harmonic mean.



$${1 over n}sum_{i=1}^n X_i geq {n over sum_{i=1}^n {1 over X_i}}$$
$$sum_{i=1}^n {1 over X_i} geq {n^2 over sum_{i=1}^n X_i}$$
$$n mathbb{E} left[ {1 over X_1} right] geq n^2 mathbb{E}left[ {1 over sum_{i=1}^n X_i} right] $$
$$mathbb{E}left[ {1 over sum_{i=1}^n X_i} right] leq {1 over n}mathbb{E} left[ {1 over X_1} right]$$






share|cite|improve this answer











$endgroup$



Arithmetic mean is greater than or equal to the harmonic mean.



$${1 over n}sum_{i=1}^n X_i geq {n over sum_{i=1}^n {1 over X_i}}$$
$$sum_{i=1}^n {1 over X_i} geq {n^2 over sum_{i=1}^n X_i}$$
$$n mathbb{E} left[ {1 over X_1} right] geq n^2 mathbb{E}left[ {1 over sum_{i=1}^n X_i} right] $$
$$mathbb{E}left[ {1 over sum_{i=1}^n X_i} right] leq {1 over n}mathbb{E} left[ {1 over X_1} right]$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 18 at 4:17

























answered Jan 17 at 20:54









Aditya DuaAditya Dua

1,15418




1,15418












  • $begingroup$
    This is not the RHS...
    $endgroup$
    – user2268997
    Jan 18 at 1:33










  • $begingroup$
    @user2268997 Better?
    $endgroup$
    – Aditya Dua
    Jan 18 at 4:18










  • $begingroup$
    Sure. All you need now is to (formally) rule-out the equal in the less-than-or-equal.
    $endgroup$
    – user2268997
    Jan 18 at 17:53










  • $begingroup$
    There is at least one degenrate case ($X_i = const$) where the equality holds. Not sure how to prove that it is the only case.
    $endgroup$
    – Aditya Dua
    Jan 18 at 19:58










  • $begingroup$
    I know. An alternative way would be to plug in some distribution for Xi and show that the equality does not hold for it.
    $endgroup$
    – user2268997
    Jan 19 at 0:12


















  • $begingroup$
    This is not the RHS...
    $endgroup$
    – user2268997
    Jan 18 at 1:33










  • $begingroup$
    @user2268997 Better?
    $endgroup$
    – Aditya Dua
    Jan 18 at 4:18










  • $begingroup$
    Sure. All you need now is to (formally) rule-out the equal in the less-than-or-equal.
    $endgroup$
    – user2268997
    Jan 18 at 17:53










  • $begingroup$
    There is at least one degenrate case ($X_i = const$) where the equality holds. Not sure how to prove that it is the only case.
    $endgroup$
    – Aditya Dua
    Jan 18 at 19:58










  • $begingroup$
    I know. An alternative way would be to plug in some distribution for Xi and show that the equality does not hold for it.
    $endgroup$
    – user2268997
    Jan 19 at 0:12
















$begingroup$
This is not the RHS...
$endgroup$
– user2268997
Jan 18 at 1:33




$begingroup$
This is not the RHS...
$endgroup$
– user2268997
Jan 18 at 1:33












$begingroup$
@user2268997 Better?
$endgroup$
– Aditya Dua
Jan 18 at 4:18




$begingroup$
@user2268997 Better?
$endgroup$
– Aditya Dua
Jan 18 at 4:18












$begingroup$
Sure. All you need now is to (formally) rule-out the equal in the less-than-or-equal.
$endgroup$
– user2268997
Jan 18 at 17:53




$begingroup$
Sure. All you need now is to (formally) rule-out the equal in the less-than-or-equal.
$endgroup$
– user2268997
Jan 18 at 17:53












$begingroup$
There is at least one degenrate case ($X_i = const$) where the equality holds. Not sure how to prove that it is the only case.
$endgroup$
– Aditya Dua
Jan 18 at 19:58




$begingroup$
There is at least one degenrate case ($X_i = const$) where the equality holds. Not sure how to prove that it is the only case.
$endgroup$
– Aditya Dua
Jan 18 at 19:58












$begingroup$
I know. An alternative way would be to plug in some distribution for Xi and show that the equality does not hold for it.
$endgroup$
– user2268997
Jan 19 at 0:12




$begingroup$
I know. An alternative way would be to plug in some distribution for Xi and show that the equality does not hold for it.
$endgroup$
– user2268997
Jan 19 at 0:12


















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