Expectation of inverse of sum of positive iid variables












4












$begingroup$


Let $(X_i)_{i}$ be a sequence of iid positive variables of mean 1 and variance $sigma^2$. Let $bar{X}_n = frac{sum_{i=1}^n X_i}{n}$.



My question is: Can we can bound $mathbb{E}(1/bar{X}_n)$ as a function of $sigma$ and $n$?



There seems to be some strategy that may work based on the taylor extension, but




  • I'm not sure about the hypothesis that need to be met;

  • if it works in this case; and

  • if we can say something definite on $bar{X}_n$ or if we need to use the central limit theorem and can only say this for the normal approximation?


More details about the Taylor expansion. According to this wikipedia article,
$$mathbb{E}(f(X)) approx f(mu_X) +frac{f''(mu_X)}{2}sigma_X^2$$



So in my case it would give something like:
$$mathbb{E}(1/bar{X}_n) approx 1 +frac{sigma^2}{4 n}$$
I'm trying to find maybe a formal proof of a similar result, or hypothesis so that it works. Maybe references?
Thanks



EDIT: if needed, we can consider that the $(X_i)_i$ are discrete, there exists $v_1<cdots<v_K$ such that $mathbb{P}(X=v_k)=p_k$ and $sum p_k = 1$. In this case we know that $bar{X}_n geq v_1$. Although I believe something can be said in the general case.



PS: this is almost a cross-post of this on Math.SE.










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  • $begingroup$
    Hello Gopi, thank you for your question. If this is ALMOST a cross-post your question is welcomed here, but if it is a cross-post you should rather put a bounty on the post in Math.SE
    $endgroup$
    – Ferdi
    yesterday










  • $begingroup$
    It's almost, the question is different: bounds on the expectation vs convergence speed. If I get the answer on MSE it probably won't help me with this question, but I've referenced it because it's likely that someone that has the answer to one has the answer to both.
    $endgroup$
    – Gopi
    yesterday










  • $begingroup$
    Perhaps Markov's inequality or Chebyshev's inequality are useful.
    $endgroup$
    – Ertxiem
    yesterday
















4












$begingroup$


Let $(X_i)_{i}$ be a sequence of iid positive variables of mean 1 and variance $sigma^2$. Let $bar{X}_n = frac{sum_{i=1}^n X_i}{n}$.



My question is: Can we can bound $mathbb{E}(1/bar{X}_n)$ as a function of $sigma$ and $n$?



There seems to be some strategy that may work based on the taylor extension, but




  • I'm not sure about the hypothesis that need to be met;

  • if it works in this case; and

  • if we can say something definite on $bar{X}_n$ or if we need to use the central limit theorem and can only say this for the normal approximation?


More details about the Taylor expansion. According to this wikipedia article,
$$mathbb{E}(f(X)) approx f(mu_X) +frac{f''(mu_X)}{2}sigma_X^2$$



So in my case it would give something like:
$$mathbb{E}(1/bar{X}_n) approx 1 +frac{sigma^2}{4 n}$$
I'm trying to find maybe a formal proof of a similar result, or hypothesis so that it works. Maybe references?
Thanks



EDIT: if needed, we can consider that the $(X_i)_i$ are discrete, there exists $v_1<cdots<v_K$ such that $mathbb{P}(X=v_k)=p_k$ and $sum p_k = 1$. In this case we know that $bar{X}_n geq v_1$. Although I believe something can be said in the general case.



PS: this is almost a cross-post of this on Math.SE.










share|cite|improve this question









New contributor




Gopi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Hello Gopi, thank you for your question. If this is ALMOST a cross-post your question is welcomed here, but if it is a cross-post you should rather put a bounty on the post in Math.SE
    $endgroup$
    – Ferdi
    yesterday










  • $begingroup$
    It's almost, the question is different: bounds on the expectation vs convergence speed. If I get the answer on MSE it probably won't help me with this question, but I've referenced it because it's likely that someone that has the answer to one has the answer to both.
    $endgroup$
    – Gopi
    yesterday










  • $begingroup$
    Perhaps Markov's inequality or Chebyshev's inequality are useful.
    $endgroup$
    – Ertxiem
    yesterday














4












4








4


1



$begingroup$


Let $(X_i)_{i}$ be a sequence of iid positive variables of mean 1 and variance $sigma^2$. Let $bar{X}_n = frac{sum_{i=1}^n X_i}{n}$.



My question is: Can we can bound $mathbb{E}(1/bar{X}_n)$ as a function of $sigma$ and $n$?



There seems to be some strategy that may work based on the taylor extension, but




  • I'm not sure about the hypothesis that need to be met;

  • if it works in this case; and

  • if we can say something definite on $bar{X}_n$ or if we need to use the central limit theorem and can only say this for the normal approximation?


More details about the Taylor expansion. According to this wikipedia article,
$$mathbb{E}(f(X)) approx f(mu_X) +frac{f''(mu_X)}{2}sigma_X^2$$



So in my case it would give something like:
$$mathbb{E}(1/bar{X}_n) approx 1 +frac{sigma^2}{4 n}$$
I'm trying to find maybe a formal proof of a similar result, or hypothesis so that it works. Maybe references?
Thanks



EDIT: if needed, we can consider that the $(X_i)_i$ are discrete, there exists $v_1<cdots<v_K$ such that $mathbb{P}(X=v_k)=p_k$ and $sum p_k = 1$. In this case we know that $bar{X}_n geq v_1$. Although I believe something can be said in the general case.



PS: this is almost a cross-post of this on Math.SE.










share|cite|improve this question









New contributor




Gopi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Let $(X_i)_{i}$ be a sequence of iid positive variables of mean 1 and variance $sigma^2$. Let $bar{X}_n = frac{sum_{i=1}^n X_i}{n}$.



My question is: Can we can bound $mathbb{E}(1/bar{X}_n)$ as a function of $sigma$ and $n$?



There seems to be some strategy that may work based on the taylor extension, but




  • I'm not sure about the hypothesis that need to be met;

  • if it works in this case; and

  • if we can say something definite on $bar{X}_n$ or if we need to use the central limit theorem and can only say this for the normal approximation?


More details about the Taylor expansion. According to this wikipedia article,
$$mathbb{E}(f(X)) approx f(mu_X) +frac{f''(mu_X)}{2}sigma_X^2$$



So in my case it would give something like:
$$mathbb{E}(1/bar{X}_n) approx 1 +frac{sigma^2}{4 n}$$
I'm trying to find maybe a formal proof of a similar result, or hypothesis so that it works. Maybe references?
Thanks



EDIT: if needed, we can consider that the $(X_i)_i$ are discrete, there exists $v_1<cdots<v_K$ such that $mathbb{P}(X=v_k)=p_k$ and $sum p_k = 1$. In this case we know that $bar{X}_n geq v_1$. Although I believe something can be said in the general case.



PS: this is almost a cross-post of this on Math.SE.







variance expected-value iid






share|cite|improve this question









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Gopi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Gopi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited yesterday







Gopi













New contributor




Gopi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked yesterday









GopiGopi

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New contributor




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New contributor





Gopi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Gopi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    Hello Gopi, thank you for your question. If this is ALMOST a cross-post your question is welcomed here, but if it is a cross-post you should rather put a bounty on the post in Math.SE
    $endgroup$
    – Ferdi
    yesterday










  • $begingroup$
    It's almost, the question is different: bounds on the expectation vs convergence speed. If I get the answer on MSE it probably won't help me with this question, but I've referenced it because it's likely that someone that has the answer to one has the answer to both.
    $endgroup$
    – Gopi
    yesterday










  • $begingroup$
    Perhaps Markov's inequality or Chebyshev's inequality are useful.
    $endgroup$
    – Ertxiem
    yesterday


















  • $begingroup$
    Hello Gopi, thank you for your question. If this is ALMOST a cross-post your question is welcomed here, but if it is a cross-post you should rather put a bounty on the post in Math.SE
    $endgroup$
    – Ferdi
    yesterday










  • $begingroup$
    It's almost, the question is different: bounds on the expectation vs convergence speed. If I get the answer on MSE it probably won't help me with this question, but I've referenced it because it's likely that someone that has the answer to one has the answer to both.
    $endgroup$
    – Gopi
    yesterday










  • $begingroup$
    Perhaps Markov's inequality or Chebyshev's inequality are useful.
    $endgroup$
    – Ertxiem
    yesterday
















$begingroup$
Hello Gopi, thank you for your question. If this is ALMOST a cross-post your question is welcomed here, but if it is a cross-post you should rather put a bounty on the post in Math.SE
$endgroup$
– Ferdi
yesterday




$begingroup$
Hello Gopi, thank you for your question. If this is ALMOST a cross-post your question is welcomed here, but if it is a cross-post you should rather put a bounty on the post in Math.SE
$endgroup$
– Ferdi
yesterday












$begingroup$
It's almost, the question is different: bounds on the expectation vs convergence speed. If I get the answer on MSE it probably won't help me with this question, but I've referenced it because it's likely that someone that has the answer to one has the answer to both.
$endgroup$
– Gopi
yesterday




$begingroup$
It's almost, the question is different: bounds on the expectation vs convergence speed. If I get the answer on MSE it probably won't help me with this question, but I've referenced it because it's likely that someone that has the answer to one has the answer to both.
$endgroup$
– Gopi
yesterday












$begingroup$
Perhaps Markov's inequality or Chebyshev's inequality are useful.
$endgroup$
– Ertxiem
yesterday




$begingroup$
Perhaps Markov's inequality or Chebyshev's inequality are useful.
$endgroup$
– Ertxiem
yesterday










2 Answers
2






active

oldest

votes


















3












$begingroup$

You cannot bound that expectation in $sigma, n$. That's because there is the distinct possibility that the expectation do not exist at all (or, is $infty$.) See I've heard that ratios or inverses of random variables often are problematic, in not having expectations. Why is that?. If the conditions given there is fulfilled for the density of $X_1$, it will so be for the density of $bar{X}_n$. If densities do not exist, but probability mass functions do, it is simpler, since your assumptions prohibit a probability atom at zero, but a probability density can still be positive at zero even if $P(X >0)=1$.



For a useful bound you will at least need to restrict the common distribution of $X_1, dotsc, X_n$ much more.



EDIT


After your new information, and with $v_1>0$, the expectation of $1/bar{X}_n$ certainly will exist (irrespective if $K$ is finite or not.) And, since the function $xmapsto 1/x$ is convex for $x>0$, we can use the Jensen Inequality to conclude that $DeclareMathOperator{E}{mathbb{E}}E 1/bar{X}_n ge 1/E bar{X}_n$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But the thinig is that this is really not a general case but a very specific case: it's very unlikely that there is a probability mass near 0 (we can even evaluate it with Markov's inequality): $bar{X}_n$ is centered around 1 and has a variance of $sigma^2 / n$.
    $endgroup$
    – Gopi
    yesterday








  • 1




    $begingroup$
    But even a tiny (but positive) probability close to zero can lead to expectation of inverse being $infty$. Can you rule out that possibility?
    $endgroup$
    – kjetil b halvorsen
    yesterday










  • $begingroup$
    In my case I can indeed (the distribution of $X_i$ is discrete). I'll add this hypothesis to the question. I'm still interested from a theoretical perspective by the general case, I believe something can still be said in this case (or I'm interested by a counter example that would show your statement).
    $endgroup$
    – Gopi
    yesterday






  • 1




    $begingroup$
    Information about discreteness is really important! And, n what support? If positive integers, $P(X=0)=0$ and $mu=1$ is really restrictive ...
    $endgroup$
    – kjetil b halvorsen
    yesterday






  • 1




    $begingroup$
    Obviously the support is not positive integers ;). The support is the set of rational numbers.
    $endgroup$
    – Gopi
    yesterday



















1












$begingroup$

I think I have the gist of it.
Given that $f(x)=1/x$ is infinitely differentiable in 1. Taylor's theorem tells us:



There exists $varepsilon>0$ such that $f(x) = f(1) + f'(1) (x-1)+ frac{f''(1)(x-1)^2}{2} + frac{f'''(varepsilon) (x-1)^2}{2}$.



In our case, if $X_i$ belongs in the domaine $[v_1;+infty[$, then $bar{X}_n$ has the same domain and we have $varepsilon geq v_1$.



Hence
$mathbb{E}(1/bar{X}_n) = mathbb{E}left (1 - (bar{X}_n-1) + frac{(x-1)^2}{4}+ frac{f'''(varepsilon) (bar{X}_n-1)^2}{2} right)$, and
begin{align*}
mathbb{E}(1/bar{X}_n) &= 1 + frac{f'''(varepsilon) mathbb{E}left ((bar{X}_n-1)^2right )}{2} = 1 +frac{ V(bar{X}_n)}{4} - frac{ V(bar{X}_n)}{12 varepsilon^4}\
end{align*}

and hence
$$1 + frac{ sigma^2}{4 n^2}- frac{sigma^2}{12 v_1^4 n^2} leq mathbb{E}(1/bar{X}_n) leq 1 + frac{ sigma^2}{4 n^2}.$$



For the case where $X_i$ do not admit a minimum but has an unlimited number of moments, one can do a similar transformation using the full taylor expansion:



begin{align*}
mathbb{E}(1/bar{X}_n) &= sum_{i=0}^{+infty} frac{f^{(i)}(1)}{i!}mathbb{E}left((bar{X}_n-1)^iright)\
&= sum_{i=0}^{+infty} frac{(-1)^i}{i!i!}mathbb{E}left((bar{X}_n-1)^iright)
end{align*}



Now if we can say something about the $k^{th}$ moment of $tilde{X}_n$ being $O(1/n^k)$ this validates that $mathbb{E}(1/bar{X}_n) approx 1 + frac{ sigma^2}{4 n^2}$.






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  • $begingroup$
    Turns out that $bar{X}_n$ does admit $n$ moments and that they are of the form $O(n^{-p/2})$: arxiv.org/pdf/1105.6283.pdf
    $endgroup$
    – Gopi
    23 hours ago











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2 Answers
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active

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votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

You cannot bound that expectation in $sigma, n$. That's because there is the distinct possibility that the expectation do not exist at all (or, is $infty$.) See I've heard that ratios or inverses of random variables often are problematic, in not having expectations. Why is that?. If the conditions given there is fulfilled for the density of $X_1$, it will so be for the density of $bar{X}_n$. If densities do not exist, but probability mass functions do, it is simpler, since your assumptions prohibit a probability atom at zero, but a probability density can still be positive at zero even if $P(X >0)=1$.



For a useful bound you will at least need to restrict the common distribution of $X_1, dotsc, X_n$ much more.



EDIT


After your new information, and with $v_1>0$, the expectation of $1/bar{X}_n$ certainly will exist (irrespective if $K$ is finite or not.) And, since the function $xmapsto 1/x$ is convex for $x>0$, we can use the Jensen Inequality to conclude that $DeclareMathOperator{E}{mathbb{E}}E 1/bar{X}_n ge 1/E bar{X}_n$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But the thinig is that this is really not a general case but a very specific case: it's very unlikely that there is a probability mass near 0 (we can even evaluate it with Markov's inequality): $bar{X}_n$ is centered around 1 and has a variance of $sigma^2 / n$.
    $endgroup$
    – Gopi
    yesterday








  • 1




    $begingroup$
    But even a tiny (but positive) probability close to zero can lead to expectation of inverse being $infty$. Can you rule out that possibility?
    $endgroup$
    – kjetil b halvorsen
    yesterday










  • $begingroup$
    In my case I can indeed (the distribution of $X_i$ is discrete). I'll add this hypothesis to the question. I'm still interested from a theoretical perspective by the general case, I believe something can still be said in this case (or I'm interested by a counter example that would show your statement).
    $endgroup$
    – Gopi
    yesterday






  • 1




    $begingroup$
    Information about discreteness is really important! And, n what support? If positive integers, $P(X=0)=0$ and $mu=1$ is really restrictive ...
    $endgroup$
    – kjetil b halvorsen
    yesterday






  • 1




    $begingroup$
    Obviously the support is not positive integers ;). The support is the set of rational numbers.
    $endgroup$
    – Gopi
    yesterday
















3












$begingroup$

You cannot bound that expectation in $sigma, n$. That's because there is the distinct possibility that the expectation do not exist at all (or, is $infty$.) See I've heard that ratios or inverses of random variables often are problematic, in not having expectations. Why is that?. If the conditions given there is fulfilled for the density of $X_1$, it will so be for the density of $bar{X}_n$. If densities do not exist, but probability mass functions do, it is simpler, since your assumptions prohibit a probability atom at zero, but a probability density can still be positive at zero even if $P(X >0)=1$.



For a useful bound you will at least need to restrict the common distribution of $X_1, dotsc, X_n$ much more.



EDIT


After your new information, and with $v_1>0$, the expectation of $1/bar{X}_n$ certainly will exist (irrespective if $K$ is finite or not.) And, since the function $xmapsto 1/x$ is convex for $x>0$, we can use the Jensen Inequality to conclude that $DeclareMathOperator{E}{mathbb{E}}E 1/bar{X}_n ge 1/E bar{X}_n$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But the thinig is that this is really not a general case but a very specific case: it's very unlikely that there is a probability mass near 0 (we can even evaluate it with Markov's inequality): $bar{X}_n$ is centered around 1 and has a variance of $sigma^2 / n$.
    $endgroup$
    – Gopi
    yesterday








  • 1




    $begingroup$
    But even a tiny (but positive) probability close to zero can lead to expectation of inverse being $infty$. Can you rule out that possibility?
    $endgroup$
    – kjetil b halvorsen
    yesterday










  • $begingroup$
    In my case I can indeed (the distribution of $X_i$ is discrete). I'll add this hypothesis to the question. I'm still interested from a theoretical perspective by the general case, I believe something can still be said in this case (or I'm interested by a counter example that would show your statement).
    $endgroup$
    – Gopi
    yesterday






  • 1




    $begingroup$
    Information about discreteness is really important! And, n what support? If positive integers, $P(X=0)=0$ and $mu=1$ is really restrictive ...
    $endgroup$
    – kjetil b halvorsen
    yesterday






  • 1




    $begingroup$
    Obviously the support is not positive integers ;). The support is the set of rational numbers.
    $endgroup$
    – Gopi
    yesterday














3












3








3





$begingroup$

You cannot bound that expectation in $sigma, n$. That's because there is the distinct possibility that the expectation do not exist at all (or, is $infty$.) See I've heard that ratios or inverses of random variables often are problematic, in not having expectations. Why is that?. If the conditions given there is fulfilled for the density of $X_1$, it will so be for the density of $bar{X}_n$. If densities do not exist, but probability mass functions do, it is simpler, since your assumptions prohibit a probability atom at zero, but a probability density can still be positive at zero even if $P(X >0)=1$.



For a useful bound you will at least need to restrict the common distribution of $X_1, dotsc, X_n$ much more.



EDIT


After your new information, and with $v_1>0$, the expectation of $1/bar{X}_n$ certainly will exist (irrespective if $K$ is finite or not.) And, since the function $xmapsto 1/x$ is convex for $x>0$, we can use the Jensen Inequality to conclude that $DeclareMathOperator{E}{mathbb{E}}E 1/bar{X}_n ge 1/E bar{X}_n$.






share|cite|improve this answer











$endgroup$



You cannot bound that expectation in $sigma, n$. That's because there is the distinct possibility that the expectation do not exist at all (or, is $infty$.) See I've heard that ratios or inverses of random variables often are problematic, in not having expectations. Why is that?. If the conditions given there is fulfilled for the density of $X_1$, it will so be for the density of $bar{X}_n$. If densities do not exist, but probability mass functions do, it is simpler, since your assumptions prohibit a probability atom at zero, but a probability density can still be positive at zero even if $P(X >0)=1$.



For a useful bound you will at least need to restrict the common distribution of $X_1, dotsc, X_n$ much more.



EDIT


After your new information, and with $v_1>0$, the expectation of $1/bar{X}_n$ certainly will exist (irrespective if $K$ is finite or not.) And, since the function $xmapsto 1/x$ is convex for $x>0$, we can use the Jensen Inequality to conclude that $DeclareMathOperator{E}{mathbb{E}}E 1/bar{X}_n ge 1/E bar{X}_n$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited yesterday

























answered yesterday









kjetil b halvorsenkjetil b halvorsen

31.4k984225




31.4k984225












  • $begingroup$
    But the thinig is that this is really not a general case but a very specific case: it's very unlikely that there is a probability mass near 0 (we can even evaluate it with Markov's inequality): $bar{X}_n$ is centered around 1 and has a variance of $sigma^2 / n$.
    $endgroup$
    – Gopi
    yesterday








  • 1




    $begingroup$
    But even a tiny (but positive) probability close to zero can lead to expectation of inverse being $infty$. Can you rule out that possibility?
    $endgroup$
    – kjetil b halvorsen
    yesterday










  • $begingroup$
    In my case I can indeed (the distribution of $X_i$ is discrete). I'll add this hypothesis to the question. I'm still interested from a theoretical perspective by the general case, I believe something can still be said in this case (or I'm interested by a counter example that would show your statement).
    $endgroup$
    – Gopi
    yesterday






  • 1




    $begingroup$
    Information about discreteness is really important! And, n what support? If positive integers, $P(X=0)=0$ and $mu=1$ is really restrictive ...
    $endgroup$
    – kjetil b halvorsen
    yesterday






  • 1




    $begingroup$
    Obviously the support is not positive integers ;). The support is the set of rational numbers.
    $endgroup$
    – Gopi
    yesterday


















  • $begingroup$
    But the thinig is that this is really not a general case but a very specific case: it's very unlikely that there is a probability mass near 0 (we can even evaluate it with Markov's inequality): $bar{X}_n$ is centered around 1 and has a variance of $sigma^2 / n$.
    $endgroup$
    – Gopi
    yesterday








  • 1




    $begingroup$
    But even a tiny (but positive) probability close to zero can lead to expectation of inverse being $infty$. Can you rule out that possibility?
    $endgroup$
    – kjetil b halvorsen
    yesterday










  • $begingroup$
    In my case I can indeed (the distribution of $X_i$ is discrete). I'll add this hypothesis to the question. I'm still interested from a theoretical perspective by the general case, I believe something can still be said in this case (or I'm interested by a counter example that would show your statement).
    $endgroup$
    – Gopi
    yesterday






  • 1




    $begingroup$
    Information about discreteness is really important! And, n what support? If positive integers, $P(X=0)=0$ and $mu=1$ is really restrictive ...
    $endgroup$
    – kjetil b halvorsen
    yesterday






  • 1




    $begingroup$
    Obviously the support is not positive integers ;). The support is the set of rational numbers.
    $endgroup$
    – Gopi
    yesterday
















$begingroup$
But the thinig is that this is really not a general case but a very specific case: it's very unlikely that there is a probability mass near 0 (we can even evaluate it with Markov's inequality): $bar{X}_n$ is centered around 1 and has a variance of $sigma^2 / n$.
$endgroup$
– Gopi
yesterday






$begingroup$
But the thinig is that this is really not a general case but a very specific case: it's very unlikely that there is a probability mass near 0 (we can even evaluate it with Markov's inequality): $bar{X}_n$ is centered around 1 and has a variance of $sigma^2 / n$.
$endgroup$
– Gopi
yesterday






1




1




$begingroup$
But even a tiny (but positive) probability close to zero can lead to expectation of inverse being $infty$. Can you rule out that possibility?
$endgroup$
– kjetil b halvorsen
yesterday




$begingroup$
But even a tiny (but positive) probability close to zero can lead to expectation of inverse being $infty$. Can you rule out that possibility?
$endgroup$
– kjetil b halvorsen
yesterday












$begingroup$
In my case I can indeed (the distribution of $X_i$ is discrete). I'll add this hypothesis to the question. I'm still interested from a theoretical perspective by the general case, I believe something can still be said in this case (or I'm interested by a counter example that would show your statement).
$endgroup$
– Gopi
yesterday




$begingroup$
In my case I can indeed (the distribution of $X_i$ is discrete). I'll add this hypothesis to the question. I'm still interested from a theoretical perspective by the general case, I believe something can still be said in this case (or I'm interested by a counter example that would show your statement).
$endgroup$
– Gopi
yesterday




1




1




$begingroup$
Information about discreteness is really important! And, n what support? If positive integers, $P(X=0)=0$ and $mu=1$ is really restrictive ...
$endgroup$
– kjetil b halvorsen
yesterday




$begingroup$
Information about discreteness is really important! And, n what support? If positive integers, $P(X=0)=0$ and $mu=1$ is really restrictive ...
$endgroup$
– kjetil b halvorsen
yesterday




1




1




$begingroup$
Obviously the support is not positive integers ;). The support is the set of rational numbers.
$endgroup$
– Gopi
yesterday




$begingroup$
Obviously the support is not positive integers ;). The support is the set of rational numbers.
$endgroup$
– Gopi
yesterday













1












$begingroup$

I think I have the gist of it.
Given that $f(x)=1/x$ is infinitely differentiable in 1. Taylor's theorem tells us:



There exists $varepsilon>0$ such that $f(x) = f(1) + f'(1) (x-1)+ frac{f''(1)(x-1)^2}{2} + frac{f'''(varepsilon) (x-1)^2}{2}$.



In our case, if $X_i$ belongs in the domaine $[v_1;+infty[$, then $bar{X}_n$ has the same domain and we have $varepsilon geq v_1$.



Hence
$mathbb{E}(1/bar{X}_n) = mathbb{E}left (1 - (bar{X}_n-1) + frac{(x-1)^2}{4}+ frac{f'''(varepsilon) (bar{X}_n-1)^2}{2} right)$, and
begin{align*}
mathbb{E}(1/bar{X}_n) &= 1 + frac{f'''(varepsilon) mathbb{E}left ((bar{X}_n-1)^2right )}{2} = 1 +frac{ V(bar{X}_n)}{4} - frac{ V(bar{X}_n)}{12 varepsilon^4}\
end{align*}

and hence
$$1 + frac{ sigma^2}{4 n^2}- frac{sigma^2}{12 v_1^4 n^2} leq mathbb{E}(1/bar{X}_n) leq 1 + frac{ sigma^2}{4 n^2}.$$



For the case where $X_i$ do not admit a minimum but has an unlimited number of moments, one can do a similar transformation using the full taylor expansion:



begin{align*}
mathbb{E}(1/bar{X}_n) &= sum_{i=0}^{+infty} frac{f^{(i)}(1)}{i!}mathbb{E}left((bar{X}_n-1)^iright)\
&= sum_{i=0}^{+infty} frac{(-1)^i}{i!i!}mathbb{E}left((bar{X}_n-1)^iright)
end{align*}



Now if we can say something about the $k^{th}$ moment of $tilde{X}_n$ being $O(1/n^k)$ this validates that $mathbb{E}(1/bar{X}_n) approx 1 + frac{ sigma^2}{4 n^2}$.






share|cite|improve this answer










New contributor




Gopi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$













  • $begingroup$
    Turns out that $bar{X}_n$ does admit $n$ moments and that they are of the form $O(n^{-p/2})$: arxiv.org/pdf/1105.6283.pdf
    $endgroup$
    – Gopi
    23 hours ago
















1












$begingroup$

I think I have the gist of it.
Given that $f(x)=1/x$ is infinitely differentiable in 1. Taylor's theorem tells us:



There exists $varepsilon>0$ such that $f(x) = f(1) + f'(1) (x-1)+ frac{f''(1)(x-1)^2}{2} + frac{f'''(varepsilon) (x-1)^2}{2}$.



In our case, if $X_i$ belongs in the domaine $[v_1;+infty[$, then $bar{X}_n$ has the same domain and we have $varepsilon geq v_1$.



Hence
$mathbb{E}(1/bar{X}_n) = mathbb{E}left (1 - (bar{X}_n-1) + frac{(x-1)^2}{4}+ frac{f'''(varepsilon) (bar{X}_n-1)^2}{2} right)$, and
begin{align*}
mathbb{E}(1/bar{X}_n) &= 1 + frac{f'''(varepsilon) mathbb{E}left ((bar{X}_n-1)^2right )}{2} = 1 +frac{ V(bar{X}_n)}{4} - frac{ V(bar{X}_n)}{12 varepsilon^4}\
end{align*}

and hence
$$1 + frac{ sigma^2}{4 n^2}- frac{sigma^2}{12 v_1^4 n^2} leq mathbb{E}(1/bar{X}_n) leq 1 + frac{ sigma^2}{4 n^2}.$$



For the case where $X_i$ do not admit a minimum but has an unlimited number of moments, one can do a similar transformation using the full taylor expansion:



begin{align*}
mathbb{E}(1/bar{X}_n) &= sum_{i=0}^{+infty} frac{f^{(i)}(1)}{i!}mathbb{E}left((bar{X}_n-1)^iright)\
&= sum_{i=0}^{+infty} frac{(-1)^i}{i!i!}mathbb{E}left((bar{X}_n-1)^iright)
end{align*}



Now if we can say something about the $k^{th}$ moment of $tilde{X}_n$ being $O(1/n^k)$ this validates that $mathbb{E}(1/bar{X}_n) approx 1 + frac{ sigma^2}{4 n^2}$.






share|cite|improve this answer










New contributor




Gopi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$













  • $begingroup$
    Turns out that $bar{X}_n$ does admit $n$ moments and that they are of the form $O(n^{-p/2})$: arxiv.org/pdf/1105.6283.pdf
    $endgroup$
    – Gopi
    23 hours ago














1












1








1





$begingroup$

I think I have the gist of it.
Given that $f(x)=1/x$ is infinitely differentiable in 1. Taylor's theorem tells us:



There exists $varepsilon>0$ such that $f(x) = f(1) + f'(1) (x-1)+ frac{f''(1)(x-1)^2}{2} + frac{f'''(varepsilon) (x-1)^2}{2}$.



In our case, if $X_i$ belongs in the domaine $[v_1;+infty[$, then $bar{X}_n$ has the same domain and we have $varepsilon geq v_1$.



Hence
$mathbb{E}(1/bar{X}_n) = mathbb{E}left (1 - (bar{X}_n-1) + frac{(x-1)^2}{4}+ frac{f'''(varepsilon) (bar{X}_n-1)^2}{2} right)$, and
begin{align*}
mathbb{E}(1/bar{X}_n) &= 1 + frac{f'''(varepsilon) mathbb{E}left ((bar{X}_n-1)^2right )}{2} = 1 +frac{ V(bar{X}_n)}{4} - frac{ V(bar{X}_n)}{12 varepsilon^4}\
end{align*}

and hence
$$1 + frac{ sigma^2}{4 n^2}- frac{sigma^2}{12 v_1^4 n^2} leq mathbb{E}(1/bar{X}_n) leq 1 + frac{ sigma^2}{4 n^2}.$$



For the case where $X_i$ do not admit a minimum but has an unlimited number of moments, one can do a similar transformation using the full taylor expansion:



begin{align*}
mathbb{E}(1/bar{X}_n) &= sum_{i=0}^{+infty} frac{f^{(i)}(1)}{i!}mathbb{E}left((bar{X}_n-1)^iright)\
&= sum_{i=0}^{+infty} frac{(-1)^i}{i!i!}mathbb{E}left((bar{X}_n-1)^iright)
end{align*}



Now if we can say something about the $k^{th}$ moment of $tilde{X}_n$ being $O(1/n^k)$ this validates that $mathbb{E}(1/bar{X}_n) approx 1 + frac{ sigma^2}{4 n^2}$.






share|cite|improve this answer










New contributor




Gopi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$



I think I have the gist of it.
Given that $f(x)=1/x$ is infinitely differentiable in 1. Taylor's theorem tells us:



There exists $varepsilon>0$ such that $f(x) = f(1) + f'(1) (x-1)+ frac{f''(1)(x-1)^2}{2} + frac{f'''(varepsilon) (x-1)^2}{2}$.



In our case, if $X_i$ belongs in the domaine $[v_1;+infty[$, then $bar{X}_n$ has the same domain and we have $varepsilon geq v_1$.



Hence
$mathbb{E}(1/bar{X}_n) = mathbb{E}left (1 - (bar{X}_n-1) + frac{(x-1)^2}{4}+ frac{f'''(varepsilon) (bar{X}_n-1)^2}{2} right)$, and
begin{align*}
mathbb{E}(1/bar{X}_n) &= 1 + frac{f'''(varepsilon) mathbb{E}left ((bar{X}_n-1)^2right )}{2} = 1 +frac{ V(bar{X}_n)}{4} - frac{ V(bar{X}_n)}{12 varepsilon^4}\
end{align*}

and hence
$$1 + frac{ sigma^2}{4 n^2}- frac{sigma^2}{12 v_1^4 n^2} leq mathbb{E}(1/bar{X}_n) leq 1 + frac{ sigma^2}{4 n^2}.$$



For the case where $X_i$ do not admit a minimum but has an unlimited number of moments, one can do a similar transformation using the full taylor expansion:



begin{align*}
mathbb{E}(1/bar{X}_n) &= sum_{i=0}^{+infty} frac{f^{(i)}(1)}{i!}mathbb{E}left((bar{X}_n-1)^iright)\
&= sum_{i=0}^{+infty} frac{(-1)^i}{i!i!}mathbb{E}left((bar{X}_n-1)^iright)
end{align*}



Now if we can say something about the $k^{th}$ moment of $tilde{X}_n$ being $O(1/n^k)$ this validates that $mathbb{E}(1/bar{X}_n) approx 1 + frac{ sigma^2}{4 n^2}$.







share|cite|improve this answer










New contributor




Gopi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this answer



share|cite|improve this answer








edited yesterday





















New contributor




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Check out our Code of Conduct.









answered yesterday









GopiGopi

1314




1314




New contributor




Gopi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





Gopi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Gopi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    Turns out that $bar{X}_n$ does admit $n$ moments and that they are of the form $O(n^{-p/2})$: arxiv.org/pdf/1105.6283.pdf
    $endgroup$
    – Gopi
    23 hours ago


















  • $begingroup$
    Turns out that $bar{X}_n$ does admit $n$ moments and that they are of the form $O(n^{-p/2})$: arxiv.org/pdf/1105.6283.pdf
    $endgroup$
    – Gopi
    23 hours ago
















$begingroup$
Turns out that $bar{X}_n$ does admit $n$ moments and that they are of the form $O(n^{-p/2})$: arxiv.org/pdf/1105.6283.pdf
$endgroup$
– Gopi
23 hours ago




$begingroup$
Turns out that $bar{X}_n$ does admit $n$ moments and that they are of the form $O(n^{-p/2})$: arxiv.org/pdf/1105.6283.pdf
$endgroup$
– Gopi
23 hours ago










Gopi is a new contributor. Be nice, and check out our Code of Conduct.










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