Timeline for Expectation of reciprocal of a variable
Current License: CC BY-SA 4.0
12 events
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| Nov 19, 2019 at 21:07 | comment | added | Ivan Svetunkov | An interesting observation is that this method works perfectly for Inverse Gaussian distribution. Tweedie (1957) mentions on page 372 this as one of the methods of obtaining negative moments of that distribution. And it seems that this is not an approximation, but the exact formula for that distribution. I cannot get, why though... | |
| Dec 3, 2018 at 21:34 | history | edited | Matteo Fasiolo | CC BY-SA 4.0 |
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| Apr 19, 2018 at 17:24 | comment | added | Thomas Ahle | @SandipanKarmakar You can use the inequality $\sqrt{E[X^2]}\ge E[|X|]\ge\frac{E[X^2]^{3/2}}{\sqrt{E[X^4]}}$. | |
| Sep 20, 2017 at 10:19 | comment | added | BloXX | I don't understand how this solution gets so many upvotes. For a single random variable $X$ there is no justificiation about the quality of this approximation. The third derivative $f(x)=1/x$ is not bounded. Moreover the remainder of the approx. is $1/6f'''(\xi)(X-\mu)^3$ where $\xi$ is itself a random variable between $X$ and $\mu$. The remainder won't vanish in general and may be very huge. Taylor approx. may only be useful if one has sequence of random variables $X_n -\mu = O_p(a_n)$ where $a_n \to 0$. Even then uniform integrability is needed additionally if interested in the expectation. | |
| Aug 1, 2017 at 13:22 | comment | added | Aaron Hendrickson | @MatteoFasiolo I see. So really one could make the case that so long as the higher order central moment are small the approximation is accurate. This is in effect making a statement about symmetry since symmetry requires all odd central moments to be zero. | |
| Aug 1, 2017 at 12:50 | comment | added | Matteo Fasiolo | @AaronHendrickson my reasoning is simply that the next term in the expansion is proportional to $E\{(X-E(X))^3\}$ which is related to the skewness of the distribution of $X$. Skewness is an asymmetry measure. However, zero skewness does not guarantee symmetry and I am not sure whether symmetry guarantees zero skewness. Hence, this is all heuristic and there might be plenty of counterexamples. | |
| Jul 31, 2017 at 11:03 | comment | added | Aaron Hendrickson | @MatteoFasiolo Can you please explain why the symmetry of the distribution of $X$ (or lack thereof) has an effect on the accuracy of the Taylor approximation? Do you have a source that you could point me to that explains why this is? | |
| Dec 20, 2016 at 16:53 | history | edited | Matteo Fasiolo | CC BY-SA 3.0 |
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| May 8, 2014 at 20:35 | comment | added | Matteo Fasiolo | I don't think you can use it for $|X|$ as that function is not differentiable. I would rather divide the problem into the cases and say $E(|X|) = E(X|X > 0)p(X>0) + E(-X|X < 0)p(X<0)$, I guess. | |
| May 8, 2014 at 11:58 | comment | added | Sandipan Karmakar | oh yes yes...I am very sorry that I could not apprehend that fact...I have one more q...Is this applicable to any kind of function???actually I am stuck with $|x|$...How can the expectation of $|x|$ can be deduced in terms of $E(x)$ and $V(x)$ | |
| Dec 30, 2013 at 16:39 | history | edited | Matteo Fasiolo | CC BY-SA 3.0 |
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| Dec 30, 2013 at 15:01 | history | answered | Matteo Fasiolo | CC BY-SA 3.0 |