Timeline for Can the error be modeled in the approximation of expectation
Current License: CC BY-SA 4.0
27 events
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| Apr 3, 2023 at 6:48 | comment | added | CfourPiO | Yes, my bad; the correct way of saying was when the limit of $M$ was infinite. | |
| Apr 3, 2023 at 1:16 | comment | added | whuber♦ | When we compute the variance, we always do it for finite $M$! It cannot be computed for infinite $M:$ you have to take a limit -- but obviously this limit is infinite in your case. | |
| Apr 2, 2023 at 18:46 | comment | added | CfourPiO | For example, something like this. $\epsilon = M \int_{-\infty}^{-\infty} \exp({-x \omega^2 })\frac{1}{2\pi\sigma^2} \exp({-\frac{x^2}{2\sigma^2}}) dx - \sum_{m=M+1}^{\infty} f(a_m, \omega)$ | |
| Apr 2, 2023 at 18:42 | comment | added | CfourPiO | Correct me if I’m wrong. When we compute the variance of the mean or the expectation, we still assume $M \to \infty$. So, if we have multiple realisations of $s$, it’s mean is given by the last expression and the variance gives the information about how much it can vary around this expectation when we have multiple copies of $s$. However, in all these computations, still $M$ is considered to be a very high number. What I want to know is that if there’s a numerical way where we can define the error in the approximation of the expectation itself when $M$ is finite. | |
| Apr 2, 2023 at 17:56 | comment | added | whuber♦ | Could you elaborate on what you mean by "model the error"? It's a random variable and surely computing its variance is basic information. Where do you wish to go beyond that? | |
| Apr 2, 2023 at 17:41 | comment | added | CfourPiO | @whuber in the link you provide, it gives an explanation of how the moments are computed. What I was interested to know is how to model the error in the approximated first moment itself. The approximation here is that $M \to \infty$. How to model the error when M is finite? | |
| Apr 2, 2023 at 14:16 | comment | added | whuber♦ | That narrows it down, because $s$ clearly is a sum of iid Lognormal values. Unfortunately, when $M\gt 1$ there is no closed form for its distribution. You can still (easily) work out the moments of $s$ and compute measures of dispersion like the variance from those: see stats.stackexchange.com/a/89973/919. | |
| Apr 2, 2023 at 8:25 | history | edited | CfourPiO | CC BY-SA 4.0 |
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| Apr 2, 2023 at 8:24 | comment | added | CfourPiO | Ah sorry. Let me edit it. | |
| Apr 1, 2023 at 23:55 | comment | added | whuber♦ | Re the edit: it doesn't make sense, but perhaps you mean $f(x,\omega)=\exp(-x\omega^2)$? | |
| Apr 1, 2023 at 21:37 | comment | added | CfourPiO | I have edited the question with more details. | |
| Apr 1, 2023 at 21:37 | history | edited | CfourPiO | CC BY-SA 4.0 |
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| Apr 1, 2023 at 21:11 | comment | added | CfourPiO | I will try to expand the question a bit more clearly with expressions of $f$ and what I have tried. | |
| Apr 1, 2023 at 15:24 | comment | added | whuber♦ | Because $f$ is arbitrary, that does not give us any additional useful information: $f(a_m,\omega)$ could have (literally) any distribution whatsoever. | |
| Apr 1, 2023 at 4:38 | comment | added | CfourPiO | All the $a_m$ s are random draws from a Gaussian distribution. | |
| Mar 31, 2023 at 19:15 | comment | added | whuber♦ | @StratosFair By simplifying the notation to use "$X$" in place of the random variable "$f(a_m,\omega)$" and without loss of generality (due to linearity of expectation) considering $m=1,$ the assertion is $E[X]=\int xp(x)\,\mathrm d x.$ In what sense is this "not true"? Have I perhaps oversimplified the question or misinterpreted it? | |
| Mar 31, 2023 at 15:41 | comment | added | Stratos supports the strike | If the $a_m$ are not identically distributed then the equality $$\mathbb{E}\left[\sum_{m = 1}^{M} f(a_m, \omega)\right] = M \int_{-\infty}^{+\infty} f(x, \omega) p(x) dx $$ is not true. Do they follow the same distribution ? If so, what distribution is it ? | |
| Mar 31, 2023 at 12:20 | comment | added | CfourPiO | Pardon! I indeed made a mistake. I wanted to know the expectation of the sum indeed and not $s$. I re-wrote it. Does it look clear now? | |
| Mar 31, 2023 at 12:18 | history | edited | CfourPiO | CC BY-SA 4.0 |
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| Mar 31, 2023 at 12:08 | history | edited | CfourPiO | CC BY-SA 4.0 |
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| Mar 31, 2023 at 12:06 | comment | added | CfourPiO | I will try to reformulate. | |
| Mar 31, 2023 at 12:04 | comment | added | Tim | I don't know what you are trying to say, so can't help. | |
| Mar 31, 2023 at 12:04 | comment | added | CfourPiO | How should it be written? | |
| Mar 31, 2023 at 11:53 | comment | added | Tim | Than your notation is not clear. | |
| Mar 31, 2023 at 11:44 | comment | added | CfourPiO | If I have different realizations of $s$, the second formula is the expectation of $s$. Am I wrong? | |
| Mar 31, 2023 at 11:23 | comment | added | Tim | Your notation is confusing. The second formula does not seem to have anything to do with expectation of a sum $s$. | |
| Mar 31, 2023 at 6:59 | history | asked | CfourPiO | CC BY-SA 4.0 |