Timeline for Weak law of large numbers in finite populations [closed]
Current License: CC BY-SA 3.0
28 events
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| Sep 1, 2016 at 16:20 | history | closed |
Dilip Sarwate COOLSerdash gung - Reinstate Monica John mpiktas |
Needs details or clarity | |
| Aug 31, 2016 at 13:38 | comment | added | whuber♦ | You appear to use the work "realisation" in an unusual way, one that likely was not intended by your reference. Your use sounds like you ought to be writing "subject" or "element" (of the sample space). Your "Edit 2" appears to answer your question: when sampling with replacement, infinitely many samples are possible. In "Edit 3," note that when sampling without replacement, the individual results are identically distributed but they cannot possibly be independent. Whence "iid" never holds in that case except for samples of size one (trivially). | |
| Aug 31, 2016 at 3:01 | review | Close votes | |||
| Sep 1, 2016 at 16:20 | |||||
| Aug 29, 2016 at 17:09 | comment | added | FredrikAa | @JuhoKokkala I see your point. I have been using a book that defined the population mean as the mean of all the realisations in the population, please see Population, sample and model . Then we should be able to get the true value of the population mean, with no error. Therefore I was wondering why the LLW could be of any use in finite populations. | |
| Aug 27, 2016 at 18:50 | comment | added | Juho Kokkala | This comment chain is getting all too long, and the issue does not seem to be very related to the law of large numbers (I suspect that if the unclarities related to finite populations, samples, etc. are resolved, it will turn out that the question as formulated is not very relevant). Perhaps you want to post a new question. If you have a real-life application in mind, it might be helpful to describe that, too, as that may make it easier to see behind the formalizations and possible errors in them. | |
| Aug 27, 2016 at 18:45 | comment | added | Juho Kokkala | You are still mixing the distribution of the underlying random variable and the distribution observed in your sample(the finite population) -- I don't know what it means to say that 'the distribution of $Y$ is truly a normal distribution'. I'd also avoid the term 'population mean' since it is often used to refer to the model parameter (mean of a hypothetical infinite population) -- if you model the $Y_i$s as iid realizations from some distribution, then in modeling terms your finite population is really a sample even if it happens to contain every entity that exist in the real world. | |
| Aug 27, 2016 at 18:33 | comment | added | FredrikAa | @JuhoKokkala, thnx for your ans.. So, what you are saying is that we should really write $$\mu_{\text{pop}}=\frac{\sum_{i=1}^nY_i}{n}$$ to denote the population mean. When we write $Y\sim \text{N}(\mu,\sigma^2)$, $\mu$ is a model parameter. We have that $\mu=\mu_{\text{pop}}$ if and only if the distribution of $Y$ is truly a normal distribution. This cannot occur if the population is finite (i.e we have $Y_1,\ldots,Y_n$, where $n$ is finite), since when we make a histogram of the realisations of $Y_1,\ldots,Y_n$, the histogram cannot be perfectly normal. Is this a correct interp of your ans? | |
| Aug 27, 2016 at 15:11 | comment | added | Juho Kokkala | It is not possible for the $Y_i$s be iid and $E(Y_1)=\mu=\frac{\sum_{i=1}^n Y_i}{n}$ to hold (except technically iid holds if $Y_i$ is a constant rv). If for example $Y_i \sim N(10,5)$, the mean of any finite number of $Y_i$s shall not be exactly $10$ (except with probability $0$). So, $\mu$ cannot be both the mean of the underlying distribution and the mean of the $n$ iid samples. | |
| Aug 26, 2016 at 20:18 | comment | added | FredrikAa | @DilipSarwate I try do my best. Hopefully the edits and the comment on VCG's answer clarify? | |
| Aug 26, 2016 at 19:06 | history | edited | FredrikAa | CC BY-SA 3.0 |
corrected edit 3
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| Aug 26, 2016 at 19:05 | comment | added | FredrikAa | @JuhoKokkala I try to do my best. Hopefully the edits clarify? | |
| Aug 26, 2016 at 18:46 | history | edited | FredrikAa | CC BY-SA 3.0 |
Added Edit 2 and Edit 3
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| Aug 26, 2016 at 18:27 | comment | added | FredrikAa | is done without replacement. You cannot make a sample of size $n+1$ without replacement from the population of size $n$. | |
| Aug 26, 2016 at 18:25 | comment | added | FredrikAa | @Juho Kokkala, assume you have a population of $n$ individuals. We consider a population characteristic with distribution $Y$. Let $Y_1,\ldots ,Y_n$ denote the random variables for the characteristics of individual $1$, $2$ etc. The population mean is then $\mu=\frac{\sum_{i=1}^n Y_i}{n}$. By definition $\mu=\text{E}(Y_1)$. We then consider a sample from the population. Let us say we consider a sample of size $n$. Let $X_1,\ldots,X_n$ denote the sample. If $X_1=Y_1,\ldots,X_n=Y_n$, and $Y_1,\ldots,Y_n$ are independent, then $X_1,\ldots,X_n$ are iid EVEN IF the sampling | |
| Aug 26, 2016 at 13:44 | comment | added | Juho Kokkala | @gung The second answer begins "So if your question is", so I don't think it is good evidence for clarity of the question. (However, no need to argue, if my close vote remains the only one, I suppose it's good evidence that the problem is with me and not with the question) | |
| Aug 26, 2016 at 13:34 | history | edited | gung - Reinstate Monica | CC BY-SA 3.0 |
light editing & formatting
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| Aug 26, 2016 at 9:56 | review | Close votes | |||
| Aug 26, 2016 at 13:34 | |||||
| Aug 26, 2016 at 9:39 | comment | added | Juho Kokkala | "Finite population" is a confusing term -- I initially interpreted "finite population" in the sense discussed in this answer stats.stackexchange.com/a/99166/24669 (linked in @VCG's answer here), but then the random variable's range would be finite. Instead the "finiteness" here may refer to there only being a finite number of samples from the "data-generating distribution" since no more individuals exist. | |
| Aug 26, 2016 at 9:33 | comment | added | Juho Kokkala | So, if the random variable's range is not finite, it clearly models some other variability in addition to that of random sampling the individuals from the finite set of individuals. But in this case you will not find out $E[X_i]$ exactly even by sampling all individuals in the finite set. Can you clarify whether you are looking for estimating $E[X_i]$ or $(X_1+\ldots+X_N)/N$? In the first case, no, you cannot physically obtain more samples than there are individuals; in the second case I don't see how LLN is supposed to be related. | |
| Aug 25, 2016 at 21:25 | answer | added | VCG | timeline score: 2 | |
| Aug 25, 2016 at 21:23 | history | edited | FredrikAa | CC BY-SA 3.0 |
Corrected clarification of "finite population"
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| Aug 25, 2016 at 21:06 | history | edited | FredrikAa | CC BY-SA 3.0 |
Clarified "finite population"
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| Aug 25, 2016 at 21:01 | history | edited | FredrikAa | CC BY-SA 3.0 |
added 221 characters in body
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| Aug 25, 2016 at 12:54 | history | edited | Tim |
edited tags
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| Aug 25, 2016 at 12:47 | vote | accept | FredrikAa | ||
| Aug 26, 2016 at 18:11 | |||||
| Aug 25, 2016 at 12:44 | history | edited | Greenparker | CC BY-SA 3.0 |
edited body
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| Aug 25, 2016 at 12:39 | comment | added | VCG | So you don't ever have infinitely many. It's that after you have a large finite number, you get `close' to the limit value of $\mu$. | |
| Aug 25, 2016 at 12:22 | history | asked | FredrikAa | CC BY-SA 3.0 |