There is a "central limit theorem" for the sample median as well. Why was video, audio and picture compression the poorest when storage space was the costliest? $$ 1) 1 E( =The OLS coefficient estimator 0 is unbiased, meaning that . It is found that among them the sample mean has the lowest variance. Hasse1987: M is the total number of samples, e.g. \operatorname{Var}(\tilde{X})=\operatorname{Var}(\bar{X})+\eta Let $Z_i$, $1 \leqslant i \leqslant n$ be independent identically distributed normal variables with mean $\mu$ and variance $\sigma^2$, and let $Z_{k:n}$ denote $k$-th order statistics. Stack Overflow for Teams is moving to its own domain! $$. Indeed, normality assumption was not used. $M = \frac{1}{2} \left( Z_{m:2m} + Z_{m+1:2m} \right)$. If the following holds, where ^ is the estimate of the true population parameter : then the statistic ^ is unbiased estimator of the parameter . Abbott PROPERTY 2: Unbiasedness of 1 and . Let $Z_i$, $1 \leqslant i \leqslant n$ be independent identically distributed normal variables with mean $\mu$ and variance $\sigma^2$, and let $Z_{k:n}$ denote $k$-th order statistics. Though the sample mean is an unbiased estimator of the unknown population mean, it cannot be optimal in general. Thanks for the proof. So Cochran's meaning may take some effort to discern. Use MathJax to format equations. Share Cite Follow answered Dec 13, 2012 at 9:18 Espen Nielsen On the other hand, you can look at the sample median (rather than the sample mean) as an estimator for the median. We have. Transcribed image text: [RBH 31.6] Problem 2: Prove that the sample mean is the best linear unbiased estimator of the population mean ji as follows: (a) If the real numbers a1, ., an satisfy the constraint a i = C, where C is a given constraint, i=1 show that ais minimized by a; = for all i. Why are taxiway and runway centerline lights off center? When the migration is complete, you will access your Teams at stackoverflowteams.com, and they will no longer appear in the left sidebar on stackoverflow.com. Hence, the probability $\Pr\{Z_i=1\}$ is equal to the number of samples of size $n$ which include $i$, given by $n-1\choose N-1$, divided by the number of size $n$ samples, given by $n\choose N$: $$ For example, the sample mean, , is an unbiased estimator of the population mean, . Either way, prove it. $$ \mathbb{E}\left( \frac{ (2\mu-Z_{m+1:2m}) + (2\mu - Z_{m:2m})}{2} \right) = \mathbb{E}(2\mu - M) How many rectangles can be observed in the grid? Stack Exchange network consists of 182 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The sample mean is the random variable All you need is that s2 = 1 n 1 n i = 1(xi x)2 is an unbiased estimator of the variance 2. What does the term "unbiased estimator" mean? $$ But for the $N=5,n=3$ case, $k=6$ as Michael Hardy's answer shows, and $10\neq 6*5$ ? Therefore, the sample mean is an unbiased estimator of the population mean. Connect and share knowledge within a single location that is structured and easy to search. $$ An unbiased estimator of can be obtained by dividing by . The best answers are voted up and rise to the top, Not the answer you're looking for? $$ with other numbers than $3$ and $5$, and also making Cochran's argument precise, but I'm not sure how best to express that right now. Is sample minimum an unbiased estimator for population mean? The estimate is usually obtained by using a predefined rule (a function) that associates an estimate to each sample that could possibly be observed The function is called an estimator. Since $\operatorname{E}(\operatorname{median})=0$, we conclude $\operatorname{E}(\operatorname{median}(X_1,\ldots,X_n))$ $=\operatorname{E}(\operatorname{median}(Y_1+\mu,\ldots,Y_n+\mu))$ $=\operatorname{E}(\mu + \operatorname{median}(Y_1,\ldots,Y_n))=\mu$. $$ Since $\operatorname{E}(\operatorname{median})=0$, we conclude $\operatorname{E}(\operatorname{median}(X_1,\ldots,X_n))$ $=\operatorname{E}(\operatorname{median}(Y_1+\mu,\ldots,Y_n+\mu))$ $=\operatorname{E}(\mu + \operatorname{median}(Y_1,\ldots,Y_n))=\mu$. $$ @DilipSarwate Thanks for the comment. To learn more, see our tips on writing great answers. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company. If this estimator is unbiased (that is, E[T] = ), then the CramrRao inequality states the variance of this estimator is bounded from below: where [5] For a more . Then the sample median corresponds to $M = Z_{m+1:2m+1}$. Why? The double sample mean estimator is like . $$ Answer (1 of 2): I have to prove that the sample variance is an unbiased estimator. Why are UK Prime Ministers educated at Oxford, not Cambridge? As grows large it approaches 1, and even for smaller values the correction is minor. There's got to be a short proof based on symmetry. Please advice how can this be proved. Can lead-acid batteries be stored by removing the liquid from them? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. (I suppose you could consider both sides as random, but the rhs is still deterministic since the population size is $N$.). E ( X ) = . Let $\mu$ be the population mean (so that's assumed to exist), and assume the distribution is symmetric and there's a density. $$ If he wanted control of the company, why didn't Elon Musk buy 51% of Twitter shares instead of 100%? = {} & \frac{1}{N} \cdot (Y_1+\cdots+ Y_N) \end{align}. Then the sample median corresponds to $M = Z_{m+1:2m+1}$. The ys on the lhs are random, those on the rhs are not. Covalent and Ionic bonds with Semi-metals, Is an athlete's heart rate after exercise greater than a non-athlete. Definition An estimator is said to be unbiased if and only if where the expected value is calculated with respect to the probability distribution of the sample . This again implies that $\mathbb{E}(M) = \mu$ as a consequence of the symmetry. Hence we must have $\sum_{i=1}^{N} \delta_{i}=0$ for our new linear estimator to be unbiased. Use MathJax to format equations. X_{i}=\mu+\varepsilon_{i} $$ & y_2 + y_4 + y_5 \\[3pt] Let $X_1,\ldots,X_n$ be the sample; let $Y_i=X_i-\mu$ for $i=1,\ldots,n$. Otherwise, ^ is the biased estimator. f_{Z_{m:2m}, Z_{m+1:2m}}(x_1,x_2) = m^2 \binom{2m}{m}f_X(x_1) f_X(x_2) \left(F_X(x_1) (1-F_X(x_2))\right) ^{m-1} [ x_1 \leqslant x_2 ] So $M$ is the total number of $y$s that show up, with repetition, in $E(y_1+\ldots+y_n)$ ? Then the sample median corresponds to Please advice how can this be proved. $$ With this the second one should also be clear. I will edit the post to reflect that. Why plants and animals are so different even though they come from the same ancestors? Clearly, again $f_{Z_{m:2m}, Z_{m+1:2m}}(x_1,x_2)=f_{Z_{m:2m}, Z_{m+1:2m}}(2\mu - x_2,2 \mu - x_1)$ by symmetry, therefore Does a beard adversely affect playing the violin or viola? Can a black pudding corrode a leather tunic? Y is distributed uniform distribution U (0,a), we construct the double sample mean estimator to estimate population parameter a. So both estimators will be able to approach to the true parameter, as long as there are enough data samples, i.e. The sampling distribution of the sample median looks asymptotically like: f_{M}(x) = (m+1) \binom{2m+1}{m} f_X(x) \left( F_X(x) (1-F_X(x)) \right)^m Making statements based on opinion; back them up with references or personal experience. \left(\frac{M}{N} \cdot Y_1+\cdots+\frac{M}{N} \cdot Y_N \right)\right/M \\[8pt] f_{Z_{m:2m}, Z_{m+1:2m}}(x_1,x_2) = m^2 \binom{2m}{m}f_X(x_1) f_X(x_2) \left(F_X(x_1) (1-F_X(x_2))\right) ^{m-1} [ x_1 \leqslant x_2 ] Why are standard frequentist hypotheses so uninteresting? \mathbb{E}\left( \frac{ (2\mu-Z_{m+1:2m}) + (2\mu - Z_{m:2m})}{2} \right) = \mathbb{E}(2\mu - M) Actually, he ought to have expressed it the way you did. For a sample of N students selected independently from the population: (e) Is the sample mean BLUE? If this is the case, then we say that our statistic is an unbiased estimator of the parameter. Would a bicycle pump work underwater, with its air-input being above water? $$ \begin{align} Let's return to our simulation. $n = 2m+1$. The best answers are voted up and rise to the top, Not the answer you're looking for? Suppose that a data set is collected with n numerical observations x1, x2 . Define linear estimator $\tilde{X}=\frac{1}{N} \sum_{i=1}^{N} w_{i} X_{i}$ with weights made up: $w_{i}=1+\delta_{i}$. is independent of , for all , where t = T(y). \operatorname{Var}(\tilde{X})=\frac{1}{N^{2}} \sum_{i=1}^{N}\left(1+\delta_{i}\right)^{2} \sigma^{2}=\frac{\sigma^{2}}{N}+\frac{\sigma^{2}}{N^{2}} \sum_{i=1}^{N}\left(2 \delta_{i}+\delta_{i}^{2}\right)=\operatorname{Var}(\bar{X})+\frac{\sigma^{2}}{N^{2}} \sum_{i=1}^{N} \delta_{i}^{2} Now consider the case of even $n$, i.e. Added: The normality assumption was not used in the above demonstration, thus the proof holds for any continuous random variable with symmetric probability density and finite mean. But actually I am trying to understand Cochran's particular proof, specifically his remark about the terms on each side. Use MathJax to format equations. On the other hand, since , the sample standard deviation, , gives a . $$ So wha. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. $$. where $m$ is the median, $f$ is the density, and $\hat{m}_n$ is the sample median with $n$ samples. Does English have an equivalent to the Aramaic idiom "ashes on my head"? Here $\mu$ is the mean number of years of education completed by parents. How can I make a script echo something when it is paused? Will it have a bad influence on getting a student visa? & y_3 + y_4 + y_5 The mode is also the same if the distribution is unimodal. Will Nondetection prevent an Alarm spell from triggering? Well, the expected deviation between any sample mean and the population mean is estimated by the standard error: 2M = / (n). Proof that the Sample Variance is an Unbiased Estimator of the Population Variance . William Cochran's book on sampling gives the following proof that a sample mean is unbiased: Since every unit appears in the same number of samples, it is clear that E [ Y 1 + + Y n] must be some multiple of y 1 + y 2 + + y N. The multiplier must be n / N since the first expression has n terms and the second has N terms. Either way, prove it. = \frac 6 {10} (y_1+y_2+y_3+y_4+y_5) = \frac 3 5 (y_1+y_2+y_3+y_4+y_5) \begin{gather} Site design / logo 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. What is this political cartoon by Bob Moran titled "Amnesty" about? $$ Does a beard adversely affect playing the violin or viola? What is is asked exactly is to show that following estimator of the sample variance is unbiased: s2 = 1 n 1 n i = 1(xi x)2. I have to prove that the sample variance is an unbiased estimator. Did the words "come" and "home" historically rhyme? The probability density of this order statistics is: An unbiased estimator of a population parameter is an estimator whose expected value is equal to that pa-rameter. How actually can you perform the trick with the "illusion of the party distracting the dragon" like they did it in Vox Machina (animated series)? Now my question is if we take say first random variable X 1 as an estimate we can go ahead and say that since E [ X 1] = , X 1 is an unbiased estimator. In randomization theory, the $y_i$'s are not random variables, but fixed quantities. Since $m=-m$, we must have $m=0$. $$ $m=\operatorname{E}(\operatorname{median})=\operatorname{E}(\operatorname{median}(Y_1,\ldots,Y_n))$, $-m=\operatorname{E}(-{\operatorname{median}})=\operatorname{E}(\operatorname{median})$, $\operatorname{E}(\operatorname{median})=0$, $\operatorname{E}(\operatorname{median}(X_1,\ldots,X_n))$, $=\operatorname{E}(\operatorname{median}(Y_1+\mu,\ldots,Y_n+\mu))$, $=\operatorname{E}(\mu + \operatorname{median}(Y_1,\ldots,Y_n))=\mu$. But he also gives this alternative proof that I was trying to make sense of. Sample mean is the maximum likelihood estimator if the data at hand follows a Gaussian distribution. Let $n$ be odd, i.e. What is the use of NTP server when devices have accurate time? Sci-Fi Book With Cover Of A Person Driving A Ship Saying "Look Ma, No Hands!". In this video I discuss the basic idea behind unbiased estimators and provide the proof that the sample mean is an unbiased estimator. & y_1 + y_3 + y_4 \\[3pt] $$ The density assumption can also be dropped: symmetry can be expressed as $P(Y_i \le c) = P(Y_i \ge -c)$ for all real $c$. $\bar{y}_{i}$ is the mean over the i th sample. Number of unique permutations of a 3x3x3 cube. Now we derive variance: $n = 2m$. Added: The normality assumption was not used in the above demonstration, thus the proof holds for any continuous random variable with symmetric probability density and finite mean. The best answers are voted up and rise to the top, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, $\varepsilon_{i} \sim i i d\left(0, \sigma^{2}\right)$, $\tilde{X}=\frac{1}{N} \sum_{i=1}^{N} w_{i} X_{i}$, $$ Answer: I expect there is a short elegant combinatorial argument showing that that will happen in general, i.e. is an unbiased estimator of . (So those are weaker assumptions than normality, and maybe the density assumption can be dropped too.) The figure shows a plot of versus sample size. What is the rationale of climate activists pouring soup on Van Gogh paintings of sunflowers? . Connect and share knowledge within a single location that is structured and easy to search. $$ I don't understand the use of diodes in this diagram. To learn more, see our tips on writing great answers. Since $m=\mu$ in the symmetric case, you can use the sample mean there. Does there exist an unbiased estimator for the absolute value of the mean? (2 points) (b) Please prove whether sample mean estimator is unbiased or not. The joint probability density is: Is there an industry-specific reason that many characters in martial arts anime announce the name of their attacks? How to go about finding a Thesis advisor for Master degree, Prove If a b (mod n) and c d (mod n), then a + c b + d (mod n). QGIS - approach for automatically rotating layout window. Typeset a chain of fiber bundles with a known largest total space. = {} & \left. X = X n = X 1 + X 2 + X 3 + + X n n = X 1 n + X 2 n + X 3 n + + X n n. Therefore, In statistics, one talks about bias in estimate as the difference between the population parameter and the expected value of the parameter being proposed as an estimate. $$, $$ The joint probability density is: Stack Exchange network consists of 182 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Site design / logo 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. Thanks for contributing an answer to Mathematics Stack Exchange! compare sample mean and sample median as estimators of . In symbols, . [Math] compare sample mean and sample median as estimators of , symmetric distribution with finite mean has median equal to its mean, mode is also the same if the distribution is unimodal, "central limit theorem" for the sample median. William Cochran's book on sampling gives the following proof that a sample mean is unbiased: Since every unit appears in the same number of samples, it is clear that $E[Y_1 + \cdots + Y_n]$ must be some multiple of $y_1+y_2+\cdots+y_N$. econometrics. For example the sample mean is an unbiased estimate . Added: The normality assumption was not used in the above demonstration, thus the proof holds for any continuous random variable with symmetric probability density and finite mean. This again implies that $\mathbb{E}(M) = \mu$ as a consequence of the symmetry. = {} & \left(\bar{y}_1 + \cdots + \bar{y}_M\right)/M \\[8pt] $$ In this video, we discuss a trait that is desirable in point estimators. \mathbb{E}(M) = \mathbb{E}\left( \frac{ Z_{m:2m} + Z_{m+1:2m}}{2} \right) = \mathbb{E}(\tilde{X})=\mathbb{E}\left(\frac{1}{N} \sum_{i=1}^{N}\left(1+\delta_{i}\right) X_{i}\right)=\mu+\frac{\mu}{N} \sum_{i=1}^{N} \delta_{i} Then use that the square root function is strictly concave such that (by a strong form of Jensen's inequality) E(s2) < E(s2) = unless the distribution of s2 is degenerate at 2. Name for phenomenon in which attempting to solve a problem locally can seemingly fail because they absorb the problem from elsewhere? Are witnesses allowed to give private testimonies? Using the Rao-Blackwell theorem one can also prove that determining the MVUE is simply a matter of finding a complete sufficient statistic for the family and conditioning any unbiased estimator on it. Prove that the sample median is an unbiased estimator. Thus "every unit appears in the same number of samples", since in this case, that "same number" is $6$. The multiplier must be $n/N$ since the first expression has $n$ terms and the second has $N$ terms. & y_1 + y_2 + y_4 \\[3pt] Suppose two estimators are unbiased, what is the intuition behind the preference of the estimator with the less variance? \mathbb{E}(M) = \mathbb{E}(2 \mu -M) \implies \mathbb{E}(M) = \mu My book says that sample median of a normal distribution is an unbiased estimator of its mean, by virtue of the symmetry of normal distribution. One useful approach to finding the MVUE begins by finding a sufficient statistic for the parameter. \end{align} MathJax reference. Since E(b2) = 2, the least squares estimator b2 is an unbiased estimator of 2. How do I prove MVUE? When the migration is complete, you will access your Teams at stackoverflowteams.com, and they will no longer appear in the left sidebar on stackoverflow.com. 0 The OLS coefficient estimator 1 is unbiased, meaning that . Asking for help, clarification, or responding to other answers. MathJax reference. What is the difference between an "odor-free" bully stick vs a "regular" bully stick? (1) Example: The sample mean X is an unbiased estimator for the population mean , since E(X) = . In more precise language we want the expected value of our statistic to equal the parameter. The sampling distribution of the sample median looks asymptotically like: $$ \hat{m}_n\sim\mathcal{N}\left( m, [4n f(m)^2]^{-1 . We separately consider the case of even $n$ and odd $n$. Movie about scientist trying to find evidence of soul. Solution: In order to show that X is an unbiased estimator, we need to prove that. That is the "multiplier". i.e., if we know T(Y ), then there is no . Can you say that you reject the null at the 95% level? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. (clarification of a documentary). @DilipSarwate Thanks for the comment. You don't need normality to prove this. Example of $\arg\min\limits_{T: E(T(X))=\theta}\mathbb{E}|T(X) - \theta| \neq \arg\min\limits_{E(T(X))=\theta}\mathbb{E}(T(X) - \theta)^2$, show that the sample median of normal distribution is median unbiased. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company. For a sample of N students selected independently from the population: (e) Is the sample mean BLUE? Why are standard frequentist hypotheses so uninteresting? $$ How many ways are there to solve a Rubiks cube? Let $\mu$ be the population mean (so that's assumed to exist), and assume the distribution is symmetric and there's a density. What is the difference between an "odor-free" bully stick vs a "regular" bully stick? Answer: Consider the class of all unbiased estimators. Why is HIV associated with weight loss/being underweight? Can you say that you reject the null at the 95% level? Definitely it's a weaker assertion than what can be proved by the same methods. Then the sample median corresponds to $M = Z_{m+1:2m+1}$. f_{M}(x) = (m+1) \binom{2m+1}{m} f_X(x) \left( F_X(x) (1-F_X(x)) \right)^m Consequences resulting from Yitang Zhang's latest claimed results on Landau-Siegel zeros, Sci-Fi Book With Cover Of A Person Driving A Ship Saying "Look Ma, No Hands! The sample mean is a random variable that is an estimator of the population mean. A random sample is determined by (dependent) indicator random variables $Z_1,\dots,Z_N\in\{0,1\}$ such that $Z_1+\dots+Z_N$ is equal to the sample size $n$. & y_1 + y_3 + y_5 \\[3pt] To prove sample mean is an unbiased estimator of the population mean, we use the following step, E [ X i n] = 1 n E [ X i] = , given that E [ X i] = . . When the migration is complete, you will access your Teams at stackoverflowteams.com, and they will no longer appear in the left sidebar on stackoverflow.com. \operatorname{Var}(\tilde{X})=\operatorname{Var}(\bar{X})+\eta Space - falling faster than light? Is opposition to COVID-19 vaccines correlated with other political beliefs? Unbiased estimator of mean of exponential distribution. Theorem. $$. A proof that the sample variance (with n-1 in the denominator) is an unbiased estimator of the population variance. & y_1 + y_4 + y_5 \\[3pt] Since $m=-m$, we must have $m=0$. 618 06 : 58. Position where neither player can force an *exact* outcome. Then the sample median corresponds to Why don't American traffic signs use pictograms as much as other countries? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. Then: X = 1 n n i = 1Xi. Can plants use Light from Aurora Borealis to Photosynthesize? explanation of proof that sample mean is unbiased, Mobile app infrastructure being decommissioned, Understanding Cochran 1977 proof of variance of the sample mean in sampling without replacement, Finding conditional probability of an individual component of a joint distribution, Unbiased estimator of the ratio of variances. In your question the mean and the median are the same. the sample median is approximately normally distributed with mean Degrees of freedom in SEM: Are we testing the models that we claim to test?. Finally, note that for non-zero weights the expression $\sum_{i=1}^{N} \delta_{i}^{2}=\eta>0$, hence we have that the variance of the new estimator is greater than that of the sample mean. On the other hand, sample median is the maximum likelihood estimator of the mean if the data follows a Laplace distribution. I will edit the post to reflect that. Answer to Prove that the sample mean is an unbiased estimator for p.. $$ Easynomics. Answer: First part of proof proves conditions for linear estimator to be unbiased. $$ Answer: I do not know what you mean by 'the sample variance is unbiased'. \mathbb{E}(M) = \mathbb{E}(2 \mu -M) \implies \mathbb{E}(M) = \mu The maximum likelihood estimator of the mean on the original scale is a function of the sample mean and sample variance both computed on the log scale. & y_1 + y_2 + y_3 \\[3pt] Answer (1 of 3): Better than that, the sample mean is unbiased for the mean (assuming it exists) if the distribution is symmetrical: see Prove that the sample median is an unbiased estimator. 12. Hence We separately consider the case of even $n$ and odd $n$. Let $y_1,\dots,y_N$ denote the values of some attribute of the population units $1,\dots,N$. \text{expected value of the sample sum} = \frac{6y_1+6y_2+6y_3+6y_4+6y_5}{10} \\[10pt] Therefore, the sample mean $\bar{Y}$ is an unbiased estimator of the population mean $\bar{y}$. (So those are weaker assumptions than normality, and maybe the density assumption can be dropped too.) By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Prove that the sample median is an unbiased estimator. How does this work in practice ? First part of proof proves conditions for linear estimator to be unbiased. Proof that sample mean is unbiased. Example :- Consider all unbiased estimators of population mean. f_{Z_{m:2m}, Z_{m+1:2m}}(x_1,x_2) = m^2 \binom{2m}{m}f_X(x_1) f_X(x_2) \left(F_X(x_1) (1-F_X(x_2))\right) ^{m-1} [ x_1 \leqslant x_2 ] My book says that sample median of a normal distribution is an unbiased estimator of its mean, by virtue of the symmetry of normal distribution. \mathbb{E}(M) = \mathbb{E}\left( \frac{ Z_{m:2m} + Z_{m+1:2m}}{2} \right) = Connect and share knowledge within a single location that is structured and easy to search. Why does sending via a UdpClient cause subsequent receiving to fail? f_{M}(x) = (m+1) \binom{2m+1}{m} f_X(x) \left( F_X(x) (1-F_X(x)) \right)^m The sample mean estimator is following (a) Please get the distribution of sample mean. medianorder-statisticsprobabilitystatistics. Did find rhyme with joined in the 18th century? It only takes a minute to sign up. Formally, an estimator for parameter is said to be unbiased if: E() = . Added: Regarding Batman's comment, one may want to see this answer as well. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields.
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