I understand that variance can be zero if all data points are equal.
Can anybody state an example, where sample variance is equal to sample mean and what - if any - is its significance to distributions like the Poisson.
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asked Apr 20, 2012 at 8:37
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4$\begingroup$ Well for Poisson distribution, you get that variance is always equal mean. For any parametric distribution which has more than one parameter, the mean and variance are usually the functions of these parameters, so it is not that hard to find such parameter values that these functions coincide. Take for example normal distribution $N(\mu,\sigma^2)$. Take $\mu=\sigma^2$ and presto, you have a distribution whose variance is equal to mean. $\endgroup$– mpiktasCommented Apr 20, 2012 at 8:51
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3$\begingroup$ Notice that in general you'd want to compare the mean to the standard deviation, not to the variance: variance and mean have different dimensions, so it seldom makes sense to say "mean equals variance" (that property would depend on scale). I can make sense for discrete distributions, as the Poisson. $\endgroup$– leonbloyCommented Apr 20, 2012 at 11:33
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3$\begingroup$ (This is a bit of a joke, but it's worth contemplating.) The datasets $\{0,0,\ldots,0\}$ have zero mean and zero variance. If such data are assumed to come from a Poisson distribution, they provide evidence of a low intensity. :-) $\endgroup$– whuber ♦Commented Apr 20, 2012 at 15:54
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2$\begingroup$ Voted to close as not a real question... $\endgroup$– MacroCommented May 8, 2012 at 13:48
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6$\begingroup$ I think there's some value in the question, from two points of view: (a) illustrating that it's not at all hard to construct samples with mean=variance by hand and (b) illustrating that even though for a Poisson mean and and variance are the same on average, that having that be true in a sample isn't an indication of its Poisson-ness (by giving an example which is pretty clearly not Poisson). There's a misconception underlying the question which is worth dealing with, I think. So - while I understand the impulse to close, I want to keep the question. $\endgroup$– Glen_bCommented Aug 23, 2013 at 6:49
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3 Answers
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I'll assume you only want samples consisting of non-negative integers (otherwise its obviously not Poisson). Indeed, without those restrictions, it's pretty trivial - though largely meaningless, because the variance-ratio changes as you change scale. (It makes more sense with counts though.)
I'll also assume you want the $n-1$ form of the sample variance (the unbiased form).
Some examples:
n=2: a) 0 1
b) 1 3
c) 3 6
d) 6 10 (you may discern a pattern here, and yes, it holds in general for n=2)
n=3: a) 0 1 2
b) 2 4 6
c) 4 6 9
n=4: a) 1 1 2 4
b) 3 3 3 7
n=7: 1 1 2 3 4 5 5
n=11: 1 1 1 2 3 3 3 4 5 5 6
These took me a few minutes to construct.
It doesn't really have much significance other than giving a lack of evidence against the Poisson (at least on the basis of the ratio of variance to mean). In no way does it tell you that the data is Poisson.
Edit: For example, here's a sample that is pretty obviously not Poisson (in the sense that the chance that you could end up with a sample like that from a Poisson is reaally small):
n=21: 2 2 2 2 2 2 2 2 2 2 4 6 6 6 6 6 6 6 6 6 6
For starters, all the values are even!
Here's another that's pretty clearly not Poisson:
n=9: 3 10 10 10 10 10 10 14 14
Edit: here's a biggish sample that's not so plainly inconsistent with Poisson:
n=101:
1 2 4 5 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 7 8 8
8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 10 10 10 10 10
10 10 10 10 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11
11 11 12 12 12 12 12 12 12 12 13 13 13 13 14 14 14 14 14 14 15 16 16 17 18
22
... though it's a bit too kurtotic to really be very consistent with a Poisson.
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for any random variable $\xi$, we can introduce a new variable $x=\xi+const$ so that $E[x]=E[\xi+const]=E[\xi]+const$
$Var[x]=Var[\xi+const]=Var[\xi]$
choosing $const=Var[\xi]-E[\xi]$ we can make $E[x]=Var[x]$.
So, you can make an example of your own using this method. There's no significance of this fact. It says that by moving the variable's mean you can make it equal to anything including its variance
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$\begingroup$ The difficulty with this answer is drawing a connection to the Poisson; you can't arbitrarily shift values around with Poisson counts. $\endgroup$– Glen_bCommented Mar 26, 2014 at 5:11
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$\begingroup$ The larger difficulty is that the difference between variance and mean is dimensionally nonsensical unless the variable concerned is dimensionless and unit-free. $\endgroup$– Nick CoxCommented Mar 26, 2014 at 9:43
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$\begingroup$ In the actual Poisson sample mean and variance are not equal usually, unless only by a chance. Distribution itself has this quality. Counts are dimensionless with caveat. Other measurements can be easily transformed like 1 meter to 100cm. However 1 count per second can't be transformed into 60 counts per minute in the sample. Though the Poisson parameter can be. $\endgroup$– AksakalCommented Mar 26, 2014 at 11:36
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3$\begingroup$ I don't think you are taking the dimensional argument seriously; it is not refuted or rebutted by changing units of measurement. Units are matter of convention, but dimensions aren't. $\endgroup$– Nick CoxCommented Mar 26, 2014 at 12:18
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yes variance can be equal to mean in Poisson distribution.
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2$\begingroup$ This doesn't appear to be directly relevant, because the question refers to "data points" and the sample variance, whereas you mention a theoretical distribution. Could you expand your answer to show its relevance? $\endgroup$– whuber ♦Commented Oct 30, 2019 at 18:11