포아송 랜덤 변수의 반올림 평균 분포는 무엇입니까?


20

변수 λ 1 , λ 2 , , λ n 으로 포아송 분포 된 랜덤 변수 X1,X2,,Xn 있으면 Y = n i = 1 X i 의 분포는 무엇입니까?λ1,λ2,,λnY=i=1nXin(즉, 평균의 정수 층)?

Poissons의 합계도 Poisson이지만 위의 경우와 동일한 지 여부를 결정할 통계에 대해서는 확신이 없습니다.


@amoeba 제목이 실제로 "반올림"이 아니기 때문에 제목 편집을 롤백했습니다. 카디널의 이전 편집은 정확하지는 않지만 정확하기 때문에 선호되는 것 같습니다.
whuber

@whuber 좋아. 이 편집을 할 때 주저하고 있었지만 현재 제목이 여기서 주요 난이도를 암시하지 않기 때문에 "반올림"이라는 단어를 포함하기로 결정했습니다. 적절한 용어는 "반올림"이어야합니다. "아마도 포아송 랜덤 변수의 평균 분포는 반올림 되었습니까?" 비록 내가 약간 성가신 소리를 인정할지라도.
amoeba는

@amoeba 추가 편집은 물론 환영합니다!
whuber

답변:


27

질문 일반화의 분배를 요청 Y=X/m 분포 X 공지 자연수에지지된다. (문제에서 Xλ=λ1+λ2++λn 의 포아송 분포를 갖습니다 m=n.)

분포 Y 쉽게의 분포에 의해 결정되는 mY 확률 생성 함수 (PGF)의 PGF의 관점에서 결정될 수있다, X . 다음은 파생 개요입니다.


쓰기 의 PGF 대한 X , 여기서 (정의에 의해) P N = ( X = N ) . m Y 는 pgf q 가 다음 과 같은 방식으로 X 로 구성됩니다p(x)=p0+p1x++pnxn+Xpn=Pr(X=n)mYXq

q(x)=(p0+p1++pm1)+(pm+pm+1++p2m1)xm++(pnm+pnm+1++p(n+1)m1)xnm+.

이것은 절대적으로 수렴하기 때문에 , 우리는 항을 형태의 합으로 재 배열 할 수 있습니다|x|1

Dm,tp(x)=pt+pt+mxm++pt+nmxnm+

for t=0,1,,m1. The power series of the functions xtDm,tp consist of every mth term of the series of p starting with the tth: this is sometimes called a decimation of p. Google searches presently don't turn up much useful information on decimations, so for completeness, here's a derivation of a formula.

하자 어떤 원시적 m 통일의 루트; 예를 들어, ω = exp ( 2 i π / m )를 사용하십시오 . 그런 다음 ω m = 1m - 1 j = 0에서 ω j = 0 을 따릅니다.ωmthω=exp(2iπ/m)ωm=1j=0m1ωj=0

xtDm,tp(x)=1mj=0m1ωtjp(x/ωj).

To see this, note that the operator xtDm,t is linear, so it suffices to check the formula on the basis {1,x,x2,,xn,}. Applying the right hand side to xn gives

xtDm,t[xn]=1mj=0m1ωtjxnωnj=xnmj=0m1ω(tn)j.

When t and n differ by a multiple of m, each term in the sum equals 1 and we obtain xn. Otherwise, the terms cycle through powers of ωtn and these sum to zero. Whence this operator preserves all powers of x congruent to t modulo m and kills all the others: it is precisely the desired projection.

A formula for q follows readily by changing the order of summation and recognizing one of the sums as geometric, thereby writing it in closed form:

q(x)=t=0m1(Dm,t[p])(x)=t=0m1xt1mj=0m1ωtjp(ωjx)=1mj=0m1p(ωjx)t=0m1(ωj/x)t=x(1xm)mj=0m1p(ωjx)xωj.

For example, the pgf of a Poisson distribution of parameter λ is p(x)=exp(λ(x1)). With m=2, ω=1 and the pgf of 2Y will be

q(x)=x(1x2)2j=021p((1)jx)x(1)j=x1/x2(exp(λ(x1))x1+exp(λ(x1))x+1)=exp(λ)(sinh(λx)x+cosh(λx)).

One use of this approach is to compute moments of X and mY. The value of the kth derivative of the pgf evaluated at x=1 is the kth factorial moment. The kth moment is a linear combination of the first k factorial moments. Using these observations we find, for instance, that for a Poisson distributed X, its mean (which is the first factorial moment) equals λ, the mean of 2(X/2) equals λ12+12e2λ, and the mean of 3(X/3) equals λ1+e3λ/2(sin(3λ2)3+cos(3λ2)):

Means

The means for m=1,2,3 are shown in blue, red, and yellow, respectively, as functions of λ: asymptotically, the mean drops by (m1)/2 compared to the original Poisson mean.

Similar formulas for the variances can be obtained. (They get messy as m rises and so are omitted. One thing they definitively establish is that when m>1 no multiple of Y is Poisson: it does not have the characteristic equality of mean and variance) Here is a plot of the variances as a function of λ for m=1,2,3:

Variances

It is interesting that for larger values of λ the variances increase. Intuitively, this is due to two competing phenomena: the floor function is effectively binning groups of values that originally were distinct; this must cause the variance to decrease. At the same time, as we have seen, the means are changing, too (because each bin is represented by its smallest value); this must cause a term equal to the square of the difference of means to be added back. The increase in variance for large λ becomes larger with larger values of m.

The behavior of the variance of mY with m is surprisingly complex. Let's end with a quick simulation (in R) showing what it can do. The plots show the difference between the variance of mX/m and the variance of X for Poisson distributed X with various values of λ ranging from 1 through 5000. In all cases the plots appear to have reached their asymptotic values at the right.

set.seed(17)
par(mfrow=c(3,4))
temp <- sapply(c(1,2,5,10,20,50,100,200,500,1000,2000,5000), function(lambda) {
  x <- rpois(20000, lambda)
  v <- sapply(1:floor(lambda + 4*sqrt(lambda)), 
              function(m) var(floor(x/m)*m) - var(x))
  plot(v, type="l", xlab="", ylab="Increased variance", 
       main=toString(lambda), cex.main=.85, col="Blue", lwd=2)
})

Plots


1
This is a great answer! It will probably take me some time to digest :)
Lubo Antonov

1
and that is why I said "Using the floor function ... affects the variance slightly too though in a more complicated manner."
Henry

1
+1 Thanks for the detailed answer. There certainly are complicated ways in which the floor function affects the variance.
Dilip Sarwate

1
+1 for simulation in R with code --- this is a very nice example of using sapply() for simulation. Thanks.
Assad Ebrahim

1
@Roberto Thank you. However, the distinction between "x" and "s", being purely a matter of notation, is utterly trivial and of no mathematical or statistical import.
whuber

12

As Michael Chernick says, if the individual random variables are independent then the the sum is Poisson with parameter (mean and variance) i=1nλi which you might call λ.

Dividing by n reduces the mean to λ/n and variance λ/n2 so the variance will be less than the equivalent Poisson distribution. As Michael says, not all values will be integers.

Using the floor function reduces the mean slightly, by about 1212n, and affects the variance slightly too though in a more complicated manner. Although you have integer values, the variance will still be substantially less than the mean and so you will have a narrower distribution than the Poisson.


thanks, not a result I can use, but at least I know now :)
Lubo Antonov

If the lambdas are not all equal, shouldn't the result be more like a negative binomial than a Poisson (ignoring the non-integer part for the moment)? What am I missing here?
gung - Reinstate Monica

2
@gung: You are missing the point that the individual λi only affect the distribution through their sum and how many there are. It doesn't matter what particular values they take: λ1=1,λ2=2,λ3=9 will give the same result as λ1=4,λ2=4,λ3=4.
Henry

10

The probability mass function of the average of n independent Poisson random variables can be written down explicitly, though the answer might not help you very much. As Michael Chernick noted in comments on his own answer, the sum iXi of independent Poisson random variables Xi with respective parameters λi is a Poisson random variable with parameter λ=iλi. Hence,

P{i=1nXi=k}=exp(λ)λkk!,  k=0,1,2,,
Thus, Y^=n1i=1nXi is a random variable taking on value k/n with probability exp(λ)λkk!. Note that Y^ is not an integer-valued random variable (though it does take on uniformly-spaced rational values). It follows easily that Y=Y^ is an integer-valued random variable taking on value m with probability
P{Y=m}=P{1ni=1nXi=m}=exp(λ)i=0n1λmn+i(mn+i)!,  m=0,1,2,,
This is not the probability mass function of a Poisson random variable. Formulas for the mean and variance can be written down using this probability mass function, but they don't obviously lead to nice simple answers in terms of λ and n. Approximate values can be obtained as pointed out by Henry.

+1 There are closed formulas for the moments of Y, though.
whuber

Thanks for the rigorous formulation! Any chance you'd like to take a crack at the formulas for mean and variance?
Lubo Antonov

2
Perhaps @whuber will post a link (or a citation of a book or journal article) where the closed-form formulas for the moments can be found, or will write an answer giving the formulas themselves, with or without a detailed derivation.
Dilip Sarwate

@Dilip My claim about closed formulas was not based on anything published, so I have posted a separate reply indicating what I had in mind and how it might be used to understand this situation.
whuber

3

Y will not be Poisson. Note that Poisson random variables take on non negative integer values. Once you divide by a constant you create a random variable that can have non-integer values. It will still have the shape of the Poisson. It is just that the discrete probabilities may occur at non-integer points.


That makes sense, but what if Y is actually discrete, for example the floor of the average? Would that make it Poisson?
Lubo Antonov

@lucas1024 I don't think so but I am not sure.
Michael R. Chernick

The shape of the sum Xi is definetevely Poisson, right? its mean and variance are identical as well. Isn't there something like an scaled Poisson ? Y is just a poisson variable (the sum) that is scaled by n1
JDav

@JDav The sum is Poisson with the rate parameter equal to the sum of the individual rate parameters. But the OP scales by 1/n and then wants to truncate the the the integer just below Y. I don't know exactly what that does to the distribution.
Michael R. Chernick

My previous comment assumed independence.
Michael R. Chernick
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