Nilai yang diharapkan dari median sampel diberikan mean sampel


16

Misalkan YY menunjukkan median dan misalkan ˉ XX¯ menunjukkan rata-rata, dari sampel acak berukuran n = 2 k + 1n=2k+1 dari distribusi yaitu N ( μ , σ 2 )N(μ,σ2) . Bagaimana saya bisa menghitung E ( Y | ˉ X = ˉ x )E(Y|X¯=x¯) ?

Secara intuitif, karena asumsi normalitas, masuk akal untuk mengklaim bahwa E ( Y | ˉ X = ˉ x ) = ˉ xE(Y|X¯=x¯)=x¯ dan memang itulah jawaban yang benar. Bisakah itu diperlihatkan dengan ketat?

Pikiran awal saya adalah untuk mendekati masalah ini menggunakan distribusi normal bersyarat yang umumnya merupakan hasil yang diketahui. Masalahnya adalah karena saya tidak tahu nilai yang diharapkan dan akibatnya varians dari median, saya harus menghitung mereka yang menggunakan statistik urutan k + 1k+1 . Tapi itu sangat rumit dan saya lebih suka tidak pergi ke sana kecuali saya benar-benar harus.


2
Saya percaya ini adalah konsekuensi langsung dari generalisasi yang baru saja saya posting di stats.stackexchange.com/a/83887 . Distribusi residu x i - ˉ xxix¯ jelas simetris sekitar 00 , di mana mediannya memiliki distribusi simetris, sehingga rata-ratanya adalah nol. Oleh karena itu harapan median itu sendiri (bukan hanya dari residu) sama dengan 0 + E ( ˉ X | ˉ X = ˉ x ) = ˉ x  0+E(X¯ | X¯=x¯)=x¯ , QED.
whuber

@whuber Maaf, residu?
JohnK

Saya mendefinisikan mereka dalam komentar saya: mereka adalah perbedaan antara masing-masing x ixi dan rata-rata mereka.
whuber

@whuber Tidak, saya mengerti tetapi saya masih berusaha memahami bagaimana jawaban Anda yang lain berhubungan dengan pertanyaan saya dan bagaimana tepatnya harapan yang Anda gunakan bekerja.
JohnK

2
@whuber Oke, maka tolong perbaiki saya. Jika saya salah, E ( Y | ˉ X ) = E ( ˉ X | ˉ X ) + E ( Y - ˉ X | ˉ X ) Dan sekarang istilah kedua adalah nol karena median simetris di sekitar ˉ x . Oleh karena itu, ekspektasinya berkurang menjadi ˉ xE(Y|X¯)=E(X¯|X¯)+E(YX¯|X¯)x¯x¯
JohnK

Jawaban:


7

Let XX denote the original sample and ZZ the random vector with entries Zk=XkˉXZk=XkX¯. Then ZZ is normal centered (but its entries are not independent, as can be seen from the fact that their sum is zero with full probability). As a linear functional of XX, the vector (Z,ˉX)(Z,X¯) is normal hence the computation of its covariance matrix suffices to show that ZZ is independent of ˉXX¯.

Turning to YY, one sees that Y=ˉX+TY=X¯+T where TT is the median of ZZ. In particular, TT depends on ZZ only hence TT is independent of ˉXX¯, and the distribution of ZZ is symmetric hence TT is centered.

Finally, E(YˉX)=ˉX+E(TˉX)=ˉX+E(T)=ˉX.

E(YX¯)=X¯+E(TX¯)=X¯+E(T)=X¯.

Thank you, this was asked almost a year ago and I am very glad that someone finally cleared it up.
JohnK

7

The sample median is an order statistic and has a non-normal distribution, so the joint finite-sample distribution of sample median and sample mean (which has a normal distribution) would not be bivariate normal. Resorting to approximations, asymptotically the following holds (see my answer here):

n[(ˉXnYn)(μv)]LN[(00),Σ]

n[(X¯nYn)(μv)]LN[(00),Σ]

with

Σ=(σ2E(|Xv|)[2f(v)]1E(|Xv|)[2f(v)]1[2f(v)]2)

Σ=(σ2E(|Xv|)[2f(v)]1E(|Xv|)[2f(v)]1[2f(v)]2)

where ˉXnX¯n is the sample mean and μμ the population mean, YnYn is the sample median and vv the population median, f()f() is the probability density of the random variables involved and σ2σ2 is the variance.

So approximately for large samples, their joint distribution is bivariate normal, so we have that

E(YnˉXn=ˉx)=v+ρσvσˉX(ˉxμ)

E(YnX¯n=x¯)=v+ρσvσX¯(x¯μ)

where ρρ is the correlation coefficient.

Manipulating the asymptotic distribution to become the approximate large-sample joint distribution of sample mean and sample median (and not of the standardized quantities), we have ρ=1nE(|Xv|)[2f(v)]11nσ[2f(v)]1=E(|Xv|)σ

ρ=1nE(|Xv|)[2f(v)]11nσ[2f(v)]1=E(|Xv|)σ

So E(YnˉXn=ˉx)=v+E(|Xv|)σ[2f(v)]1σ(ˉxμ)

E(YnX¯n=x¯)=v+E(|Xv|)σ[2f(v)]1σ(x¯μ)

We have that 2f(v)=2/σ2π2f(v)=2/σ2π due to the symmetry of the normal density so we arrive at

E(YnˉXn=ˉx)=v+π2E(|Xμσ|)(ˉxμ)

E(YnX¯n=x¯)=v+π2E(Xμσ)(x¯μ)

where we have used v=μv=μ. Now the standardized variable is a standard normal, so its absolute value is a half-normal distribution with expected value equal to 2/π2/π (since the underlying variance is unity). So

E(YnˉXn=ˉx)=v+π22π(ˉxμ)=v+ˉxμ=ˉx

E(YnX¯n=x¯)=v+π22π(x¯μ)=v+x¯μ=x¯

2
As always, nice answer +1. However, since we have no information about the sample size, the asymptotic distribution might not hold. If there is no way to obtain the exact distribution though, I suppose I'll have to make do. Thank you very much.
JohnK

6

The answer is ˉxx¯.

Let x=(x1,x2,,xn)x=(x1,x2,,xn) have a multivariate distribution FF for which all the marginals are symmetric about a common value μμ. (It does not matter whether they are independent or even are identically distributed.) Define ˉxx¯ to be the arithmetic mean of the xi,xi, ˉx=(x1+x2++xn)/nx¯=(x1+x2++xn)/n and write xˉx=(x1ˉx,x2ˉx,,xnˉx)xx¯=(x1x¯,x2x¯,,xnx¯) for the vector of residuals. The symmetry assumption on FF implies the distribution of xˉxxx¯ is symmetric about 00; that is, when ERnERn is any event,

PrF(xˉxE)=PrF(xˉxE).

PrF(xx¯E)=PrF(xx¯E).

Applying the generalized result at /stats//a/83887 shows that the median of xˉxxx¯ has a symmetric distribution about 00. Assuming its expectation exists (which is certainly the case when the marginal distributions of the xixi are Normal), that expectation has to be 00 (because the symmetry implies it equals its own negative).

Now since subtracting the same value ˉxx¯ from each of a set of values does not change their order, YY (the median of the xixi) equals ˉxx¯ plus the median of xˉxxx¯. Consequently its expectation conditional on ˉxx¯ equals the expectation of xˉxxx¯ conditional on ˉx, plus E(ˉx | ˉx). The latter obviously is ˉx whereas the former is 0 because the unconditional expectation is 0. Their sum is ˉx, QED.


Thank you for posting it as a full answer. I now understand the essence of your argument but I might ping you if something is still unclear.
JohnK

5
JohnK, I need to alert you to be cautious. A counterexample to this argument has been brought to my attention. I have encouraged its originator to post it here for further discussion, but briefly it concerns a discrete bivariate distribution with symmetric marginals but asymmetric conditional marginals. Its existence points to a flawed deduction early in my argument. I currently hope that the argument might be rescued by imposing stronger conditions on the xi, but my attention is presently focused elsewhere and I might not get to think about this for awhile.
whuber

4
In the meantime I would encourage you to unaccept this answer. I would ordinarily delete any answer of mine known to be incorrect, but (as you might be able to tell) I like solutions based on first principles rather than detailed calculations, so I hope this argument can be rescued. I therefore intend to leave it open for criticism and improvement (and therefore made it CW); let the votes fall as they may.
whuber

Of course, thanks for letting me know. We will discuss it further when you have time. In the meantime I will settle for the asymptotic argument proposed by @Alecos Papadopoulos.
JohnK

6

This is simpler than the above answers make it. The sample mean is a complete and sufficient statistic (when the variance is known, but our results do not depend on the variance, hence will be valid also in the situation when the variance is unknown). Then the Rao-Blackwell together with the Lehmann-Scheffe theorems (see wikipedia ...) will imply that the conditional expectation of the median, given the arithmetic mean, is the unique minimum variance unbiased estimator of the expectation μ. But we know that is the arithmetic mean, hence the result follows.

We did also use that the median is an unbiased estimator, which follows from symmetry.


1
By symmetry E[Y]=μ, indeed. Then from these two theorems we know that E[Y|ˉX] is the Unique Minimum Variance Unbiased Estimator for μ which we already know to be equal to ˉX. This is a brilliant answer, thank you very much. I would have marked it as the correct one, had I not done that already for another answer.
JohnK
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