Misalkan Y
Secara intuitif, karena asumsi normalitas, masuk akal untuk mengklaim bahwa E ( Y | ˉ X = ˉ x ) = ˉ x
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 + 1
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Jawaban:
Let XX denote the original sample and ZZ the random vector with entries Zk=Xk−ˉXZk=Xk−X¯ . 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(Y∣ˉX)=ˉX+E(T∣ˉX)=ˉX+E(T)=ˉX.
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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]
with
Σ=(σ2E(|X−v|)[2f(v)]−1E(|X−v|)[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(Yn∣ˉXn=ˉx)=v+ρσvσˉX(ˉx−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(|X−v|)[2f(v)]−11nσ[2f(v)]−1=E(|X−v|)σ
So E(Yn∣ˉXn=ˉx)=v+E(|X−v|)σ[2f(v)]−1σ(ˉx−μ)
We have that 2f(v)=2/σ√2π2f(v)=2/σ2π−−√ due to the symmetry of the normal density so we arrive at
E(Yn∣ˉXn=ˉx)=v+√π2E(|X−μσ|)(ˉx−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(Yn∣ˉXn=ˉx)=v+√π2√2π(ˉx−μ)=v+ˉx−μ=ˉx
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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 x−ˉx=(x1−ˉx,x2−ˉx,…,xn−ˉx)x−x¯=(x1−x¯,x2−x¯,…,xn−x¯) for the vector of residuals. The symmetry assumption on FF implies the distribution of x−ˉxx−x¯ is symmetric about 00 ; that is, when E⊂RnE⊂Rn is any event,
PrF(x−ˉx∈E)=PrF(x−ˉx∈−E).
Applying the generalized result at /stats//a/83887 shows that the median of x−ˉxx−x¯ 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 x−ˉxx−x¯ . Consequently its expectation conditional on ˉxx¯ equals the expectation of x−ˉxx−x¯ 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.
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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.
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