10.2. Sample Median and the Sign Test 583whose solution is the sample medianQ 2 , (4.4.4).
Because our observations are continuous random variables, we have the identity∑ni=1sgn(Xi−θ)=2S(θ)−n.Hence the sample median also solvesS(θ)≈n/2. Consider again Figure 10.2.2.
Imaginen/2 on the vertical axis. This is halfway in the total drop ofS(θ), fromn
to 0. The order statistic on the horizontal axis corresponding ton/2 is essentially
the sample median (middle order statistic). In terms of testing, this last equation
says that, based on the data, the sample median is the “most acceptable”hypothesis,
becausen/2 is the null expected value of the test statistic. We often think of this
as estimation by the inversion of a test.
We now sketch the asymptotic distribution of the sample median. Assume with-
out loss of generality that the true median ofXiis 0. Suppose−∞<x<∞.
Using the fact thatS(θ) is nonincreasing and the identityS(θ)≈n/2, we have the
following equivalences:
{√
nQ 2 ≤x}⇔{
Q 2 ≤x
√
n}
⇔{
S(
x
√
n)
≤n
2}
.Hence we have
P 0 (√
nQ 2 ≤x)=P 0[
S(
x
√
n)
≤n
2]= P−x/√n[
S(0)≤n
2]= P−x/√n[
S(0)−(n/2)
√
n/ 2≤ 0]→ Φ(0−xτS−^1 )=P(τSZ≤x),whereZhas a standard normal distribution, Notice that the limit was obtained
by invoking the Asymptotic Power Lemma withα=0.5 and hencezα=0. Rear-
ranging the last term earlier, we obtain the asymptotic distribution of the sample
median, which we state as a theorem:
Theorem 10.2.3. For the random sample X 1 ,X 2 ,...,Xn, assume that Model
(10.2.1) holds. Suppose thatf(0)> 0 .LetQ 2 denote the sample median. Then
√
n(Q 2 −θ)→N(0,τS^2 ), (10.2.35)whereτS=(2f(0))−^1.
In Section 6.2 we defined the ARE between two estimators to be the reciprocal
of their asymptotic variances. For the sample median and mean, this is the same
ratio as that based on sample size determinations of their respective tests given
earlier in expression (10.2.27).