Örnek varyansını düşündüğümüzde ortaya çıkan yan adım bağımlılıklarına,
(n−1)s2=∑i=1n((Xi−μ)−(x¯−μ))2
=∑i=1n(Xi−μ)2−2∑i=1n((Xi−μ)(x¯−μ))+∑i=1n(x¯−μ)2
ve küçük bir manipülasyondan sonra,
=∑i=1n(Xi−μ)2−n(x¯−μ)2
bu nedenle
n−−√(s2−σ2)=n−−√n−1∑i=1n(Xi−μ)2−n−−√σ2−n−−√n−1n(x¯−μ)2
kurgulama,
n−−√(s2−σ2)=n−−√n−1∑i=1n(Xi−μ)2−n−−√n−1n−1σ2−nn−1n−−√(x¯−μ)2
=nn−−√n−11n∑i=1n(Xi−μ)2−n−−√n−1n−1σ2−nn−1n−−√(x¯−μ)2
=nn−1[n−−√(1n∑i=1n(Xi−μ)2−σ2)]+n−−√n−1σ2−nn−1n−−√(x¯−μ)2
terimi asimptotik olarak birlik haline gelir. √ terimin/(n−1)determinitiktir ven→∞olarak sıfıra gider.n√n−1σ2n→∞
Ayrıca n−−√(x¯−μ)2=[n−−√(x¯−μ)]⋅(x¯−μ). The first component converges in distribution to a Normal, the second convergres in probability to zero. Then by Slutsky's theorem the product converges in probability to zero,
n−−√(x¯−μ)2→p0
We are left with the term
[n−−√(1n∑i=1n(Xi−μ)2−σ2)]
Alerted by a lethal example offered by @whuber in a comment to this answer, we want to make certain that (Xi−μ)2 is not constant. Whuber pointed out that if Xi is a Bernoulli (1/2) then this quantity is a constant. So excluding variables for which this happens (perhaps other dichotomous, not just 0/1 binary?), for the rest we have
E(Xi−μ)2=σ2,Var[(Xi−μ)2]=μ4−σ4
and so the term under investigation is a usual subject matter of the classical Central Limit Theorem, and
n−−√(s2−σ2)→dN(0,μ4−σ4)
Note: the above result of course holds also for normally distributed samples -but in this last case we have also available a finite-sample chi-square distributional result.