Tuesday, 23 July 2019

probability theory - Convergence of Truncated Expectation of Order Statistics E[Yk:N|Yk:N>v]rightarrowv



Setting
Let (Xi)iN be a set of i.i.d. random variables, with Xi mapping to some interval [a,b].
Let Yk:N be the kth order statistic of this set and v[a,b].
Denote by fX,FX the continuous pdf and the continuous CDF of Xi and by fYk:N the pdf of Yk:N



Quantity of interest
I am interested in the truncated expectation of the order statistic E[Yk:N|Yk:N>v].
This can be written as E[Yk:N|Yk:N>v]=vyfYk:N(y)dyvfYk:N(y)dy.
Conjecture
Computing this quantity in MATLAB, suggests that
E[Yk:N|Yk:N>v]Nv.
Also my intuition is in line with this conjecture: For growing N, the support of fYk:N shrinks to a small region and we can predict E[Yk:N|Yk:N>v] better. Furthermore, the probability of the next value being close to v is large.




However, I am missing a formal proof.
Any ideas?


Answer



We need to assume something. Assume E|X|< and F(v) is increasing, such that for all u>v, F(u)>F(v)



For u>v we have,
P(Yk:n>u|Yk:n>v)=P(Yk:n>u)P(Yk:n>v).



Now P(Yk:n>x) is asking for the probability that out of n tries at most k1 of the Xi is below or equal to x. So if Nn,xBin(F(x),n) (binomial distributed) we have,

P(Yk:n>x)=P(Nn,x<k).
Now this probability is decreasing in x and it is not hard to see that we can write for a fixed k,
P(Nn,x<k)=C(x,n)nk1(1F(x))nk, with $C(x,n)C_0ifF(v) > 0,Hence,$
\frac{P(Y_{k:n}>u)}{P(Y_{k:n}>v)} = \frac{P(N_{n,u}
$$ if F(v)>0, with 0p<1 due to the fact that F(v) is monotonically increasing at v. Hence, this goes to zero as n goes to infinity. If F(v)=0, then we note, E(Yk:n|Yk:n>v)=E(Yk:n) and it enough to observe that still we have $C(x,n)$$
P(Y_{k:n}>u) = P(N_{n,u}$$ Again with 0p<1 due to the fact that F(v) is increasing at v. Again, because the geometric decrease is faster than the polynomial increase in nk1 this goes to zero as n goes to infinity.



This shows the most probability mass lies at v so expectation over any finite region above an u will have a value that goes to zero and because of E|X| is finite, the tail goes to zero and we are left with essentially a delta measure on v and the expectation is indeed v.


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find lim without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...