# Convergence of Truncated Expectation of Order Statistics $E[Y_{k:N}|Y_{k:N}>v]\rightarrow v$

Setting
Let $(X_i)_{i\leq N}$ be a set of i.i.d. random variables, with $X_i$ mapping to some interval $[a,b]$.
Let $Y_{k:N}$ be the $k$th order statistic of this set and $v\in[a,b]$.
Denote by $f_X,F_X$ the continuous pdf and the continuous CDF of $X_i$ and by $f_{Y_{k:N}}$ the pdf of $Y_{k:N}$

Quantity of interest
I am interested in the truncated expectation of the order statistic $$E[Y_{k:N}|Y_{k:N}>v].$$ This can be written as $$E[Y_{k:N}|Y_{k:N}>v]=\frac{\int_v^\infty yf_{Y_{k:N}}(y)dy}{\int_v^\infty f_{Y_{k:N}}(y)dy}.$$ Conjecture
Computing this quantity in MATLAB, suggests that $$E[Y_{k:N}|Y_{k:N}>v]\underset{N\rightarrow\infty}{\rightarrow}v.$$ Also my intuition is in line with this conjecture: For growing $N$, the support of $f_{Y_{k:N}}$ shrinks to a small region and we can predict $E[Y_{k:N}|Y_{k:N}>v]$ better. Furthermore, the probability of the next value being close to $v$ is large.

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

• You need to assume $X_1$ has a continuous CDF. It is not necessarily true otherwise, for example take $X_i$ Bernoulli $\{0,1\}$ and $v=1/2$. Then $E[Y_{k:n}|Y_{k:n}>1/2]=1$ for all $k, n$. – Michael May 24 '18 at 17:05
• Here is one way to go (for the continuous case): Suppose that for all $\epsilon>0$ we can prove $\lim_{n\rightarrow\infty}P[Y_{k:n} \in (v, v+\epsilon]|Y_{k:n}>v]=1$. Can you prove $\lim_{n\rightarrow\infty} E[Y_{k:n}|Y_{k:n}>v]=v$? If so, then it remains only to prove that limit. – Michael May 24 '18 at 17:15
• By the way, how did you simulate this in matlab? It is nontrivial. For example if you assume $X_i$ are uniform over $[0,1]$ and let $v=1/2$, when $n$ is siginificantly larger than $k$, it will take forever to randomly generate instances of $\{X_1, ..., X_n\}$ for which $Y_{k:n}>1/2$ (since that is an extremely rare event when $n$ is large). So likely you did something else. It seems easier to do the numerical integration of that PDF than to simulate. – Michael May 24 '18 at 17:25
• You can write $E[Y|A] = E[Y|Y \leq v+\epsilon, A]P[Y\leq v+\epsilon|A] + E[Y|Y>v+\epsilon, A]P[Y>v+\epsilon|A]$. Stefan is correct that when I wrote "continuous CDF" I should have written "a distribution that gives positive probability to any positive-sized interval in $[a,b]$" which holds when $F_X(x)$ is increasing over $[a,b]$. You can modify my above Bernoulli counter-example to also work with $X$ uniform over the disjoint intervals $[0, 0.1] \cup [0.9,1]$ and that has a continuous CDF. – Michael May 25 '18 at 14:27
• I think Stefan did a good job on this. Here is a way to proceed in my way, which may overlap some of his approaches as well. Assume $n>2k$. Let $A_i$ be the event that exactly $i$ of the random variables are larger than $v$. So then $$P[Y_{k:n}>v\} = \cup_{i=n-(k-1)}^n A_i$$ and $$\cup_{i=n-(k-1)}^n\{ A_i \cap \{\mbox{at least k are in (v, v+\epsilon]}\}\} \subseteq \{Y_{k:n} \in (v, v+\epsilon]\}$$ – Michael May 26 '18 at 23:09

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

For $u > v$ we have, $$P(Y_{k:n} > u | Y_{k:n} > v) = \frac{P(Y_{k:n}>u)}{P(Y_{k:n}>v)}.$$

Now $P(Y_{k:n}>x)$ is asking for the probability that out of $n$ tries at most $k-1$ of the $X_i$ is below or equal to $x$. So if $N_{n,x} \in Bin(F(x),n)$ (binomial distributed) we have, $$P(Y_{k:n}>x) = P(N_{n,x} < k).$$ Now this probability is decreasing in $x$ and it is not hard to see that we can write for a fixed $k$, $$P(N_{n,x} < k) = C(x,n)n^{k-1}(1-F(x))^{n-k},$$ with $C(x,n)<C_1$ and $C(v,n)>C_0$ if $F(v) > 0$, Hence, $$\frac{P(Y_{k:n}>u)}{P(Y_{k:n}>v)} = \frac{P(N_{n,u}<k)}{P(N_{n,v}<k)}=\frac{C(u,n)n^{k-1}(1-F(u))^{n-k+1}}{C(v,n)n^{k-1}(1-F(v))^{n-k+1}}=\frac{C(u,n)}{C(v,n)}p^{n-k+1} < \frac{C_1}{C_0}p^{n-k+1},$$ if $F(v) > 0$, with $0\leq p <1$ 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, $E(Y_{k:n}|Y_{k:n}>v)=E(Y_{k:n})$ and it enough to observe that still we have $C(x,n)<C_1$ and $$P(Y_{k:n}>u) = P(N_{n,u}<k) = C(u,n)n^{k-1}(1-F(u))^{n-k+1} < C_1 n^{k-1} p^{n-k+1}$$ Again with $0\leq p <1$ due to the fact that $F(v)$ is increasing at $v$. Again, because the geometric decrease is faster than the polynomial increase in $n^{k-1}$ 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 $v$.

• Thanks for your answer. I am not sure, how you come to to the representation of $P(N_{n,x}<k)$ as a product of constants and polynomials. I think $k(n)$ does not behave like a polynomial but grows larger as it involves factorials. Could you please elaborate on that? I tried rewriting $\sum_{j=0}^{k-1}\binom{n}{j}F(x)^j[1-F(x)]^{n-j}$ into the form you proposed, but I can't see the polynomial part. Furthermore, I think it should be $[1-F(x)]^{n-k-1}$ in the representation, as we are only summing up to $k-1$. What do you think? – MathProb May 25 '18 at 9:31
• Yes it is k-1, and n - (k - 1) = n - k + 1 else we sum up to k+1. note that $\binom{n}{k}=n(n-1) \cdots (n-k+2) / k!$ which is a polynomial (actually k is $n^{k-1}$) – Stefan May 25 '18 at 10:03
• oops the expansion is to (n-k+1) and not (n-k+2) and I replaced $k(n)$ with $n^{k-1}$. – Stefan May 25 '18 at 10:10
• In the sum of probabilities the powers of $(1-F)$ is $n,n-1,\ldots ,n-k+1$ you don't want to divide by $1-F$ so the best common factor is $n-k+1$ – Stefan May 25 '18 at 10:19
• consider the n part in the sum, the first term has 1, the second one is $n$, the third one is a poynomial of the second order etc. the highest order is $n^{k-1}$ bringing that out that, the factors goes to zero or is bounded below and above (the last term) this is all bounded by above F and (1-F) are all below 1 and hence any power is below 1 and we get an upper bound. For the lower bound the last term is independant of $1-F$ and $F(u)>F(v)>0$ do you follow? – Stefan May 25 '18 at 13:15