Expected Value of Difference between Sequences of Independent Trials

Question

I am trying to prove the following:

Let $X_{1}...X_{n}$ be i.i.d. random variables which admit a density $f$.

Let $Y_{n}=\min(X_{i},i=1,...,n)$ and let $Z_{n}$ be the number of r.v. $X_{i}$ which exceed $Y_{n}$ by more than 2. How do I evaluate $E(Z_{n})?$.

Now assume that each $X_{i}$ have the uniform distribution on [0,4]. How do I find $\lim_{n \to \infty} E(Z_{n})?$.

Answer 1

Let $Y_{-i,n}=\min_{j\ne i}\{Y_j\}$. First, $Z_n=\sum_{i=1}^n1\{X_i>Y_n+2\}$. Taking expectations on both sides we get,

$$ \mathbb{E}Z_n=\sum_{i=1}^n\mathbb{P}\{X_i>Y_n+2\} \\ =\sum_{i=1}^n\mathbb{P}\{X_i>Y_n+2,X_i\le Y_{-i,n}\}+\mathbb{P}\{X_i>Y_n+2,X_i> Y_{-i,n}\} \\ =\sum_{i=1}^n\mathbb{P}\{X_i>Y_{-i,n}+2\}=n\mathbb{P}\{X_1>Y_{-1,n}+2\}, $$

where the last equality follows by symmetry. Now, $$ \mathbb{P}\{X_1>Y_{-1,n}+2\}=\mathbb{E}\left[\mathbb{P}\{X_1>Y_{-1,n}+2\mid Y_{-1,n}\}\right] $$

and since $X_1$ is independent of $Y_{-1,n}$,

$$ \mathbb{P}\{X_1>Y_{-1,n}\mid Y_{-1,n}\}=g(Y_{-1,n}), $$

where $g(y)=\mathbb{P}\{X_1>y+2\}$. Consequently, since $f_{Y_{-1,n}}(y)=(n-1)f(y)\left(1-F(y)\right)^{n-2}$, where $F$ is the CDF of $X$,

$$ \mathbb{E}Z_n=n(n-1)\int g(y)f(y)(1-F(y))^{n-2}dy. \quad(*) $$

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For $X_1\sim U[0,4]$, $g(y)=1-\frac{y+2}{4}$ for $0\le y\le 2$, and $0$, otherwise. Thus, using $(*)$, we get

$$ \mathbb{E}Z_n=n(n-1)\int_0^2 \left(1-\frac{y+2}{4}\right)\frac{1}{4}\left(1-\frac{y}{4}\right)^{n-2}dy=\frac{n}{2}-1+\frac{1}{2^n}. $$

Answer 2

Let $F$ be the CDF of the $X$'s and $F_n$ that of the $Y_n$'s. Then $$ F_n(y)=P(Y_n<y)=1-P(Y_n>y)=1-P(X_1>y,\, \ldots Y_n\geq y)=1-(1-F(y))^n $$ and the density of $Y_n$ is $$ f_n(y)= \frac{d}{dy}F_n(y)= n\,(1-F(y))^{n-1}\,f(y). $$ You want to compute $$ E(Z_n)=\int_{-\infty}^{\infty}E(Z_n|Y_n=y)\, f_n(y)\, dy $$ where $E(Z_n|Y_n=y)$ is the expected number of $X_i$'s that are $>y+2$, i.e. $$ E(Z_n|Y_n=y) =(n-1)\, \frac{P(X_i>y+2)}{P(X_i>y)} =(n-1)\, \frac{1-F(y+2)}{1-F(y)} $$ (putting it all together yields a nasty long expression).

In the simpler case where the $X_i$'s are uniform on $[0,4]$, you get \begin{align*} f(x) & = 1/4\quad (0<x<4)\\ F(x) & = x/4\quad (0<x<4)\\ f_n(y) & = \frac{n}{4} \, (1-y/4)^{n-1}\quad (0<x<4)\\ E(Z_n|Y_n=y) & =(n-1)\, \frac{1-(y+2)/4}{1-y/4} =(n-1)\, \frac{y-2}{y-4}\quad (0<x<4)\\ E(Z_n) & = \int_0^2 (n-1)\, \frac{y-2}{y-4} \, \frac{n}{4}\,(1-y/4)^{n-1}\, dy = \frac{n}{2}-1+ \frac{1}{2^n}. \end{align*} Therefore (not surprisingly) $\lim_{n\to\infty} Z_n=\infty$.