The expected ratio of 2nd largest to largest number in uniform distribution

Question

Just some random question that I think of.

Suppose $X_1, \ldots, X_n\sim U[0,1]$ follows, independently, uniform distribution from $0$ to $1$. What is the expected ratio of the second largest number over the largest number. In other words, $x_p:=\arg\max_iX_i$ and $x_q=\arg\max_{i\neq p} X_i$. What is $\mathbb{E}[\frac{x_q}{x_p}]$

Answer 1

$\require{begingroup}\begingroup\renewcommand{\dd}[1]{\,\mathrm{d}#1}$Denote the common density as $f$ and cumulative distribution function as $F$ for the i.i.d. $X_i~$, then the joint density of two of the order statistics $X_{(j)}$ and $X_{(k)}$, where $1 \leq j < k \leq n$, is well known: $$\begin{aligned} & \phantom{ {}= {} } f_{jk}( X_{(j)} = r,\, X_{(k)} = s ) \\ &= \frac{ n! }{ (j-1)!(k-j-1)! (n-k)!} F(r)^{j-1} f(r) \bigl( F(s) - F(r) \bigr)^{k-j-1} f(s) \bigl(1 - F(s) \bigr)^{n-k} \mathbb{1}_{0 \leq r \leq s} \end{aligned}$$ where $\mathbb{1}_{\text{blah}}$ is the indicator function (equivalently the Iverson bracket).

Here we have the uniform density $f(t) = \mathbb{1}_{0 \leq t \leq 1}$ and the cumulative distribution function $F(t) = t \cdot \mathbb{1}_{0 \leq t \leq 1}~$. The index for the $2$nd largest $x_q$ it is $j = n-1$, and for the max $x_p$ it is $k = n$. The joint density is thus $$f_{qp}( x_q = r,\, x_p = s ) = \frac{ n! }{ (n-2)!\, 0!\, 0!} r^{n-2} \mathbb{1}_{0 \leq r \leq s} $$ Wikipedia also has a section specifically for the uniform distribution.

The desired expectation "$2$nd over max" is an integration over a triangular region within the unit square $(r, s) \in [0,1]^2$. $$\begin{aligned} \mathbb{E}\left[ \frac{x_q}{x_p} \right] &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{r}s f_{qp}(r,s) \dd{r} \dd{s} \\ &= \int_{s = 0}^1 \int_{r = 0}^s \frac{ r }s \cdot n(n-1) r^{n-2} \dd{r} \dd{s} \\ &= n(n-1) \int_{s = 0}^1 \frac1s \cdot \left[ \int_{r = 0}^s r^{n-1} \dd{r} \right] \dd{s} \\ &= (n - 1) \int_{s = 0}^1 \frac1s \cdot s^n \dd{s} \\ &= (n - 1) \int_{s = 0}^1 s^{n-1} \dd{s}\\ &= \frac{ n - 1}n \end{aligned}$$ Note that marginally the individual expectation is $\mathbb{E}[x_q] = \frac{ n - 1}{n + 1}$ for the $2$nd largest and $\mathbb{E}[x_p] = \frac{ n }{n + 1}$ for the max. This phenomenon that "jointly-expected-ratio is the same as ratio-of-the-marginal-expectations" is in general not true. The uniform distribution is special in many regards, including this one.$\endgroup$

Answer 2

Let $Y$ be the largest of $X_1,\ldots,X_n$. wlog suppose $Y = X_n$. Then conditional on $Y=y$, $X_1,\ldots,X_{n-1}$ are uniformly distributed on $[0,y]$. They break the interval $[0,y]$ into $n$ pieces of average size $y/n$. Conditional on $Y = y$, the expectation of the second largest $X_i$ is then $y-y/n$. Thus the expectation of the second largest divided by the largest is $(n-1)/n$.