Order statistics of uniform random variables and exponential random variables

Question

Let $(U_{(1)},\dots,U_{(n)})$ be the first order statistic of $n$ i.i.d. uniform random variables in $(0,1)$. Let $(E_i)_i$ be $n$ i.i.d. exponential random variables of parameter 1, also independent from everything else.

Is it possible to write a formula for $\mathbb{P}(E_1 U_{(i)}\geq E_i U_{(1)}, \forall i\geq2)$? Is there a nice formula at least when $n$ goes to infinity?

Answer 1

I suspect you could do something like $$ \int_{x=0}^1 \int_{k=0}^\infty \left(\int_{y=x}^1 (1-e^{-yk/x}) \, dy\right)^{n-1} ne^{-k}\, dk \, dx $$ where $x$ is your $U_{(1)}$, $k$ is your $E_1$ and $y$ represents the higher values of $U_i$. I doubt this is easy to calculate beyond $n=2$ where it gives $\log_e(2) \approx 0.693$.

But I think simulation should work. The probability seems to reduce as $n$ increases, quickly getting below $0.56$ but never below $0.55$. Using R where the errors with a million simulations each should usually be smaller than $\pm 0.001$:

probinequality <- function(n){
   u <- runif(n)
   ei <- rexp(n)
   minu <- min(u)
   argmin <- which(u == min(u)) 
   return(all(ei[argmin] * u >= ei * minu))
   }

set.seed(2023) # for replication - use different seed for different values

sims <- replicate(10^6, probinequality(n=2))
mean(sims)
# 0.693554     # note: 0.693147 is about the true value 

sims <- replicate(10^6, probinequality(n=3))
mean(sims)
# 0.618249

sims <- replicate(10^6, probinequality(n=4))
mean(sims)
# 0.589736

sims <- replicate(10^6, probinequality(n=10))
mean(sims)
# 0.557648

sims <- replicate(10^6, probinequality(n=1000))
mean(sims)
# 0.55486

Answer 2

We can use conditional expectation to transform this probability as follows $$\begin{align} L&:=\mathbb{P}\left(E_i \le E_1\frac{U_{(i)}}{U_{(1)}},\forall i\ge2 \right)\\ &=\color{blue}{\frac{1}{2}}\mathbb{P}\left( \left\{ E_i \le E_1\frac{U_{(i)}}{U_{(1)}},\forall i\ge2 \right\} \cap \color{blue}{\underbrace{\left\{E' \le E_1\frac{U_{(1)}}{U_{(1)}} \right\}}_{\text{this is a trick to use the beautiful formula in $(5)$}}} \right )\\ &=\color{blue}{\frac{1}{2}}\mathbb{E}\left(\mathbb{E}\left(\prod_{\color{blue}{1}\le i \le n}\mathbf{1}_{\left\{E_i \le E_1\frac{U_{(i)}}{U_{(1)}}\right\}} |E_i,(U_{(i)})_{2\le i \le n}\right)\right)\\ &=\color{blue}{\frac{1}{2}}\mathbb{E}\left(\prod_{\color{blue}{1}\le i \le n}\mathbb{P}\left(E_i \le E_1\frac{U_{(i)}}{U_{(1)}} |E_i,(U_{(i)})_{2\le i \le n}\right)\right)\\ &=\color{blue}{\frac{1}{2}}\mathbb{E}\left(\prod_{\color{blue}{1}\le i \le n}\left(1-\exp\left(-E_1\frac{U_{(i)}}{U_{(1)}} \right) \right)\right)\\ &=\color{blue}{\frac{1}{2}}\int_{0\le x \le +\infty}\mathbb{E}\left(\prod_{\color{blue}{1}\le i \le n}\left(1-\exp\left(-x\frac{U_{(i)}}{U_{(1)}} \right) \right)\right)e^{-x}dx\\ &=\color{blue}{\frac{1}{2}}\int_{0\le y \le 1}\mathbb{E}\left(\prod_{\color{blue}{1}\le i \le n}\left(1-y^{\frac{U_{(i)}}{U_{(1)}} } \right)\right)dy \tag{1} \end{align}$$

I doubt we can compute (semi-) analytically $(1)$.

There is a result concerning the distribution of $\left(\frac{U_{(i)}}{ U_{(1)}}\right)_{i=2,...,n}$. From this, we can prove easily that, with $(Z_i)_{i=1,...,n-1}$ iid and following the uniform distribution, we have $$\frac{U_{(i)}}{ U_{(1)}} \stackrel{\mathcal{D}}{=} Z_1^{-1}\cdot Z_2 ^{-1/2}...Z_{i-1}^{-1/{(i-1)}} \hspace{1cm} \forall i=2,...,n \tag{2}$$ We can use $(2)$, then $(1)$ requires $(n+1)$ integrals for an exact calculation by Monte Carlo simulation for example.

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Another approach is to approximate $\left(\frac{U_{(i)}}{ U_{(1)}}\right)_{i=\color{blue}{1},...,n}$ by their expected values. From $(2)$, we can prove easily that $$\mathbb{E}\left(\frac{U_{(i)}}{ U_{(1)}}\right) = i \hspace{1cm} \forall i=\color{blue}{1},...,n$$ then we assume that $$\mathbb{E}\left(\prod_{2\le i \le n}\left(1-y^{\frac{U_{(i)}}{U_{(1)}} } \right)\right) \approx \prod_{2\le i \le n}\left(1-y^{\mathbb{E}\left(\frac{U_{(i)}}{U_{(1)}}\right) } \right)=\prod_{2\le i \le +\infty}\left(1-y^{i } \right) \tag{3}$$

Applying $(3)$ to $(1)$, we have $$L \xrightarrow{n\to+\infty} \frac{1}{2}\int_{0\le y \le 1}\left(\prod_{2\le i \le +\infty}\left(1-y^{i } \right)\right)dy := M\tag{4}$$

The integral $M$ of $(4)$ has closed form expression according to this question $$\int_0^1\prod_{n=1}^\infty(1-x^n)dx=\frac{4\pi\sqrt3}{\sqrt{23}}\frac{\sinh\frac{\pi\sqrt{23}}3}{\cosh\frac{\pi\sqrt{23}}2}\tag{5}$$

By consequence, we can approximate $L$ by

$$\color{red}{L \approx \frac{1}{2}\frac{4\pi\sqrt3}{\sqrt{23}}\frac{\sinh\frac{\pi\sqrt{23}}3}{\cosh\frac{\pi\sqrt{23}}2}}$$