Solved – Show convergence of the first order statistic of independent uniform$(0,n)$ distributed random variables

convergencemathematical-statisticsorder-statisticsuniform distribution

Let $X_1, X_2, \ldots, X_n$ be independent random variables, where $X_k \sim Unif(0,k)$.

Let $Y_n = X_{(1)} =\min\{X_1, X_2, \ldots, X_n\}$

Show that $Y_n$ converges in distribution to $Y \sim exp(1)$ (the exponential distribution with mean 1).

I will let $F_k(y) = \mathbb{P}(X_k \leq y)$ be the CDF of $X_k$.
Then, I know the following:
$$F_{Y_n}(y) = \mathbb{P}(\text{At least one } X_k \leq y) = 1-\prod_{k = 1}^n(1-F_k(y)).$$

As far as I can tell, this can be rewritten as
$$F_{Y_n}(y) = \cases{0~~\text{ if } y < 0\\\\ 1-\prod_{k =1}^n(1-y/k)~~\text{ if } 0 \leq y < 1\\\\1 ~~\text{ if } y \geq 1}$$

If $y \geq 1$, then $F_1(y) = 1$, which makes the product $\prod_{k = 1}^n(1-F_k(y)) = 0.$

This is where I have been stuck now. How can I proceed to show that $F_{Y_n}(y) \rightarrow Y \sim \text{exp}(1)$ in distribution?

ADDED As we have somewhat concluded via the comments and answers, it seems that perhaps each $X_k$ is mean to be independent AND identically distributed as Unif(0,n).

The text I pulled this question from is unpublished (course notes for some class I believe), so occasionally the notation is a bit ambiguous.

Best Answer

If you mean $X_1, \ldots, X_n \overset{iid}{\sim} \text{Uniform}(0,n)$, then $1-\prod_{k = 1}^n(1-F_k(y))$ turns into $$ 1-\prod_{k =1}^n(1-y/n) = 1 - (1-y/n)^n \to 1 - e^{-y}. $$ This is the CDF of an Exponential(1) random variable. Perhaps this is what you meant? They are independent, but they are also identical. And their distribution does depend on how big your sample size is.

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Solved – Sufficient order statistics

Your confusion stems from the loose notations I presume: when writing $$f(\mathbf{x}) = \prod_{i=1}^n f(x_i) = \prod_{i=1}^n f(x_{(i)}),$$ George Casella and Roger Berger first use $f$ for the density of the vector, $f(\mathbf{x})$, then second as the density of the components, $f(x_i)$, and then again as the density of the components, $f(x_i)$, rather than as the densities of the order statistics. Distinguishing between all three as in \begin{align*}\mathbf X &\sim f_{1:n}(\mathbf x) \\ X_i &\sim f_1(x_i)\quad i=1,\ldots,n\\ X_{(i)} &\sim f_{(i)}(x_{(i)}) \end{align*} the identity looks like $$\underbrace{f_{1:n}(\mathbf{x})}_{\text{joint}} = \prod_{i=1}^n \underbrace{f_1(x_i)}_{\text{individual}} = \overbrace{\prod_{i=1}^n \underbrace{f_1(x_{(i)})}_{\text{individual}}}^{\text{reordering}\\\text{by rank}} \ne \overbrace{\prod_{i=1}^n\underbrace{f_{(i)}(x_{(i)})}_{\text{ordered}}}^{\text{incorrect due}\\\text{to dependence}},$$

Probability – Expectation of Square Root of Sum of Independent Squared Uniform Random Variables

One approach is to first calculate the moment generating function (mgf) of $Y_n$ defined by $Y_n = U_1^2 + \dotsm + U_n^2$ where the $U_i, i=1,\dotsc, n$ is independent and identically distributed standard uniform random variables.

When we have that, we can see that $$ \DeclareMathOperator{\E}{\mathbb{E}} \E \sqrt{Y_n} $$ is the fractional moment of $Y_n$ of order $\alpha=1/2$. Then we can use results from the paper Noel Cressie and Marinus Borkent: "The Moment Generating Function has its Moments", Journal of Statistical Planning and Inference 13 (1986) 337-344, which gives fractional moments via fractional differentiation of the moment generating function.

First the moment generating function of $U_1^2$, which we write $M_1(t)$. $$ M_1(t) = \E e^{t U_1^2} = \int_0^1 \frac{e^{tx}}{2\sqrt{x}}\; dx $$ and I evaluated that (with help of Maple and Wolphram Alpha) to give $$ \DeclareMathOperator{\erf}{erf} M_1(t)= \frac{\erf(\sqrt{-t})\sqrt{\pi}}{2\sqrt{-t}} $$ where $i=\sqrt{-1}$ is the imaginary unit. (Wolphram Alpha gives a similar answer, but in terms of the Dawson integral.) It turns out we will mostly need the case for $t<0$. Now it is easy to find the mgf of $Y_n$: $$ M_n(t) = M_1(t)^n $$ Then for the results from the cited paper. For $\mu>0$ they define the $\mu$th order integral of the function $f$ as $$ I^\mu f(t) \equiv \Gamma(\mu)^{-1} \int_{-\infty}^t (t-z)^{\mu-1} f(z)\; dz $$ Then, for $\alpha>0$ and nonintegral, $n$ a positive integer, and $0<\lambda<1$ such that $\alpha=n-\lambda$. Then the derivative of $f$ of order $\alpha$ is defined as $$ D^\alpha f(t) \equiv \Gamma(\lambda)^{-1}\int_{-\infty}^t (t-z)^{\lambda-1} \frac{d^n f(z)}{d z^n}\; dz. $$ Then they state (and prove) the following result, for a positive random variable $X$: Suppose $M_X$ (mgf) is defined. Then, for $\alpha>0$, $$ D^\alpha M_X(0) = \E X^\alpha < \infty $$ Now we can try to apply these results to $Y_n$. With $\alpha=1/2$ we find $$ \E Y_n^{1/2} = D^{1/2} M_n (0) = \Gamma(1/2)^{-1}\int_{-\infty}^0 |z|^{-1/2} M_n'(z) \; dz $$ where the prime denotes the derivative. Maple gives the following solution: $$ \int_{-\infty}^0 \frac{n\cdot\left(\erf(\sqrt{-z})\sqrt{\pi}-2e^z\sqrt{-z} \right)e^{\frac{n(-2\ln 2 +2 \ln(\erf(\sqrt{-z}))-\ln(-z) +\ln(\pi))}{2}}}{2\pi(-z)^{3/2}\erf(\sqrt{-z})} \; dz $$ I will show a plot of this expectation, made in maple using numerical integration, together with the approximate solution $A(n)=\sqrt{n/3-1/15}$ from some comment (and discussed in the answer by @Henry). They are remarkably close:

As a complement, a plot of the percentage error:

Above about $n=20$ the approximation is close to exact. Below the maple code used:

int( exp(t*x)/(2*sqrt(x)), x=0..1 ) assuming t>0;
int( exp(t*x)/(2*sqrt(x)), x=0..1 ) assuming t<0;
M := t -> erf(sqrt(-t))*sqrt(Pi)/(2*sqrt(-t))
Mn := (t, n) -> exp(n*log(M(t)))
A  :=  n -> sqrt(n/3 - 1/15)
Ex :=  n ->   int( diff(Mn(z, n), z)/(sqrt(abs(z))*GAMMA(1/2) ), 
                   z=-infinity..0 , numeric=true)

plot([Ex(n), A(n)], n=1..100, color=[blue, red], legend= 
      [exact, approx], labels=[n, expectation], 
      title="expectation of sum of squared uniforms")
plot([((A(n)-Ex(n))/Ex(n))*100], n=1..100, color= 
      [blue], labels=[n, "% error"], 
      title="Percentage error of approximation")

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Solved – Sufficient order statistics

Probability – Expectation of Square Root of Sum of Independent Squared Uniform Random Variables

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