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Let $(X_i)$ be i.i.d. Unif$(0,1)$. Define $M_n=\min\{X_1,\ldots,X_n\}$ and $T_n=\sum\limits_{k=1}^n M_k$ for every $n\ge1$. Show that $$\dfrac{T_n}{E(T_n)}\stackrel{P}{\to} 1$$

The problem I am encountering is that the random variables $M_n$ are not independent or uncorrelated so $\operatorname{Var}(T_n)$ does not seem to have a nice expression.

Davide Giraudo
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Andrew Richards
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1 Answers1

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This is not the answer; it is just to save the time of a person who would like to try the approach with correlation. It turns out that the sequence we want to prove the convergence in probability is bounded in $\mathbb L^2$ hence uniformly integrable, hence it should also converge in $\mathbb L^1$ to $0$.

First let us extract some useful information from this thread: the expectation of $M_n$ is $1/(n+1)$ and its density is $n\left(1-u\right)^{n-1}\mathbf 1_{(0,1)}(u)$. As a consequence, $\mathbb E\left[T_n\right]=\sum_{i=1}^n1/(i+1)$ and it would suffices to prove that $$\lim_{n\to +\infty} \frac 1{\left(\ln n\right)^2}\mathbb E\left[\left(\sum_{i=1}^nM_i-\mathbb E\left[M_i\right]\right)^2\right]=0. $$ To this aim, we would like to show that $$\tag{1} \lim_{n\to +\infty} \frac 1{\left(\ln n\right)^2}\sum_{i=1}^n\mathbb E\left[\left(M_i-\mathbb E\left[M_i\right]\right)^2\right]=0; $$ $$\tag{2}\lim_{n\to +\infty} \frac 1{\left(\ln n\right)^2}\mathbb E\left[ \sum_{1\leqslant i\lt j\leqslant n}M_iM_j-\mathbb E\left[M_iM_j\right] \right]=0. $$

  • Let us show (1). Using the expression for the density of $M_i$, we get $$ \mathbb E\left[M_i^2\right]=\int_0^1u^2i\left(1-u\right)^{i-1}\mathrm du =\int_0^1\left(1-v\right)^2iv^{i-1}\mathrm dv=1-2\frac{i}{i+1}+\frac{i}{i+2} $$ which can be simplified as $$ \mathbb E\left[M_i^2\right]=\frac 2{i+1}-\frac 2{i+2}, $$ hence $$\sum_{i=1}^n\mathbb E\left[\left(M_i-\mathbb E\left[M_i\right]\right)^2\right]=1-\frac{2}{n+2}-\sum_{i=1}^n\frac 1{(i+1)^2}\leqslant 1, $$ from which (1) follows.

  • Let us look at (2). Observe that (2) can be rewritten as $$ \lim_{n\to +\infty} \frac 1{\left(\ln n\right)^2} \sum_{i=1}^n \sum_{j=1}^{n-i}c_{i,j} =0, $$ where $c_{i,j}:=\mathbb E\left[M_iM_{i+j}\right]-\mathbb E\left[M_i\right]\mathbb E\left[M_{i+j}\right] $. Observing that $M_{i+j}$ has the same law as $\min\left\{M_i,M'_j\right\}$, where $M'_j$ has the same law as $M_j$ and is independent of $M_i$, we get $$ c_{i,j}=\int_0^1\int_0^1u\min\left\{u,v\right\}i\left(1-u\right)^{i-1}j\left(1-v\right)^{j-1}\mathrm du\mathrm dv-\frac 1{\left(i+1\right)\left(j+1\right)}. $$ A computation shows that $$ c_{i,j}=\frac 1{j+1}\left(\frac{i}{i+j+2}-\frac{i}{i+j+1}\right), $$ which can be transformed as $$ c_{i,j}=\frac 1{j+1}\frac{i+1}{i+j+2}-\frac 1{j+1}\frac{i}{i+j+1}+\frac 1{(j+1)(i+j+2)}. $$ For the first two terms, we switch the sum and use telescoping in $i$ to find $$\sum_{j=1}^{n-1}\sum_{i=1}^{n-j}\frac 1{j+1}\frac{i+1}{i+j+2}-\frac 1{j+1}\frac{i}{i+j+1}=\sum_{j=1}^{n-1}\frac 1{j+1}\frac{n-j+1}{n+2}-\frac 1{j+1}\frac{2}{j+3} $$ hence $$ \left\lvert \sum_{j=1}^{n-1}\sum_{i=1}^{n-j}\frac 1{j+1}\frac{i+1}{i+j+2}-\frac 1{j+1}\frac{i}{i+j+1}\right\rvert\leqslant 3\sum_{j=1}^{n-1}\frac 1{j+1} $$ which is negligible with respect to $(\ln n)^2$. For the last term, it behaves like $$\sum_{j=1}^{n-1}\sum_{i=j+1}^n\frac 1{ij}=\sum_{i=2}^n\frac 1i \sum_{j=1}^{i-1}\frac 1j,$$ which is exactly the failure of (2), since the latter quantity is of order $(\ln n)^2$.

Davide Giraudo
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