Prove the convergence of $\sum_{n=1}^\infty P(n^a X_n \geq \epsilon)$ where $X_n=\min \{Y_1,…,Y_n \}$, $Y_i$ i.i.d.

Good day,

I have an old exam without solutions and there is the following exercise:

Let $(Y_n)_{n \in \mathbb{N}}$ be a sequence of independent identically distributed (i.i.d.) random variables which are uniformely distributed on $[0,1]$. Further define $X_n :=\min\{Y_1,...,Y_n\}$. Show:

(i) $F_n(t)=[1-(1-t)^n] 1_{[0,1]}(t)+1_{(1,\infty)}(t)$ is the distribution function of $X_n$ - got this one

(ii) $X_n$ is integrable and compute $E(X_n)$ - got this one

(iii) Prove the convergence of $\sum_{n=1}^\infty P(n^a X_n \geq \epsilon)$ for all $a \in [0,1)$, $\epsilon>0$

Hint: You can make use of the fact that $\log(1-x) \leq -x$ holds for all $x<1$

(iv) $n^a X_n \to 0$ almost surely - got this one if I use (iii)

Normally I'd show exercises like (iii) with Borel-Cantelli-Lemma but here it seems not useful (further I use Borel-Cantelli in (iv) to show almost sure convergence).

Therefore I want to compute $P(n^a X_n \geq \epsilon)$ directly and hope I get sth like $\frac{1}{n^2}$ to get convergence of the sum.

Okay, let's try:

$$P(X_n \geq n^{-a} \epsilon)=1-F_n(n^{-a} \epsilon)=\left(1-\frac{\epsilon}{n^a} \right)^n$$

It seems a bit like $(1+\frac{x}{n})^n \to e^x$. I don't know if this is correct up to now since I didn't use the hint.

I'd have to check: $\sum \left(1-\frac{\epsilon}{n^a} \right)^n < \infty$

Maybe with the root test: $\sqrt[n]{ |a_n|}=1-\frac{\epsilon}{n^a}<1=C$ and this implies convergence. But this feels not right.

Can somebody please tell me how to solve this (maybe as the hint implies with the logarithm.)

Thanks a lot, Marvin

• Maybe you can use $$\left(1-\frac{\epsilon}{n^a} \right)^n =\exp\left[n \log\left(1-\frac{\epsilon}{n^a}\right)\right]\leq \exp\left[-n\frac{\epsilon}{n^a}\right]\ ,$$ using the hint and the fact that the exponential is an increasing function... – Pierpaolo Vivo Sep 25 '16 at 9:22
• @PierpaoloVivo Thanks for your hint. I also thought of this but I didn't know how to finish this. I have a sum like this $$\sum_{n=1}^\infty \exp(-\epsilon n^{1-a} )$$ for $1-a \in (0,1]$. I just checked WolframAlpha and it seems like for example $\sum_{n=1}^\infty \exp(-n^{1/2})$ converges by the comparison test. So it really seems like the way to go, overlooked this before. – Fritz Sep 25 '16 at 9:32

By the elementary estimate $\log(1-x) \leq -x$, $x \in (0,1)$, we have

$$\log \left( 1- \frac{\epsilon}{n^a} \right) \leq -\frac{\epsilon}{n^a}<0$$

implying

\begin{align*} \left( 1- \frac{\epsilon}{n^a} \right)^n &= \exp \left( n \log \left[1- \frac{\epsilon}{n^a} \right] \right) \\ &\leq \exp \left( - n \frac{\epsilon}{n^a} \right) = \exp(-\epsilon n^{1-a}). \end{align*}

Since the exponential function is growing faster than any polynomial, we can find for any $N \in \mathbb{N}$ some constant $c>0$ such that $x^N \leq c \exp(x)$ for all $x > 0$, or, equivalently,

$$\exp(-x) \leq \frac{1}{c} \frac{1}{x^N}$$

for all $x > 0$. Applying this for $x = \epsilon n^{1-a}$ and choosing $N \in \mathbb{N}$ sufficiently large, it is not difficult to see that

$$\sum_{n \geq 0} \exp(-\epsilon n^{1-a}) <\infty$$

for any $a \in [0,1)$. Consequently, we find

$$\sum_{n \geq 0} \mathbb{P}(X_n \geq \epsilon n^{-a}) \leq \sum_{n \geq 0} \exp(-\epsilon n^{1-a}) < \infty.$$