Exponential(1) distributed random variable convergence I am stuck with convergency in probability... I have the following exercise: Let $(X_k)_{k\ge1}$ be a sequence of independent exponential-(1) distributed random variables. Show that $n^\alpha \min\limits_{k\le n} X_k \overset{P}{\to} 0$ for $\alpha<1$ and $n\to \infty$ I don't even know, how to start. I would really appreciate, if one of you could explain to me, how to approach this. Thanks a lot!

Let $X\sim Exp(1)$. So for $x\geq0$, $\mathbb{P}(X> x)=e^{-x}$. Let $X_i$ be iid to $X$ for $i=1,2,...$ and let $Y_n=n^\alpha \min\\{X_1,...,X_n\\}$ for some $\alpha<1$.

Then, $$\mathbb{P}(Y_n> x)=\mathbb{P}(\min\\{X_1,...,X_n\\}> xn^{-\alpha})=\mathbb{P}(X_1> xn^{-\alpha},...,X_n> xn^{-\alpha})$$

As the $X_i$ are identically distributed, and independent, this is:

$$\mathbb{P}(X_1> xn^{-\alpha})\mathbb{P}(X_2> xn^{-\alpha})\cdot...\cdot\mathbb{P}(X_n> xn^{-\alpha})=\mathbb{P}(X> xn^{-\alpha})^n=(e^{-xn^{-\alpha}})^n=e^{-xn^{1-\alpha}}$$

Then, for any $\varepsilon>0$,

$$\mathbb{P}(|Y_n-0|> \varepsilon)=\mathbb{P}(Y_n>\varepsilon)=e^{-\varepsilon n^{1-\alpha}}\to 0 \text{ as }n\to \infty$$ Where here we are using that $\alpha<1$, so that $n^{1-\alpha}\to \infty$.

Therefore, $Y_n\overset{p}{\to} 0$.