Let $(Y_n)$ be i.i.d random variables following exponential distribution with parameter $1$.
Let $X_n=\min(Y_1,\dotsc, Y_n)$
Prove that $ X_n \xrightarrow{P} 0$
It's easy to prove that $P(Y_1<0)=0$. Nevertheless my proof of convergence is lengthy because I can't assume that $Y_i\geq 0$. Here's how I would write down the proof:
\begin{align}P(|X_n|>\epsilon)&=P(\{\omega, \exists i, Y_i<0\}\cap|X_n|>\epsilon)+P(\{\omega, \forall i, Y_i\geq 0\}\cap|X_n|>\epsilon)\\ &=P\left(\bigcup_{i=1}^n (Y_i<0) \cap|X_n|>\epsilon\right) +P\left(\bigcap_{i=1}^n (Y_i\geq0) \cap|X_n|>\epsilon\right) \\ &\le \sum_{i=1}^nP((Y_i<0) \cap|X_n|>\epsilon))+P(X_n>\epsilon) \\ &\le e^{-\epsilon n} \end{align}
because $\displaystyle \sum_{i=1}^nP((Y_i<0) \cap|X_n|>\epsilon))\leq \sum_{i=1}^nP(Y_i<0 )=0$
Can you come up with a more concise proof? I'm new to probabilities, so I'm trying to brush on my proof-writing.
There is actually a fairly easy way to show the statement. But first things first, your first question whether or not you can assume that an exponentially distributed random variable $X$ is positive is pretty much clear by definition of the distribution itself, we have for $\lambda >0$ $$ F(x;\lambda) =\mathbf{P}(X\leq x)= \begin{cases} 1-e^{-\lambda x} & x \ge 0, \\ 0 & x < 0. \end{cases} $$ which means nothing less than that $X$ is distributed above the non-negative reals.
Now let's tackle your second question. We want to show that we have with $X_n=\min(Y_1,\ldots, Y_n),(Y_i)\sim Exp(\lambda=1)$ iid $$ X_n \xrightarrow{P} 0 $$ First we notice, that \begin{align} \mathbf{P}(X_n>\epsilon)=&\mathbf{P}(\min(Y_1,\ldots, Y_n)>\epsilon)\\ =& \mathbf{P}(Y_1>\epsilon,Y_2>\epsilon,\ldots,Y_n>\epsilon) \\ =& \prod_{i=1}^n\mathbf{P}(Y_i>\epsilon) \\ =& \prod_{i=1}^n e^{-\epsilon}=e^{-n\epsilon} \end{align} so we indeed have, that $$ \forall \epsilon>0:\lim_{n\to\infty}\mathbf{P}(X_n>\epsilon)=0\iff X_n \xrightarrow{P} 0 $$ We actually only used the distributional iid-property of the $Y_i$ and the fact that for $Y\sim Exp(\lambda)$ we have $$ \mathbf{P}(Y>x)=1-\mathbf{P}(Y\leq x)=1 -(1-e^{-\lambda x} )=e^{-\lambda x} $$ EDIT: However, I find a bit hard to read your proof, you definitely need to add some more information, e.g.: