Question on proving tight sequences. I was just wondering how you would go about showing that a sequence of random variables is a tight sequence. For example suppose $X_{n}$ is distributed Exponentially($\lambda_n$) how would I show that
          {$X_{n}$}$_{n}$ is a tight sequence. 
Is it enough to show that if there exsists some M and $\epsilon > 0$, then if $P(|X_{n}| \ge M) < \epsilon$ it is a tight sequence?
$P(|X_{n}| \ge M) < \epsilon$ $\implies$ $1-(1-\rm{e}^{-\lambda_{n}M}) = \rm{e}^{-\lambda_{n}M} < \epsilon$
Then if $M < \infty$ and take $\epsilon = 1$, then as long as $\lambda_{n}$ is bounded away from zero it is a tight sequence. Is this correct?
Thank you for any help. 
 A: Your attempt is almost correct. You have to show that there exists for each $\epsilon>0$ a compact set $K_\epsilon$ such that for all $n$, the probability that $X_n\in K$is at least $1-\epsilon$. Since every compact subset of $\mathbb{R}$ is contained in some bounded closed interval, and since the probability that an exponentially distributed random variable is negative is cero, one can always take $K_\epsilon$ to be $[0,M_\epsilon]$ for some $M_\epsilon>0$.
Let $F_n$ be the cdf of $X_n$. Let $\epsilon>0$. We have to show that for some $M_\epsilon>0$, and for all $n$ one has $F_n(M_\epsilon)>1-\epsilon$. So for all $n$ one needs, plugging in the cdf of the exponential distribution,
$$1-e^{-\lambda_n M_\epsilon}>1-\epsilon$$
$$\epsilon>e^{-\lambda_n M_\epsilon}.$$
Now let $0<b\leq \lambda_n$ for all $n$. Then 
$$\epsilon>e^{-b M_\epsilon}\geq e^{-\lambda_n M_\epsilon}$$
for $M_\epsilon$ large enough, since $\lim_{k\to\infty }e^{-bk}=0$. So the family of random variables is indeed tight if all $\lambda_n$ are uniformly bounded away from zero. It is also clear that this condition is necessary.
