If $X_n$ has a $\mathrm{Uniform}(0,n)$, $Y_n$ has $\mathrm{Geometric}(1/n)$ and $Z_n$ has $\frac{1}{n}Y_n$ distribution how would you show whether or not each one is a tight sequence or not?
Additionally how would you find the weak distributional limit for those which have one?
I calculated the characteristic function of $X_n$ to be $$\frac{e^{int} -1}{int}$$ but i dont know the limit of this and if it converges to the characteristic function of a distribution.
For $Y_n$ I calculated the characteristic function as $$\frac{e^{it}}{n(1-e^{it})+e^{it}}$$ which I think the limit is zero as n goes to infinity.
for $Z_n$ I calculated the characteristic function as $$\frac{e^{\frac{it}{n}}}{n(1-e^{\frac{it}{n}})+e^{\frac{it}{n}}}$$ Which the limit as as n goes to infinity is 1.
Looking for some insight into how you show whether these distributions are tight or not and if anyone can tell me the weak distributional limits if they exist. I don't know where to start to show if they are tight or not. I understand the Continuity lemma for characteristic functions.
The exercise can be done without characteristic functions, but the opening poster seems to require to use this way.
In the first two cases, we do not have the convergence in distribution of any subsequence to a random variable. Indeed, the value of a characteristic function at $0$ is $1$ and if $t\in (0,2\pi)$, then the characteristic function of $X_n$ and $Y_n$ at $t$ goes to $0$ as $n$ goes to infinity.
In the last case, we have the pointwise convergence to a function which is continuous at $0$, hence we can use Levy's theorem.