Let $X_n$ have Poisson distribution with parameter $5n$.
Find all real numbers $\alpha$ such that the following sequence is tight:
$$Z_n=n^{\alpha}(X_n – 5n)$$.
I managed to calculate the distribution of $Z_n$ and write the condtion for tightness that is: $$P(Z_n=k)=\frac{5n^{5n+kn^{-\alpha}}}{(5n+kn^{-\alpha})!}e^{-5n}:=\beta_{n,k,\alpha}$$ and the tightness means that for every $\epsilon$ there is an $M>0$ such that for each $n$
$$P(Z_n\in[-M,M])>1-\epsilon \iff P(Z_n>M)<\epsilon \iff \sum_{k=⌊M+1⌋}^\infty \beta_{n,k,\alpha}<\epsilon$$
Could you tell me how to solve this problem?
Best Answer
Let $Z_n(\alpha) = n^\alpha (X_n - 5n)$
Note that when $\alpha = -\frac{1}{2}$, then we can apply CLT (since $Poiss(5n)$ is a sum of $n$ independent copies $Y_k$ of $Poiss(5)$ and $Z_n(-\frac{1}{2}) = \frac{\sum_{k=1}^n Y_k - \mathbb E[Y_k]}{\sqrt{n}}$ converges weakly to $\mathcal N(0,5),$ which gives us that sequence $Z_n(-\frac{1}{2})$ is tight.
Now note, that $Z_n(\alpha) = Z_n(-\frac{1}{2}) \cdot n^{\alpha+\frac{1}{2}} = Z_n(-\frac{1}{2})\cdot W_n(\alpha)$
When $\alpha + \frac{1}{2} < 0$ then $W_n(\alpha)$ converges to $0$ a.s so in distribution, too. Now use fact, that when you have two sequences of random variables, one of which converge in distribution to constant, second one converges in distribution, then their product converges in distribution, too. By that we get $\forall_{\alpha < -\frac{1}{2}} Z_n(\alpha)$ converges in distribution to $0$, which again gives us tightness.
Now, the last case: $\alpha > -\frac{1}{2}$. We should intuitively see, that we have a problem now, since $W_n(\alpha)$ tends to infinity. Let's prove it formally. Take any $M \in \mathbb R_+$
$\mathbb P( |Z_n(\alpha)| > M) = \mathbb P( |Z_n(-\frac{1}{2})| > \frac{M}{W_n(\alpha)}) $
Now, take any $\delta > 0$. Since $W_n(\alpha) \to \infty$ as $n \to \infty$, we have that $\frac{M}{W_n(\alpha)}$ tends to $0$, so there is $N \in \mathbb N$ such that $\forall_{n > N}: \frac{M}{W_n(\alpha)} < \delta$. Taking only those $n$, we have:
$ \mathbb P(|Z_n(\alpha)| > M) \ge \mathbb P(|Z_n(-\frac{1}{2})| > \delta)$, which due to fact that $Z_n(-\frac{1}{2})$ converges in distribution to $\mathcal N(0,5)$ and function $x \to |x|$ is continuous, gives us $ \liminf \mathbb P(|Z_n(\alpha)| > M) \ge \mathbb P(|\mathcal N(0,5)| > \delta)$, and since $\delta >0$ was arbitrary, we have $\liminf \mathbb P(|Z_n(\alpha)| > M) = 1$ (so $\lim$ to, since bound by $1$ from above is trivial.) Now, since $M \in \mathbb R_+$ was arbitrary, we cannot find one $M$ which would work for all $n \in \mathbb N$ simultaneously.
Conclusion: Tightness iff $\alpha \le -\frac{1}{2}$