Sum of squared Poisson random variables Suppose $X_1,\dots,X_n$ are i.i.d. $\text{Poisson}(\lambda)^2$. Can we say that with high probability, $\sum_i X_i$ is bounded by a linear function (linear in $n$)?
I was told it should follow from Bernstein's inequality (Theorem 1 from here), yet I couldn't quite show the condition (2) (from the paper) for $X_i$'s (from all bounds for Poisson moments, I was getting $(k!)^2$). Does this inequality hold? If so, how does condition (2) follow for $X_i$'s?
If not, is the sum of $X_i$'s linear?
Edit: By "with high probability" I mean that the probability goes to $0$ as $n\rightarrow\infty$.
 A: Let $Y = X^2,\;\;X \sim  Poisson(\lambda)$
From the algebra of random variables, we note the formula for the variance of a random variable:
$V[Z] = E[Z^2]-E[Z]^2$
Therefore,
$E[Y] = E[X^2] = V[X] + (E[X])^2$
$E[Y]=\lambda + \lambda^2 := m$
Another fact from algebra of random variables is that expectation is a linear operator:
$$E\left[\sum Z_i\right] = \sum E[Z_i]$$
So here we have:
$$E\left[\sum_{i=1}^n Y_i\right] = \sum_{i=1}^n E[Y_i] =  \sum_{i=1}^n m = nm$$
Let $S_n = \sum_{i=1}^n Y_i$ so $E[S_n]=nm$
Since $S_n \geq 0$, we apply Markov's inequality to get:
$$P(S_n \geq a) \leq \frac{E[S_n]}{a} = \frac{nm}{a}\;\;\forall a>0$$
Let's define "$S_n$ is bounded with a high-probability by a function $f(n)$ in n" means this:
$$P(S_n \geq f(n)) \leq \alpha,\;\;0<\alpha \ll 1$$
Then it follows that
$$\exists a_n>0: P(S_n \geq a_n) \leq \alpha=\frac{E[S_n]}{a_n} \implies$$
$$a_n = \frac{E[S_n]}{\alpha} = \frac{nm}{\alpha} \propto n$$
EDIT: I had the wrong definition of whp -- below is correct proof
From above, we see that $E[Y]=\lambda + \lambda^2 := m$, also we know that $V[Y] < \infty$. The $Y_i$ being summed to get $S_n$ are all iid. Therefore, we satisfy the requirements of the weak law of large numbers:
$$P\left(|S_n -nE[Y]|>xn\right)\leq \frac{V[Y]}{x^2n}\;\;\forall x>0$$
Now,
$$|S_n -nE[Y_1]|> xn \implies S_n-nE[Y] < -xn\text{ or } S_n-nE[Y] > xn$$
These are two disjoint subsets of the range of $S_n-nE[Y]$ so
$$P\left(|S_n -nE[Y]|>xn\right) \geq P\left(S_n -nE[Y]>xn\right) = P\left(S_n >xn+ nE[Y]\right)$$
Therefore,
$$P\left(S_n >xn+ nE[Y]\right) \leq P\left(|S_n -nE[Y]|>xn\right) \leq \frac{V[Y]}{x^2n}\xrightarrow{n\to \infty}0\;\;\;\square$$
