Uniform integrability for a Martingale (defined via Poisson process) I'm trying to solve the following exercise:

Let us consider the martingale $M_t = (N_t - \theta t)^2 - \theta t$. where $(N_t)_t$ is a homogeneous Poisson process with parameter $\theta$. 
  Decide if the family $(M_t)_t$ is uniformly integrable a


My attempt
I tried to use the definition  of uniform integrable familiy, but I got stuck, so I tried to show it's actually not a uniformly integrable family by showing that $$\sup_{t \geq 0}E[|M_t|] \rightarrow +\infty$$
Indeed: $$E[|M_t|]\geq|E[M_t]| = |E[N_t^2 - 2 \theta t N_t + \theta^2 t^2 - \theta t]| = |\theta t (\theta t -1) + t(t+1)- 2 \theta t^2| = |t^2(\theta - 1 )^2 + t(1-\theta)|$$
Where I used linearity of expectation and $E[N_t^2] = t(t+1)$ and $E[N_t] = t$ since $N_t$ is a Poisson process.
The supremum of that thing is $\infty$ so the family is not uniformly integrable.
Is it okay?
 A: I can't really follow your computations. Since $N_t$ is Poisson distributed with parameter $\theta t$, we have
$$\mathbb{E}(N_t^2) = \theta^2 t^2 + \theta t$$
and $$\mathbb{E}(N_t) = \theta t.$$
This gives
\begin{align*} \mathbb{E}((N_t-\theta t)^2) &= \mathbb{E}(N_t^2) -2 \theta t \mathbb{E}(N_t)+\theta^2 t^2 \\ &= \theta^2 t^2 + \theta t - 2 \theta^2 t^2 + \theta^2 t^2 \\ &= \theta t \end{align*}
and so $$\mathbb{E}(M_t) = \mathbb{E}((N_t-\theta t)^2) - \theta t = 0.$$ In particular, your assertion that $\sup_t |\mathbb{E}(M_t)|=\infty$ is wrong.
To show that $(M_t)_{t \geq 0}$ is not uniformly integrable, we can argue as follows: By the central limit theorem, we have
$$\frac{N_t-\theta t}{\sqrt{\theta t}} \to U \sim N(0,1) \quad \text{in distribution}.$$
In particular,
$$\mathbb{P}(|N_t-\theta t| \geq 2 \sqrt{\theta t}) \to \mathbb{P}(|U| \geq 2)>0.$$
Hence,
\begin{align*} \mathbb{E}(|M_t|) &\geq \mathbb{E} \bigg( \big[ |N_t-\theta t|^2 - \theta t\big] 1_{\{|N_t-\theta t| \geq 2 \sqrt{\theta t}\}} \bigg)  \\ &\geq 3 \theta t \mathbb{P}(|N_t-\theta t| \geq 2 \sqrt{\theta t}) \\ &\xrightarrow[]{t \to \infty} \infty.\end{align*}
This shows that $\sup_{t \geq 0} \mathbb{E}(|M_t|)=\infty$, and so $(M_t)_{t \geq 0}$ is not uniformly integrable.
