Given that $X_1,X_2 \dots X_n$ random variables distributed $exp(\theta)$.
let $T>0$ constant , and we define instead of $X_i$ , $(Z_i,\delta_i)$ such as : $$Z_i = min\{Z_i,T\}$$ $$\delta_i = \mathbb{I}\{X_i \le T\}$$
Show that $Q\left(\theta\middle|\theta^{(m)}\right)$ equals
$\hat{\theta}=\frac{1}{n}\left[\sum_{i=1}^{n}{\delta_iZ_i+\left(1-\delta_i\right)(T+\theta^{(m)})}\right]$
My attempt :
I showed that $\mathbb{E}_{\theta^{(m)}}\left(X_i|Z_i,\delta_i\right)$ equals
$\mathbb{E}_{\theta^{(m)}}\left(X_i|Z_i,\delta_i\right)=\ \delta_iZ_i+\left(1-\delta_i\right)(T+\theta^{\left(m\right)}),$ but I couldn't continue from here.