Given mean $\lambda = 4$, $R_1 = 0.5389$, $R_2 = 0.0532$, $R_3 = 0.3492$

Iteration 1:

1. Set $x=0$ and $P = 1$

2. Generate a random number $R_1 \sim \cup (0,1)$

   $\begin{aligned} \text { compure } P \cdot R_{1} & \Rightarrow 1(0.5389) \\ &=0.5389 \end{aligned}$

   Set $P=p \cdot R_{1}=0.5389$

3. check $P \lt e^{-\lambda}$

   i.e. $0.5389 \lt e^{-4}$

   i.e. $0.5389 \lt 0.0183, \text{ false } \Rightarrow \text{reject x and} $

   set $x=x+1=0+1,$ go to step 2

Iteration 2:

1. Generate a random number $R_2 \sim \cup (0,1)$

   $\begin{aligned} \text { Compute } P \cdot R_{2} & \Rightarrow(0.5389) - (0.0532) \\ &=0.0286 \end{aligned}$

2. Check $P \lt e^{-\lambda}$

   i.e, $0.0286 \lt e^{-4}$

   $0.0286 \lt 0.0183, false \Rightarrow \text{reject x and}$

   set $x=x+1=1+1=2$ go to step 2

Iteration 3:

1. Generate a random number $R_3 \sim \cup (0,1)$

   $\begin{aligned} \text { Compute } P \cdot R_{3} & \Rightarrow(0.0286) - (0.3492) \\ &=0.0099\end{aligned}$

2. Check $P \lt e^{-4}$

   i.e, $0.0099 \lt e^{-4}$

   $0.0099 \lt 0.0183, true \Rightarrow \text{ accept } x - 2$

The whole process is summarized in table below:

Table: Acceptance - Rejection process for Poisson variable