import numpy as np
import matplotlib
import matplotlib.pyplot as plt
plt.rcParams['figure.figsize'] = [8, 8]
Let's simulate a Poisson point process on $[0,1]$: first, with uniform intensity, and then with nonuniform intensity.
To do this, we will
lam = 70
num_points = np.random.poisson(lam=lam)
xy = np.array([np.random.uniform(size=num_points),
np.random.uniform(size=num_points)])
fig, ax = plt.subplots()
ax.scatter(xy[0], xy[1])
ax.set_xlim(0,1)
ax.set_ylim(0,1)
Now let's simulate from a Poisson point process with mean intensity $$ x^y dx dy . $$
The total number of points is Poisson with mean $$\begin{aligned} \int_0^1 \int_0^1 x^y dx dy &= \int_0^1 \frac{1^{y+1}}{y+1} dy \\ &= \log(2) . \end{aligned}$$
Note that $$ \int_0^1 x^y dx = \frac{1}{y+1} ,$$ while $$ \int_0^1 x^y dy = \int_0^1 \exp(y\log x) dy = \frac{x-1}{\log x} . $$
This means that the marginal distribution of the $y$ coordinate has density $1/(y+1)$, and so if a point $(X, Y)$ is chosen form the distribution with this density, $$ \mathbb{P}\{ Y < u \} = \frac{1}{\log2} \int_0^u \frac{1}{y+1} dy = \frac{ \log(u+1)}{\log2} , $$ and therefore, $$ \mathbb{P}\{ Y < 2^{y}-1 \} = y, \qquad \text{for } 0 \le y \le 1 .$$ So, if we let $U$ be Uniform on $[0,1]$ then we can simulate $Y$ by taking $$ Y = 2^U-1 ,$$ since then $$\begin{aligned} \mathbb{P}\{ Y < y \} &= \mathbb{P}\{ 2^U - 1 < y \} \\ &= \mathbb{P}\{ U < \log(y+1) \} \\ &= \log(y+1)/\log(2), \end{aligned}$$ as required.
Similarly, given $Y=y$, the distribution of $X$ has density equal to $$ \frac{x^y}{y+1} dx, $$ so $$\begin{aligned} \mathbb{P}\{ X < x \vert Y=y \} &= \int_0^x \frac{u^y}{y+1} du \\ &= x^{y+1} . \end{aligned}$$ As before, if $V$ is Uniform on $[0,1]$, then we can let $$ X = V^{1/(y+1)} . $$
lam = 2000
num_points = np.random.poisson(lam * np.log(2))
Y = 2 ** np.random.uniform(size=num_points) - 1
X = np.random.uniform(size=num_points) ** (1/(Y+1))
xy = np.array([X, Y])
fig, ax = plt.subplots()
ax.scatter(xy[0], xy[1])
ax.set_xlim(0,1)
ax.set_ylim(0,1)