Uniform Sampling From the Region Bounded by $\sqrt{x}$, $x=3$, and $y=0$

Question

I want to sample uniformly from the area bounded by $y=0$, $x=3$, and $y=\sqrt{x}$:

If I draw $x$ from $U[0, 3]$ and $y$ from $U[0, \sqrt{x}]$, the density will be higher in the bottom left corner:

I tried to fix this by making the number of points on a region bounded by $x=a \in [0, 3]$ on the right, the square root curve, and the $x$-axis proportional to its area (relative to that of the entire figure).

So, the total area should be: $$\int_{0}^{3}\sqrt{x}dx = \frac{2}{3}x^{\frac{3}{2}}\bigg\rvert_{0}^{3}=2\sqrt{3}$$

The area bounded by $x=a$ on the right is $\frac{2}{3}a^{\frac{3}{2}}$, so the pdf of $x$ should be proportional to the ratio, i.e.: $$ pdf(x) = C \times \frac{\frac{2}{3}a^{\frac{3}{2}}}{2\sqrt{3}} = \frac{C}{3\sqrt{3}}a^{\frac{3}{2}} $$

The corresponding CDF is then:

$$ CDF(x) = \int_{0}^{x} pdf(u)du = \int_{0}^{x} \frac{C}{3\sqrt{3}}u^{\frac{3}{2}}du = \frac{C}{3\sqrt{3}} \times \frac{2}{5} u^{\frac{5}{2}} \bigg\rvert_{0}^{x} $$

$$ CDF(x) = \frac{C}{3\sqrt{3}} \times \frac{2}{5} x^{\frac{5}{2}} = \frac{2C}{5\times3\sqrt{3}} x^{\frac{5}{2}} $$

I get $C$ from the condition $CDF(3)=1$: $$ CDF(3) = \frac{2C}{5\times3\sqrt{3}} 3^{\frac{5}{2}} = \frac{2C}{5}\times 3 = \frac{6C}{5} = 1 $$ So: $$ C = \frac{5}{6} \implies \quad CDF(x) = \left(\frac{x}{3}\right)^{\frac{5}{2}} $$

Therefore, I should sample $x$ by sampling a random number $r \in Uniform[0, 1]$ and solving $CDF(x)=r$ for $x$: $$ x = 3 \times r^{\frac{2}{5}} \quad (r \in Uniform[0,1]) $$

and get $y$ by sampling from $Uniform[0, \sqrt{x}]$. However, this approach doesn't make the distribution uniform: The sampled points are now concentrated away from the bottom left corner.

Can someone help me? :)

Answer 1

The distribution of $y$ conditional on $x$ is uniformly distributed between $0$ and $\sqrt x.$

Within any narrow vertical strip of width $\mathrm dx$ around $x$ you want the frequency of points to be proportional to $\sqrt{x}\,\mathrm d x$. Adding these up (integrating) tells us the cumulative distribution of the $x$ coordinates must be proportional to $x^{3/2}.$ Since the full proportion of $1$ is attained at $x=3,$ the distribution function must be

$$F_X(x) = \left(\frac{x}{3}\right)^{3/2}.$$

The most direct way to sample from this distribution is by generating uniform quantiles. That is, when $U$ has a uniform distribution on $[0,1],$

$$F_x^{-1}(U) = 3U^{2/3}$$

has the distribution $F_X.$

Thus, by generating $2n$ uniform random variates you can obtain $n$ points uniformly drawn from this region.

Here is an implementation in R.

n <- 1e3; y <- sqrt(x <- 3 * runif(n) ^ (2/3)) * runif(n)

A scatterplot might look like this:

One way to check the validity is to slice a set of points horizontally in a thin band and check for a uniform distribution, except near the smallest values (where there will an edge effect). After generating a million points, for instance, I plotted a thin band around $y=1.5$ with the commands

dy <- 0.01
y0 <- 1.5
hist(x[abs(y - y0) <= dy/2])

There were 2140 points with this nearly uniform histogram:

Similar histograms emerge at other values of y0.