Analysis for the Study "The Presence of Atmospheric Microplastics on Quezon City, Philippines"

Monteron, De Lara, Peregrino, Ferrer, Primavera (2024)

Background

To study the atmospheric MPs in Quezon City, we used total active samplers with a specified volume rate (67.96 cubic meters per hour) where the particles, including MPs deposit on the filter. We then aim to use this information assess the concentrations of MP from the total number of MPs found in the entire filter divided by the total volume of the air sampled. However due to logistical problems, we do not have the access to the entire filter which corresponds to the entire number of MPs with the volume of air sampled. Hence, we have to use estimate techniques.

Estimation

Single Filter

Let $A$ be the area of a given filter. We will then sample a section of the filter with an area $a$ such that $A = ak$ where $k$ is the number of equal-sized sections that the filter can be decomposed into.

We will then obtain the number of MPs in $a$ and call that $x$. Under the assumption that the MPs are uniformly deposited on the entire filter, any such $a$ is an accurate estimator for the number of MPs on any similar-sized $a$. In another words, if $X$ is the true number of MPs in a given filter, then, $X/k$ is the number of MPs in subsections with size $a$, and if the MPs are uniformly deposited, the number of MPs in any such chosen $a$ is given by

$$ X/k \sim \mathrm{Poisson}(x) $$

To take account of the inherent uncertainty of our $X/k$ estimate, we will construct two-tailed $\alpha$-confidence intervals around our obtained measure of $x$ as follows

$$\mathbb{P}(\phi < X/k < \varphi) = 1 - \alpha$$

Since we now have our $\alpha$-intervals around $X/k$ we can now use the obtained estimates, $k\phi$ and $k\varphi$ as endpoints of our interval and $kx$ for the mean, as the estimates for $X$. Since we now have estimates for $X$ we can also obtain concentration estimates by dividing our estimates by the total volume associated with the specified filter.

Multiple Filters

Now, consider a case where we have multiple with filters with each having MPs $X_1, X_2, ..., X_n$. Suppose we have obtained a corresponding $x_1, x_2, ..., x_n$ from each filter and each $x_i$ has the same area. Then,

$$ \frac{1}{k}\sum_{i} X_i \sim \mathrm{Poisson}\left(\sum_{i}x_i\right) $$

We can also construct confidence intervals as follows

$$\mathbb{P}\left(\phi < \frac{1}{k}\sum_{i} X_i < \varphi\right) = 1 - \alpha$$

From this, we can obtain a pooled estimate for the number of MPs using the same estimation process outlined above.

Systematic Error

We can imagine that before measuring the number of microplastics in the specified section of the filter, the MPs are already deposited uniformly and their distribution was described according to the Poisson process described above. However, due to several instrumental, human errors and other factors, the true number of MPs in that specified section $x$ that we aim to analyze, is scaled by some factor $p$ that we'll call detection rate.

$$ \begin{align*} Y_n &\sim \mathrm{B}(n, p) \\ x \mid X/k &\sim \mathrm{Poisson}(X/k) \end{align*} $$

$$ \begin{align*} \mathbb{P}(x \mid X/k, p) &= \sum_{n \geq x} \mathbb{P}(Y_n = x) \mathbb{P}(x \mid X/k) \\ &= \sum_{n \geq x} {n \choose x} p^x(1 - p)^{n - x} \frac{(X/k)^n e^{-X/k}}{n!} \\ &= e^{-X/k} \sum_{n \geq x} p^x(1 - p)^{n - x} \frac{n!\cdot (X/k)^n}{x!(n - x)! \cdot n!} \\ &= e^{-X/k} \sum_{n \geq x} \frac{p^x \cdot (1 - p)^{n - x} \cdot (X/k)^n}{x!(n - x)!} \\ &= e^{-X/k} \sum_{n \geq 0} \frac{p^x \cdot (1 - p)^{n} \cdot (X/k)^{n + x}}{x!n!} \\ &= \frac{e^{-X/k} p^x}{x!} \sum_{n \geq 0} \frac{(1 - p)^{n} \cdot (X/k)^{n + x}}{n!} \\ &= \frac{e^{-X/k} (Xp/k)^x}{x!} \sum_{n \geq 0} \frac{(1 - p)^{n} \cdot (X/k)^{n}}{n!} \\ &= \frac{e^{-X/k} (Xp/k)^x}{x!} \sum_{n \geq 0} \frac{((1 - p)(X/k))^{n}}{n!} \\ &= \frac{e^{-X/k} (Xp/k)^x \cdot e^{(1 - p)(X/k)}}{x!} \\ &= \frac{e^{-X/k + (1 - p)(X/k)} (Xp/k)^x}{x!} \\ &= \frac{e^{-Xp/k} (Xp/k)^x}{x!} \end{align*} $$

We can realize that this is also a Poisson distribution.

$$ x \mid X/k, p \sim \mathrm{Poisson}(Xp/k) $$

By symmetry, we can also consider when $p > 1$ as overdetection by the change of variables from $x$ to $n$ and vice versa, as well as transforming $p$ to $p - 1$. This can also be viewed as detecting $n$ particles when $x$ is the true number.