Chapter 3: Poisson Distribution

The Poisson distribution models the probability of a given number of events occurring in a fixed interval of time or space, assuming that the events occur independently and at a constant average rate.

The probability mass function is:

\[ P(X = k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!}, \quad k = 0, 1, 2, ... \]

Where:

  • \(X\) is the number of events (e.g., target copies per droplet),
  • \(\lambda\) is the average number of events per interval (copy per droplet in ddPCR),
  • \(e\) is Euler’s number, approximately 2.71828.

When to Use the Poisson Distribution

Poisson distribution is appropriate under the following conditions:

  • Events are rare and occur independently.

  • The probability of two or more events happening simultaneously is negligible.

  • The average event rate (\(\lambda\)) is constant.

  • The number of possible trials is large, and the success probability is small.

Poisson Approximation to Binomial

When:

  • \(n\) is large (e.g., > 20),

  • \(p\) is small (e.g., < 0.01), and

  • \(\lambda = n \cdot p\) is moderate (e.g., ≤ 5),

Then the binomial distribution \(B(n, p)\) can be approximated by a poisson distribution with parameter \(\lambda = np\).

Example 1: Rare Event Detection

Probability that at least 4 sets of triplets are born, assuming a triplet birth probability of 0.0001 and 10,000 births:

In R:

lambda <- 10000 * 0.0001  # λ = np = 1
1 - ppois(3, lambda)       # P(X ≥ 4)
## [1] 0.01898816

Example 2: Defective Products

If 2% of products are defective, and we sample 100 products, what is the probability that exactly 3 are defective?

lambda <- 100 * 0.02
dpois(3, lambda)
## [1] 0.180447

Example 3: Typos on a Page

If the number of typos per page follows Poisson(λ = 1), what’s the probability of at least one typo?

1 - dpois(0, 1) 
## [1] 0.6321206

Example 4: No Accidents Today

If the average number of accidents per day on a highway is λ = 3, the probability of no accidents today is:

dpois(0, 3)
## [1] 0.04978707

Interpretation in ddPCR

In ddPCR, the number of target molecules per droplet follows a Poisson distribution with parameter \(\lambda\). This is because each droplet is randomly and independently assigned zero or more DNA molecules. The key relationship used in ddPCR is:

\[ P(\text{no target in droplet}) = P(X=0) = e^{-\lambda} \] Which leads to the estimate:

\[ \lambda = -ln(\frac{\text{number of negative droplets}}{\text{total number of droplets}}) \]

This formula forms the foundation of absolute quantification in ddPCR, as we will explore in the next chapter.