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Rmd 2dceb55 Dave Tang 2024-07-30 The Poisson distribution

Introduction

A Poisson distribution is the probability distribution that results from a Poisson experiment. A probability distribution assigns a probability to possible outcomes of a random experiment. A Poisson experiment has the following properties:

  1. The outcomes of the experiment can be classified as either successes or failures.
  2. The average number of successes that occurs in a specified region is known.
  3. The probability that a success will occur is proportional to the size of the region.
  4. The probability that a success will occur in an extremely small region is virtually zero.

A Poisson random variable is the number of successes that result from a Poisson experiment. Given the mean number of successes that occur in a specified region, we can compute the Poisson probability based on the following formula:

\[ P(x; \mu) = \frac{(e^{-\mu})(\mu^x)}{x!} \]

which is also written as:

\[ Pr(X = k) = e^{-\lambda} \frac{\lambda^k}{k!} \ \ k = 0, 1, 2, \dotsc \]

Examples

Homes sold

The average number of homes sold is 2 homes per day. What is the probability that exactly 3 homes will be sold tomorrow?

\[ P(3; 2) = \frac{(e^{-2}) (2^3)}{3!} \]

Calculating this manually in R:

e <- exp(1)
((e^-2)*(2^3))/factorial(3)
[1] 0.180447

Using dpois():

dpois(x = 3, lambda = 2)
[1] 0.180447

RNA-seq

The Poisson distribution can be used to estimate the technical variance in high-throughput sequencing experiments. My basic understanding is that the variance between technical replicates can be modelled using the Poisson distribution. Check out Why Does Rna-Seq Read Count Fit Poisson Distribution? on Biostars.

From Chris Miller:

Picture a process whereby you take the genome and choose a location at random to produce a read. This is a Poisson process. If you plot the depth of sequence along this theoretical genome, it will be a poisson distribution.

Expanding on the idea above in Modern Statistics for Modern Biology:

Consider a sequencing library that contains \(n_1\) fragments corresponding to gene 1, \(n_2\) fragments for gene 2, and so on, with a total library size of \(n = n_1 + n_2 + \ldots\). This library is then sequenced and the identity of \(r\) randomly sampled fragments are determined. To paint a better mental picture, the following are some typical numbers. The number of genes will be in the order of tens of thousands; the value of \(n\) depends on the amount of cells that were used to prepare the library and typically this is in the order of billions or trillions; and the number of reads \(r\) is usually in the tens of millions, which is much smaller than \(n\). Sequencing is sampling from \(n\).

From this we can conclude that the probability that a given read maps to the \(i\) th gene is \(p_i = n_i/n\) (ratio of a specific fragment to all fragments) and this is independent of the outcomes for all the other reads. So we can model the number of reads for gene \(i\) by a Poisson distribution, where the rate of the Poisson process is the product of \(p_i\), the initial proportion of fragments for the \(i\) th gene, times \(r\), the number of reads sequenced; that is \(\lambda_i = rp_i\).

In practice, we are usually not interested in modeling the read counts within a single library, but in comparing the counts between libraries. That is, we want to know whether any differences that we see between different biological conditions are larger than what we might expect even between biological replicates. Empirically, it turns out that replicate experiments vary more than the Poisson distribution predicts.

Intuitively, what happens is that \(p_i\), and therefore also \(\lambda_i\), varies even between biological replicates. To account for that variation, we need to add another layer of modeling on top and it turns out that the gamma-Poisson (a.k.a. negative binomial) distribution suits our modeling requirements. Instead of a single \(\lambda\), which represents both mean and variance, this distribution has two parameters. In principle, these can be different for each gene and we can estimate them from the data.

Calculating confidence intervals

Calculate the confidence intervals using R. Create data with 1,000,000 values that follow a Poisson distribution with lambda = 20.

set.seed(1984)
n <- 1000000
data <- rpois(n, 20)

Functions for calculating the lower and upper tails.

poisson_lower_tail <- function(n) {
   qchisq(0.025, 2*n)/2
}
poisson_upper_tail <- function(n) {
   qchisq(0.975, 2*(n+1))/2
}

Lower limit for lambda = 20.

poisson_lower_tail(20)
[1] 12.21652

Upper limit for lambda = 20.

poisson_upper_tail(20)
[1] 30.88838

How many values in data are lower than the lower limit?

table(data<poisson_lower_tail(20))

 FALSE   TRUE 
961213  38787

How many values in data are higher than the upper limit?

table(data>poisson_upper_tail(20))

 FALSE   TRUE 
986239  13761

What percentage of values were outside of the 95% CI?

(sum(data<poisson_lower_tail(20)) + sum(data>poisson_upper_tail(20))) * 100 / n
[1] 5.2548

Plot.

hist(data)
abline(v=poisson_lower_tail(20))
abline(v=poisson_upper_tail(20))

Version Author Date
e49a794 Dave Tang 2024-07-30

Webtool

Using the Poisson Confidence Interval Calculator and lambda = 20 returns:

  • 99% confidence interval: 10.35327 - 34.66800
  • 95% confidence interval: 12.21652 - 30.88838
  • 90% confidence interval: 13.25465 - 29.06202

which matches our 95% CI values.


sessionInfo()
R version 4.4.0 (2024-04-24)
Platform: x86_64-pc-linux-gnu
Running under: Ubuntu 22.04.4 LTS

Matrix products: default
BLAS:   /usr/lib/x86_64-linux-gnu/openblas-pthread/libblas.so.3 
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time zone: Etc/UTC
tzcode source: system (glibc)

attached base packages:
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other attached packages:
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[49] knitr_1.47        htmltools_0.5.8.1 rmarkdown_2.27    compiler_4.4.0   
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