Question

A team of ornithologists in southeast Peru is studying the relative (geographical) distributions of two subspecies of Rupicola peruvianus, R. p. saturatus and R. p. aequatorialis, in a forest where the ranges of the two overlap. In doing so, they set up an observation post in the forest and make observations for six days.

(a) It is determined that a reasonable probability model for the number of R. p. aequatorialis observed over this time is Poisson, with λ1 > 0 sightings on average. In terms of λ1, what is the probability that they have k sightings?

(b) It is further determined that the number of R. p. saturatus observed over the same time is independently Poisson, with λ2 > 0 sightings on average. What is the (probability) distribution of the total number of birds that will be sighted? Hint: the sum X + Y of two independent Poisson random variables X and Y is also a Poisson random variable. Furthermore, E(X + Y ) = E(X) + E(Y ).

(c) In terms of λ1 and λ2, what is the probability that the total number of birds sighted will be exactly n?

(d) Suppose that, at the end of the six days, the total number of birds that have been sighted is exactly n. Conditional on this event, what is the probability that exactly k of these n are R. p. aequatorialis? Hint: P(A|B) = P(A∩B) P(B)

(e) Conditional on the event of observing n birds in total, what is the (named) distribution of the number of R. p. aequatorialis? Hint: write the probability in part (d) in terms of λ1 λ1+λ2

Answer 1

Solution

Back-up Theory

If a random variable X ~ Poisson (λ), i.e., X has Poisson Distribution with mean λ then

probability mass function (pmf) of X is given by P(X = x) = e ^{– λ}.λ^x/(x!) ….............................................………..(1)

where x = 0, 1, 2, ……. , ∞

If X₁ ~ Poisson (λ₁), X₂ ~ Poisson (λ₂) and X₁, X₂ are independent, then (X₁ + X₂) ~ Poisson (λ……….....…..(2)

If A and B are two events such that probability of B is influenced by occurrence or otherwise of A, then

Conditional Probability of B given A, denoted by P(B/A) = P(B ∩ A)/P(A)……...............................................….(3)

Now, to work out the solution,

Part (a)

Let

X = number of sightings of R. p. aequatorialis. Given, X ~ Poisson ( λ₁) ....................................................... (4)

the probability that they have k sightings

= P(X = k)

= e ^{– λ1}.λ₁^k/(k!)   [vide (1)] Answer 1

Part (b)

Let

Y = number of sightings of R. p. saturatus . Given, Y ~ Poisson ( λ₂)......................................................... (5)

Also given that the number of R. p. saturatus observed over the same time is independently Poisson,

vide (2), (4) and (5),

X + Y ~ Poisson (λ₂ + λ₂).

Thus, the (probability) distribution of the total number of birds that will be sighted is Poisson (λ₂ + λ₂) Answer 2

Part (c)

The probability that the total number of birds sighted will be exactly n

= P(X + Y = n)   [vide (4) and (5)]

= e ^{– (λ1+ λ1)}.(λ₁ + λ₂)ⁿ/(n!)   [vide (1) and Answer 2]   Answer 3

Part (d)

Let A represent the event that at the end of the six days, the total number of birds that have been sighted is exactly n and B represent the event that exactly k of these n are R. p. aequatorialis. Then,

A and B would mean that exactly k of these n are R. p. aequatorialis and (n - k) are R. p. saturates.

The required probability is:

P(B/A)

= P(B ∩ A)/P(A) [vide (3)]

= P(exactly k R. p. aequatorialis and (n - k) R. p. saturates)/P(total number of sighting is exactly n)

= P[X = k and Y = (n - k)]/P(X + Y = n)

= {e ^{– λ1}.λ₁^k/(k!)}x{e ^{– λ2}.λ₂^{n - k}/{(n – k)!}/ {e ^{– (λ1+ λ1)}.(λ₁ + λ₂)ⁿ/(n!)}   [vide Answers 1 and 3]

= {e ^{– (λ1+ λ1)} λ₁^k λ₂^{n – k}/(k!) (n – k)!}/ {e ^{– (λ1+ λ1)}.(λ₁ + λ₂)ⁿ/(n!)}

= {(n!)/(k!) (n – k)!} λ₁^k λ₂^{n – k}/{(λ₁ + λ₂)ⁿ }

= (ⁿC_k) λ₁^k λ₂^{n – k}/{(λ₁ + λ₂)ⁿ } Answer 4

DONE