Stochastic fluctuations in population dynamics

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1 Stochastic fluctuations in population dynamics Blue tit (Parus caeruleus) Great tit (Parus maior)

2 Fluctuations in population dynamics Variation coefficient CV = σn/n σlogn CV if CV<30% Swan (Cygnus olor)

3 Fluctuations in population dynamics Seychelles warbler (Acrocephalus sechellensis)

4 Fluctuations in population dynamics Albatross (Diomedea exulans)

5 Fluctuations in population dynamics Chamois (Rupicapra rupicapra) Soay sheep (Ovis aries)

6 CV of population fluctuations

7 Deterministic and stochastic components Fluctuations in population dynamics can be due to Deterministic drivers: - High fertility; - Seasonality; - Interactions (predation ) N t +1 = λ t N t Stochastic drivers: - demographic (Ovis canadensis) - environmental (Flamingoes in South Africa);

8 Examples Bighorn sheep (Ovis canadensis)

9 Examples Flamingo (Phoenicopterus ruber ) Phoeniconaias minor

10 Extinctions: deterministic or stochastic?

11 Discrete vs. continuous demography N t +1 = λ t N t Individual fitness Contribution of individual i at time t to next generation t + 1 N t w i,t N t +1 = w i,t i=1 Is there a relationship?

12 Environmental vs. demographic stochasticity N t +1 = λ t N t = where w i,t = w t + δ i N t λ t = i=1 w i,t N t i=1 w i,t N t E δ i = 0 2 Var δ i = σ d E w t = w 2 Var w t = σ e E λ t N t = N =? Var λ t N t = N =?

13 Environmental vs. demographic stochasticity N t+1 = λ t N = N t w i,t i=1 Typical values of σ e 2 and σ d 2 N N w λ t = i,t wt + δ = i N = wt + N E λ t i=1 Var λ t = w i=1 = Var wt Var + N i=1 N 2 δ i N i=1 δ i N = σ 2 e + σ 2 d N N >> σ 2 d σ e 2 e.g. N = 10 σ 2 d σ e 2 = N c Calculate critical population

14 Two remarkable results N t ( ) = E N ( t) p 0 ( t) = Demographic stochasticity Malthusian case - Variables p i (t) = probability that population contains i individuals at time t - Demographic parameters (birth rate ν and mortality rate µ) - Kolmogorov equations describe dynamics of p i (t) µexp rt ν exp rt = ip ( t) dn t i dt i=0 ( ) µ ( ) µ N0 ( ) if r < 0, lim if r = 0, lim if r > 0, lim = ( ν µ ) N ( t) t p 0 t p 0 t p 0 ( t) = ( t) = ( t) =

15 Realizations of birth-death stochastic process

16 Extinction probability (demographic stochasticity) Average and median extinction time

17 Demographic stochasticity and densitydependence In nonmalthusian case Extinction is certain in the long term, however lim t p 0 ( t) = 1 Extinction probability p 0 (t) K = 10 K = 15 K = 20 K = 30 K = 40 K = Tempo Time tt

18 N t +1 = where N t w i,t i=1 Environmental stochasticity = w t N t = λ t N t λ t = δ t Λ( N ) t = exp( ε t )Λ( N ) t White noise E ε t = 0 2 E ε t = Var ε 2 t = σ ε E ε t ε t τ = 0 ( ) > 1 2 E δ t = exp σ ε 2 Median δ t = exp Median ε t Mode δ t < 1 ( ) = 1

19 Environmental stochasticity: the Malthusian case λ t = δ t Λ( N ) t = exp( ε t )λ λ = median! where N t = λ t 1 λ t 2 λ 0 N 0 = λ t exp( ε t 1 + ε t ε ) 0 N 0 Take logarithms t 1 log( N ) t = t log(λ) + log( N ) 0 + ε i i=0 Result E Var t 1 i=0 ε i t 1 i=0 t 1 = E ε i = 0 i=0 t 1 2 = Var ε i = tσ ε i=0 ε i E log( N t ) = t log λ Var log( N t ) = E log N t t 1 ( ) + log( N 0 ) + E ε i = rt + log N 0 i=0 ( ) E log( N t ) ( ) 2 = E t 1 i=0 ( ) ε 2 i = t 1 2 Var ε i = tσ ε i=0

20 Environmental stochasticity: the Malthusian case

21 Environmental stochasticity: the Malthusian case Therefore, if log( λ) = r > 0 then population increases. Remark: logarithm of median must be positive not logarithm of average! In fact, with lognormal noise E λ t = E λδ t = λe δ t = λ exp σ 2 ε 2 > λ Therefore, using the average of λ t s can be misleading and produce an overestimation of the population growth rate Example Annual population with finite rate of increase equal to 1.1 in normal years and 0.3 in critical years that occur once in a decade. What is the fate of the population?

22 Environmental stochasticity: the Malthusian case E λ t = 1.02 Median λ t = 0.966

23 and environmental stochasticity Whenever environmental stochasticity brings population below a critical threshold it can be captured by an extinction vortex : N < N c quasiextinction How to estimate quasiextinction probability? Based on r and σ 2 ε How to estimate σ 2 ε? ( N ) t = r + ε t log N t +1 Use estimate of r Tail of Normal ˆε t = log N t +1 ˆσ ε 2 = ( ) ˆr 1 m 1 N t m 1 2 ˆε i i=1

24 Normal distribution table

25 Environmental stochasticity and density dependence log N t +1 = log λ βn + ε t t N t Red Deer abundance in Yellowstone Park, and logarithmic rate of increase vs deer abundane (data and linear regression)

26 Environmental stochasticity and density dependence N t+1 = λ N t exp(-β N t +ε t ) λ = 4, β = 0.001, σ ε = average, standard deviation and median of population abundance; extinction probability vs time; average and median extinction time

27 Extinction vortices

28 Population Viability Analysis (PVA) Software: VORTEX, RAMAS, ALEX

29 Equal risk curves How to classify threatened species A B C D E

30 IUCN (International Union for Conservation of Nature) classification: Mace and Lande slightly modified Ex. Category CR Criterion D: Population size of mature individuals N < 50 Criterion E: Extinct. Probab. 50%within next 10 years or next 3 generations

31 Endangered plants in Italy Endangered species in USA

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