Statistical Models for Rainfall with Applications to Index Insura

Statistical Models for Rainfall with Applications to April 21, 2008

Overview The idea: Insure farmers against the risk of crop failure, like drought, instead of crop failure itself. It reduces moral hazard and adverse selection.

Overview The idea: Insure farmers against the risk of crop failure, like drought, instead of crop failure itself. It reduces moral hazard and adverse selection. (Rainfall, Other Factors) Crop Yield Income. Statistically speaking: Crop Yield i = f (Rainfall i ) + ɛ i, Income i = f (Crop Yield i )

Stabilizing Annual Income Without Insurance: Net Income = Yield ($) With Insurance: Net Income = Yield ($) + Insurance Payout - Cost of Insurance Without Insurance With Insurance Income ($) 1300 1200 1100 1000 900 800 700 Income ($) 1300 1200 1100 1000 900 800 700 0 100 200 300 400 500 0 100 200 300 400 500 Year Year

Contract Design Net Income i = Yield i + Insurance Payout i Cost of Insurance = f (R i ) + ɛ i + g(r i ) c Payout = a(rainfall) + b, a < 0 Minimizing Variance 1400 1200 220 Payout ($) 1000 800 600 400 SD of Net Income ($) 200 180 160 200 0 140 0 200 400 600 800 1000 1400 2.0 1.5 1.0 0.5 0.0 Rainfall (mm) Slope = a

The Data 70 Daily Rainfall 60 50 40 mm 30 20 10 0 1961 1962 1963 1964 1965 Year Figure: Daily Rainfall, Lilongwe, Malawi, 1961-1965.

For most data sets, we have between 5 and 45 years of daily rainfall data. Wet Day Indicator X = {X mtd } = 1{Y mtd > 0}, the indicator of a wet day on day d in year t of month m. Rainfall Y = {Y mtd } denote the amount of rainfall on day d in year t of month m. We fit a model whose likelihood can be factored into two parts: the likelihood of the frequency of wet days, X, and the likelihood of the intensity of rainfall on wet days, Y : P(X, Y θ) = P(X θ 1 ) P(Y θ 2 ).

The model for rainfall intensity (Y ) Y mtd X mtd = 1 Gamma(α mt, β mt ),

The model for rainfall intensity (Y ) Y mtd X mtd = 1 Gamma(α mt, β mt ), ( log(αmt ) log(β mt ) ) ) N (µ m, Σ m,

The model for rainfall intensity (Y ) Y mtd X mtd = 1 Gamma(α mt, β mt ), ( log(αmt ) log(β mt ) ) ) N (µ m, Σ m, p(µ m, Σ m ) Σ m 3 2, for t = 1,..., 45, and d = 1,..., n d.

Illustration Rainfall Intensity by Month 20 15 mm 10 5 0 JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC Month

The model for rainfall frequency (X ) ( ) p1mt 1 p X mtd X mt(d 1) P mt = 1mt 1 p 2mt p 2mt

The model for rainfall frequency (X ) ( ) p1mt 1 p X mtd X mt(d 1) P mt = 1mt 1 p 2mt p 2mt p jmt Beta(µ jm, s jm )

The model for rainfall frequency (X ) ( ) p1mt 1 p X mtd X mt(d 1) P mt = 1mt 1 p 2mt p 2mt p jmt Beta(µ jm, s jm ) µ jm U(0, 1) s jm U(0, ) for d = 2,.., n d, m = 1,..., 12, t = 1,..., 45, and j = 1, 2.

Fit the model

Fit the model www.r-project.org

Model Checking - Posterior Predictive Checks 1. 1 data set = 45 years of daily rainfall. Measure ˆθ data. ex) ˆθ data = SD of annual rainfall 2. Simulate 500 copies of the data (45 years of daily rainfall each). Measure ˆθ rep for each copy from rep = 1,...,500. 3. Plot the distribution of ˆθ rep (a histogram), and then plot the observed value ˆθ data and see if it falls within the distribution of simulated statistics.

Posterior Predictive Check *Could this model have produced this data? Interannual Variability 50 40 30 20 10 0 120 140 160 180 200 220 mm

Interpreting Posterior Predictive Checks Interpret it like a p-value. 1 out of 20 values of ˆθ data will lie in the tail of the distribution of ˆθ rep just by random chance.

When a bad result is good: (ˆθ data = Y mt1 Y (m 1)t(31) ) This model exaggerates the difference in rainfall between the last day of one month and the first day of the next month. AUG SEP OCT NOV Frequency 0 50 100 150 200 250 Frequency 0 50 100 150 200 Frequency 0 50 100 150 Frequency 0 50 100 150 0.2 0.1 0.0 0.1 0.2 0.3 0.4 0.2 0.0 0.2 0.4 0.6 0.8 0.5 0.0 0.5 1.0 1.5 2 0 2 4 6 mean.last.sim[, i] mean.last.sim[, i] mean.last.sim[, i] mean.last.sim[, i] DEC JAN FEB MAR Frequency 0 20 40 60 80 100 Frequency 0 20 40 60 80 Frequency 0 20 40 60 80 100 140 Frequency 0 50 100 150 2 0 2 4 6 8 10 6 4 2 0 2 4 6 8 5 0 5 10 5 0 5 mean.last.sim[, i] mean.last.sim[, i] mean.last.sim[, i] mean.last.sim[, i] APR MAY JUN JUL Frequency 0 20 40 60 80 100 140 Frequency 0 50 100 150 Frequency 0 50 100 150 Frequency 0 50 100 150 200 250 8 6 4 2 0 2 3 2 1 0 1 2 1.0 0.8 0.6 0.4 0.2 0.0 0.2 0.4 0.6 0.4 0.2 0.0 0.2 mean.last.sim[, i] mean.last.sim[, i] mean.last.sim[, i] mean.last.sim[, i]

The Frequency Domain Rainfall-generating process is smooth through time: Y t gamma(α t, β t ), for t = 1,..., (45 365). α t N(X t η α + a sin (2π 1 1 365t) + b cos (2π 365 t), σ2 α), β t N(X t η β + c sin (2π 1 1 365t) + d cos (2π 365 t), σ2 β ). Non-informative priors on η α, η β, a, b, c, d.

Thanks : Dan Osgood (IRI) & colleagues Andy Robertson (IRI) Thanks. www.stat.columbia.edu/ shirley