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Probabilistic QPF. Steve Amburn, SOO WFO Tulsa, Oklahoma. Preview. Product review Graphic and text Forecaster involvement Examples of rainfall distributions Theory and Definitions Comparisons To previous work To real events. How do we distribute PQPF information? .
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Probabilistic QPF Steve Amburn, SOO WFO Tulsa, Oklahoma
Preview • Product review • Graphic and text • Forecaster involvement • Examples of rainfall distributions • Theory and Definitions • Comparisons • To previous work • To real events
Distribute of text information Probability to exceed 0.10” = 67% Probability to exceed 0.50” = 46% Probability to exceed 1.00” = 30% Probability to exceed 2.00” = 12% 6-hr PoP = 70% 6-hr QPF = 0.82
Premise • Rainfall events are typically characterized by a distribution or variety of rainfall amounts. • Every distribution has a mean. • Data indicates most rainfall events have exponential or gamma distributions. • Forecast the mean, and you forecast the distribution. • That forecast distribution lets us calculate exceedance probabilities (POEs), i.e., the probabilistic QPF.
Characteristics of POE Method for Probabilistic QPF (exponential distributions) • As mean increases POE increases
Forecaster Involvement • No additional workload • No extra grids to edit • No extra time to create the products • Forecasters simply issue their QPF, given the more specific definition. • New QPF definition ~ Old QPF definition
Gamma Distribution • “There are a variety of continuous distributions that are bounded on the left by zero and positively skewed. One commonly used choice, used especially often for representing precipitation data, is the gamma distribution. ...” D.S. Wilks (1995), Statistical Methods in the Atmospheric Sciences, Academic Press, pp. 467.
Gamma Distributions exponential distribution (special case of gamma) blue line => α =1red line => α=2black line => α=3
Gamma Distribution? YesUsually, the Exponential Distribution • Over time at a point (TUL, FSM...) • For individual events (nearly 700) • Virtually all were exponential distributions • A few were other forms of gamma distribution • Let’s look at the math...
Gamma Function The gamma function is defined by: Г(α) = ∫ x (α-1) e-x dx ,for α > 0 , for 0 → ∞.
Gamma Distributions exponential distribution (special case of gamma) blue line => α =1red line => α =2black line => α =3
Gamma Function for α = 1 • Г(α) = ∫ x (α-1) e-x dx ,for α > 0 , for 0 → ∞. • So, for α = 1, • Г(1) =∫X(1-1)e-x dx • =∫e-x dx = - e-∞ - (-e-0) = 0 + 1 • Г(1) = 1. ∞ 0
Gamma PDF Exponential PDF Now, for α = 1, gamma distribution PDF simplifies. f(x) = { 1 / [ βα Γ(α) ] } • xα-1 e-x/β, (gamma density function) So, substituting α = 1 yields: f(x) ={ 1 / [β1 Γ(1) ] } • x1-1 e-x/β or, f(x) = (1/β) • e-x/β, (exponential density function)
POE for theExponential Distribution So, to find the probability to exceed a value “x”, we integrate from x to infinity (∞). f(x) = POE(x) = ∫ (1/β) • e-x/β , from x → ∞. = -e-∞/β - (-e-x/β ) = 0 + e-x/β so, POE(x) = e-x/β,where β = mean
Unconditional POE • Simply multiply conditional POE by PoP, • uPOE(x) = PoP x (e-x/μ) • , where μ is the conditional QPF. • Consider, the NWS PoP = uPOE (0.005)
Individual event examples • Data from ABRFC • 4578 HRAP grid boxes in TSA CWFA. • Rainfall distributions created • Means calculated • POEs calculated from observed data • Actual means used to calculate POEs from PDFs (formula)
Actual POE = computed from the observed data. Perfect POE = computed using the exponential equation and the observed mean
Actual POE = computed from the observed data Perfect POE = computed using the exponential equation and the observed mean Climo POE = exp equation with climatological mean
Actual POE = computed from the observed data Exponential Eq = computed using the exponential equation and the observed mean
Actual POE computed from frequency data shown below. Using integrated exponential distribution (gamma distribution where alpha = 1) to compute POE. POE using integrated exponential distribution and climatological mean precipitation.
Actual POE computed from frequency data shown below. POE using integrated gamma distribution (alpha=3) POE from exponential distribution (gamma distribution, alpha = 1). POE using integrated exponential distribution and climatological mean precipitation.
Gamma Distribution Gamma function is defined by: Г(α) = x (α-1) e-x dx ,for α > 0 , for 0 → ∞. Gamma density function is given by: 1 ] _________ [ f(x) = • (xα-1 • e–x/β ) β • Γ(α)
Gamma Distribution, A = 3 1 _________ [ ] • (xα-1 • e–x/β ) f(x) = β • Γ(α) Then, integrate, for α =3, from x to infinity to obtain the probability of exceedance for x. POE(x) =(0.5)•(e –x/β)•(x2/β + 2x/β + 2) ,where β = µ/α = (qpf / 3)
Error in Exceedance Calculations (419 events) • Calculated at: 0.10, 0.25, 0.50, 1.0, 1.5” Areal Coverage
Error in Exceedance Calculations (279 events) • Calculated at: 0.10, 0.25, 0.50, 1.0, 2.0, 3.0, 4.0 inches Areal Coverage
Summary • Rainfall events have distributions • Typically exponential • Gamma for coverage > 90% • Each distribution has a mean. • Mean can be spatial for a single event • Mean can be at a point over a period of time • Forecast the mean (QPF) and you have effectively forecast the distribution. • So, QPF lets us calculate probabilities of exceedance, or PQPF.
Questions? Steve Amburn, SOO WFO Tulsa, Oklahoma