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?

Bar Chart – Osage County Average

Distribute of text information 6-hr PoP = 70% 6-hr QPF = 0.82 Probability to exceed 0.10” = 67% Probability to exceed 0.50” = 46% Probability to exceed 1.00” = 30% Probability to exceed 2.00” = 12%

Premise 1.Rainfall events are typically characterized by a distribution or variety of rainfall amounts. 2.Every distribution has a mean. 3.Data indicates most rainfall events have exponential or gamma distributions. 4.Forecast the mean, and you forecast the distribution. 5.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

What is QPF? The forecast rainfall amount in the “gage.” An average? At a point, over a “basin” or area? To be consistent with PoP, QPF should be for a point.

QPF Definition – Current (NWSI) (NWSI ) - QPF 12HR. This parameter, quantitative precipitation forecast (QPF) represents the total amount of liquid precipitation, in inches, expected at the specific point during a 12-hour period ending at 6:00 a.m., or 6:00 p.m. local time. The QPF is presented in locally defined ranges, (e.g., ), or single values. (NWSI ) - Quantitative Precipitation Forecast (QPF) - the total amount of expected liquid precipitation (in hundredths of inches).

QPF Definition - Enhanced The mean of the precipitation distribution (liquid) at a point that a forecaster would be expected from an infinite number of similar events over time Or... The mean of the expected distribution of precipitation (liquid) over an area for a single specific event. These two are virtually the same. (See Hughes NOAA tech memo, NWS FCST 24, 1980)

So, lets talk distributions... Selecting distribution for weather parameters –The way they are measured –The values that are observed Gaussian or normal distribution –Temperature and temperature climatologies –Many annual climatologies Gamma distribution –Best for skewed distributions (bounded on left by zero) –Precipitation events –Exponential distribution (special case of gamma distribution)

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 blue line => α =1 red line => α =2 black line => α =3 exponential distribution (special case of gamma)

A look at point rainfall distributions ( ) Distributions for FSM, MLC, FYV are virtually identical.

Individual Spring Events

Individual Summer Events

Individual Autumn Events

Individual Winter Events

Gamma Distribution? Yes Usually, 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 blue line => α =1 red line => α =2 black line => α =3 exponential distribution (special case of gamma)

Gamma Function for α = 1 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 ) = Г(1) = 1. 0 ∞

Gamma PDF  Exponential PDF Now, for α = 1, gamma distribution PDF simplifies. f(x) = { 1 / [ β α Γ(α) ] } x α-1 e -x/β, (gamma density function) So, substituting α = 1 yields: f(x) = { 1 / [ β 1 Γ(1) ] } x 1-1 e -x/β or, f(x) = (1/β) e -x/β, (exponential density function )

POE for the Exponential 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, 6) uPOE(x) = PoP x (e -x/μ ), where μ is the conditional QPF. Consider, the NWS PoP = uPOE (0.005)

Does POE equation really work? Compare POE formula to climatological calculations using observed data (Jorgensen, Klein and Roberts, 1969) Use J&K69 seasonal means in POE equation for comparison.

Washington, D.C. – J&K vs Equation

Other point comparisons similar

Atlanta – J&K69 vs Equation

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

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 from exponential distribution (gamma distribution, alpha = 1). Actual POE computed from frequency data shown below. POE using integrated gamma distribution (alpha=3) 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: f(x) = 1 β Γ(α) _________ ( x α-1 e –x/β ) [ ]

Gamma Distribution, A = 3 f(x) = 1 β Γ(α) _________ ( x α-1 e –x/β ) [] Then, integrate, for α =3, from x to infinity to obtain the probability of exceedance for x. POE(x) =(0.5)(e –x/β )(x 2 /β + 2x/β + 2), where β = µ/α = (qpf / 3)

Different Distributions? Perhaps For events near 100% coverage –Assuming exponential not completely accurate –Gamma function with another alpha may be better More research may be needed Forecaster training needed

Gamma, where a=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 1.Rainfall events have distributions 1.Typically exponential 2.Gamma for coverage > 90% 2.Each distribution has a mean. 1.Mean can be spatial for a single event 2.Mean can be at a point over a period of time 3.Forecast the mean (QPF) and you have effectively forecast the distribution. 4.So, QPF lets us calculate probabilities of exceedance, or PQPF.

Questions? Steve Amburn, SOO WFO Tulsa, Oklahoma