Probabilistic QPF Steve Amburn, SOO WFO Tulsa, Oklahoma

Probabilistic QPFProbabilistic 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

Probability to Exceed rainfall amounts for a mean value Mu

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4

Rainfall (tenths of inches)

Pro

ba

bil

ity Mu=.1

Mu=.2

Mu=.5

Mu=1

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

blue line => α =1

red line => α =2 black line => α =3

exponential distribution (special case of gamma)

Individual Spring Events Spring Precip Frequencies Data

81 rainfall events

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500

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1500

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2500

3000

0.05 inch bins

Freq

uenc

y pe

r bin

Individual Summer EventsSummer Precip Frequencies Data

122 rainfall events

0200

400600

8001000

12001400

16001800

0.05 inch bins

Freq

uenc

y pe

r bin

Individual Autumn Events

Autumn Precip Frequencies Data92 rainfall events

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500

1000

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3000

0.05 inch bins

Freq

uenc

y pe

r bin

Individual Winter EventsWinter Precip Frequencies Data

81 rainfall events

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500

1000

1500

2000

2500

3000

3500

0.05

0.15

0.25

0.35

0.45

0.55

0.65

0.75

0.85

0.95

1.05

1.15

1.25

1.35

1.45

1.55

1.65

1.75

1.85

0.05 inch bin categories

Freq

uenc

y

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

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) = 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,

6) 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

uPOE for 8/22/06

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

00.

15 0.3

0.45 0.

60.

75 0.9

1.05 1.

21.

35 1.5

1.65 1.

81.

95

0.05 in bins

UP

OE Actual POE

Climo POE

Perfect POE

Rainfall Frequency by Bin

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700

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1000

0.05

0.15

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0.45

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0.75

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0.95

1.05

1.15

1.25

1.35

1.45

1.55

1.65

1.75

1.85

1.95

0.05 inch bins

# G

rid

s

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.

uPOE for 9/18/06

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

00.

15 0.3

0.45 0.

60.

75 0.9

1.05 1.

21.

35 1.5

1.65 1.

81.

95

0.05 in bins

UP

OE

Actual POE

Perfect POE

Climo POE

POE from exponential distribution (gamma distribution, alpha = 1).

Actual POE computed from frequency data shown below.

POE using integrated gamma distribution (alpha=3)

uPOE for 9/18/06

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

00.

15 0.3

0.45 0.

60.

75 0.9

1.05 1.

21.

35 1.5

1.65 1.

81.

95

0.05 in bins

UP

OE

Actual POE

Perfect POE

Gamma (a=3)

Climo POE

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)

_________ • (xα-1 • e –x/β )[ ]

Gamma Distribution, A = 3

f(x) = 1

β • Γ(α)

_________ • (xα-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)•(x2/β + 2x/β + 2)

,where β = µ/α = (qpf / 3)

Gamma, where a=3

Gamma, where a=3

Frequency Distribution, 4/1/2008, ending 7 pmUncond Mean QPE = 1.01", Coverage = 97%

0

50

100

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0.7 1

1.2

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Rainfall bins (0.05" each)

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er b

ins

4/1/2008

Exceedance Probabilities, 4/1/2008, ending 7 pm Uncond Mean QPE = 1.01", Coverage = 97%

0.0000

0.1000

0.2000

0.3000

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1.0000

00.

30.

60.

91.

21.

51.

82.

12.

42.

7 33.

33.

63.

94.

24.

54.

8

Rainfall bins (0.05" each)

Pro

ba

bili

ty

Actual POE

Perfect POE

Gamma (a=3)

Gamma, where a=3

Frequency Distribution, 4/1/2008, ending 7 pmUncond Mean QPE = 1.01", Coverage = 97%

0

20

40

60

80

100

120

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160

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1.2

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4.7 5

Rainfall bins (0.05" each)

Nu

mb

er b

ins

3192008

Exceedance Probabilities, 4/1/2008, ending 7 pm Uncond Mean QPE = 1.01", Coverage = 97%

0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

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3.3

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3.9

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4.5

4.8

Rainfall bins (0.05" each)

Pro

bab

ilit

y

Actual POE

Perfect POE

Gamma (a=3)

Gamma, where a=3

Frequency Distribution, 4/1/2008, ending 7 pmUncond Mean QPE = 1.01", Coverage = 97%

0

20

40

60

80

100

120

140

160

0

0.2

0.5

0.7 1

1.2

1.5

1.7 2

2.2

2.5

2.7 3

3.2

3.5

3.7 4

4.2

4.5

4.7 5

Rainfall bins (0.05" each)

Nu

mb

er b

ins

3182008

Exceedance Probabilities, 4/1/2008, ending 7 pm Uncond Mean QPE = 1.01", Coverage = 97%

0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

0

0.3

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0.9

1.2

1.5

1.8

2.1

2.4

2.7 3

3.3

3.6

3.9

4.2

4.5

4.8

Rainfall bins (0.05" each)

Pro

bab

ilit

y

Actual POE

Perfect POE

Gamma (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

Summary1. 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