Probabilistic QPF 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
0
500
1000
1500
2000
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
0
500
1000
1500
2000
2500
3000
0.05 inch bins
Freq
uenc
y pe
r bin
Individual Winter EventsWinter Precip Frequencies Data
81 rainfall events
0
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
0
100
200
300
400
500
600
700
800
900
1000
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
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
150
200
250
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
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
0.4000
0.5000
0.6000
0.7000
0.8000
0.9000
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
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
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
0
0.3
0.6
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)
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
0.6
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