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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

Probabilistic QPF

Steve Amburn, SOO

WFO Tulsa, Oklahoma

preview
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
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
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
Characteristics of POE Method for Probabilistic QPF (exponential distributions)
  • As mean increases POE increases
forecaster involvement
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
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
Gamma Distributions

exponential distribution (special case of gamma)

blue line => α =1red line => α=2black line => α=3

gamma distribution yes usually the exponential distribution
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
Gamma Function

The gamma function is defined by:

Г(α) = ∫ x (α-1) e-x dx

,for α > 0 , for 0 → ∞.

gamma distributions1
Gamma Distributions

exponential distribution (special case of gamma)

blue line => α =1red line => α =2black line => α =3

gamma function for 1
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
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 the exponential distribution
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
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
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)
slide23

Actual POE = computed from the observed data.

Perfect POE = computed using the exponential equation and the observed mean

slide24

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

slide25

Actual POE = computed from the observed data

Exponential Eq = computed using the exponential equation and the observed mean

slide26

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.

slide27

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 distribution1
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
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
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
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
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

Questions?

Steve Amburn, SOO

WFO Tulsa, Oklahoma