6
$\begingroup$

I'm trying to understand what weather forecasts mean more precisely. As I understand it from reading Wikipedia, blogs, etc., the percentage value for rain/precipitation that you see in a forecast is technically called the "probability of precipitation". To quote the National Weather Service webpage:

"Mathematically, PoP is defined as follows: PoP = C x A where "C" = the confidence that precipitation will occur somewhere in the forecast area, and where "A" = the percent of the area that will receive measureable precipitation, if it occurs at all. So... in the case of the forecast above, if the forecaster knows precipitation is sure to occur ( confidence is 100% ), he/she is expressing how much of the area will receive measurable rain. ( PoP = "C" x "A" or "1" times ".4" which equals .4 or 40%.)"

This definition does not seem well stated to me, for the reason that "confidence" is (presumably) not uniform across an area. For instance, the phrase "how much of the area will receive measurable rain" seems odd, since a forecast would (presumably) only be able to give a probabilistic estimate for this area.

Let's cook up an example. Consider a town (the forecast area) consisting of two parts of equal area (called north side and south side). Each point in the north side will be rained on with 100% probability tomorrow, and each point in the south side will be rained on with 50% probability (at every point) tomorrow. What is the PoP in this example? At face value, the definition could be interpreted as implying that the PoP is 100%, since precipitation will occur somewhere in the forecast area. However, this value seems intuitively unsatisfying, since some people might not get any rain.

Here's what I might expect a more precise definition to be. If $A$ is the area and $C(x)$ is the "pointwise confidence function" depending on a point (location) $x$, then define $$PoP = \frac{1}{area(A)}\int_{A} C(x)\, dx.$$ In words, this is just the expected value of $C(x)$, or the probability that a randomly located person would see rain over the specified time interval. In practice, of course the integral would be estimated based on actual measurement sites. If such a formula is indeed an accurate definition, then I'd be satisfied. In the above example, the PoP would be 75%. (The official definition could in essence be viewed as a shorthand that is more useful for those without any background in calculus.) If this definition is not correct, then some explanation would be helpful.

I've read web articles with statements like the following: "As a student and observer of meteorology, it constantly bums me out that people do not understand what it means when someone says there’s an “X% chance of rain” tomorrow. A 50 percent chance of rain does not mean there’s a 1-in-2 chance that you’re going to get wet."

It's not clear to me why "A 50 percent chance of rain means there’s a 1-in-2 chance that you’re going to get wet" would be an inaccurate interpretation of PoP. If the definition I suggested above is correct, then it is perfectly correct to say that a stationary observer placed at a random location in this scenario would have a 50% chance of getting wet. Am I missing something, or is this author being careless?

I don't have any background in meteorology, and in particular I don't have much of a sense of how PoP is actually computed in practice.

$\endgroup$
  • $\begingroup$ With weather forecasting, meteorologists collect a lot of weather data & feed it into a number of computer models. The algorithms for each model varies, based on formulae & assumptions used. Sometimes the models gives the same results, sometimes slightly differing results & at other times very different results. The results of all the model runs are analyzed to give the most probably forecast. The probability of rain will depend on how may models indicate rain. The resolution of the grid used in the models will give the precision of the forecast. $\endgroup$ – Fred Jul 16 at 19:41
  • $\begingroup$ A related thing I've wondered about ever since xkcd brought it up: when we're given a probability of precipitation for a series of hours, are they independent values? Or does having rain in hour 1 lower the chance of rain in hour 2? (xkcd.com/1985) $\endgroup$ – Semidiurnal Simon Jul 16 at 22:55
  • $\begingroup$ (if nobody happens to touch on it in an answer here, I might ask as another question) $\endgroup$ – Semidiurnal Simon Jul 16 at 22:56
  • $\begingroup$ I've always taken the chance of precipitation as meaning that that percentage of the forecast area will experience rain. For example, if the chance of rain is, say, 70% and the forecast area is 1,000 square units, then 700 square units of the forecast area will experience rain. This, however, does not include geographical uniqueness, such as orthographic lifting and mountain rain shadows. $\endgroup$ – BillDOe Jul 17 at 20:25
  • $\begingroup$ @BillDOe I understand the basic concept, but what I'm looking for is a precise definition. For instance, your explanation uses the unclear word "will" in the same way that I complained about in the National Weather Service definition. How precise is "precise" enough? At a minimum, it should be able to tell me what the PoP is in my contrived north/south town example. I'd guess 75%, but this is exactly what I'd like to know. $\endgroup$ – mdr Jul 18 at 7:01
2
$\begingroup$

I agree that the PoP = C x A that we see on a lot of websites leaves something to be desired. It gets across the loose idea that the definition involves the area affected as well as the probability of occurrence, which is fine for most casual readers but can be frustrating for more inquisitive minds. How much are the confidence and the area estimated from ensemble forecasts versus expert judgement? How was it done before ensemble NWP started in the 1990s?

Anyway, there’s an interesting survey by Stewart et al (2016) that gives a broader overview of how different meteorologists use the phrase “probability of precipitation”. They categorise the PoP = C x A usage as following the NWS Operations Manual (1984), which is nicely summarized in a comment by Schaefer and Livingstone (1990). In short, your interpretation in terms of expectations is correct and, as you mention, they describe it in terms of a sum over a hypothetical network of points rather than a continuous integral:

$\textrm{PoP} = \frac{1}{N} \sum_{i=1}^N E[R_i]$,

where $R_i$ equals 1 if it rains at station $i$ and 0 if it doesn't. In your Pluieville example, this could be approximated as a town with one rain gauge in the north and one in the south, giving PoP = (1.0 + 0.5)/2 = 0.75.

I don't know for sure, but it wouldn't surprise me if these things are still calculated by subsampling gridded NWP models at actual rain gauge network locations, which would allow for comparable long-term verification of the probabilistic forecasts.

Stewart, A.E., C.A. Williams, M.D. Phan, A.L. Horst, E.D. Knox, and J.A. Knox, 2016: Through the Eyes of the Experts: Meteorologists’ Perceptions of the Probability of Precipitation. Wea. Forecasting, 31, 5–17, https://doi.org/10.1175/WAF-D-15-0058.1

National Weather Service, 1984: Zone and local forecasts. NWS Operations Manual W/OM15.

Schaefer, J.T. and R.L. Livingston, 1990: Operational Implications of the “Probability of Precipitation”. Wea. Forecasting, 5, 354–356, https://doi.org/10.1175/1520-0434(1990)005<0354:OIOTOP>2.0.CO;2

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.