3
$\begingroup$

I have rewritten the old question (below) to hopefully make things a bit clearer.

Basically I think that the temperature of the earth should be normally distributed but is not due to the ‘seasonal tilt’ and curvature of the earth’s surface. I have added some histograms for different ‘latitude bands’ of the temperatures and rainfall in mm here (I have included rainfall here but please ignore it for now). For example, [-10 ... 10] corresponds to latitude band (-)10 degrees south to 10 degrees north inclusively.

Of course temperature depends on latitude but this is due to the fact that it actually depends on seasonal tilt and curvature IMHO.

At the end of the day I am after some transformation which removes the effect of seasonal tilt/curvature to arrive at some global normal distribution for all four seasons. Does this make sense?

Thanks.

Christian

Old question:

I have some data of the earth's surface (temperature + rainfall in mm):

enter image description here

Clearly the earth's tilt affects the modality/normality of, for example, the temperature distributions. I am just wondering whether there is a way to adjust for this to make the (combined?) data more normal/less modal?

I am not a statistician ... so not sure whether this is possible? Thanks in advance.

Christian

$\endgroup$
  • $\begingroup$ Do you mean "latitude" instead of "tilt" (which is, for these purposes, essentially constant)? I suppose the x-axes of the upper row of graphs (temperature) could represent latitude in degrees, but the units on the x-axes on the lower row (rainfall) are unrecognizable; they range from 0 to 8000. What do these graphs actually represent? These graphs are not scatterplots of rainfall or temperature against latitude, which is what one would expect from the question, so it's difficult or impossible to tell what you are really asking here. $\endgroup$ – whuber Nov 25 '11 at 15:15
  • $\begingroup$ These are histograms which plot the frequency of each x value. x value above: temperature -40 ... 40 below rainfall 0 ... 2500 millimeter. I can do the latitude stuff as well if you think that's helpful. $\endgroup$ – cs0815 Nov 25 '11 at 15:58
  • $\begingroup$ Well, without any of the "latitude stuff" those figures are completely irrelevant to your question! I believe they may have misled the response you have already received, because absent the latitude information, any advice concerning data transformations is baseless. $\endgroup$ – whuber Nov 25 '11 at 16:01
  • $\begingroup$ I have rewritten my original question. $\endgroup$ – cs0815 Nov 28 '11 at 11:18
4
$\begingroup$

Unfortunately, temperature depends upon much more than latitude; elevation counts for a whole lot, as well as forestation and nearby bodies of water and local geography. Similarly with rainfall, which depends enormously upon local geography, for example, the Hoh rainforest in Olympic National Park gets ~150 inches of rain a year, but 30 miles due east gets ~15 inches of rain a year.

Having said that, if you have a collection of particular sites, you could try fitting a cosine with period 1 year to a multiyear series of daily mean temperatures - different fits for each site. I recall I did something like this some years ago, and it worked pretty well. IIRC, the minimum temperature date (the point at which the argument to the cos function should equal $\pi$) is Jan 7th over pretty much the entire U.S., including Alaska, but you should estimate this too. Something like this:

$Temp_i = a + b*cos(c + 2*\pi * i/365.25)+e_i$

where $i = 1$ on Jan. 1 and 365 (or 366) on Dec. 31.

The $a$ and $b$ coefficients will vary with location. $a$ will be the mean annual temperature at the location, and $b$ will equal, more or less, 1/2 the difference between the average max and average min temperatures.

Something like this may provide you with a model that will allow you to remove the systematic component due to earth tilt and the varying distance of the earth from the sun (minor) of temperature at a given site. Of course, the errors will be autocorrelated due to persistence of local weather patterns, e.g., high pressure systems. Also of course, it doesn't really help with your stated goal, but may get you part way towards your underlying objective.

$\endgroup$
1
$\begingroup$

I doubt there is much that will help with the idiosyncratic, jagged shape of temperature, but with rainfall there are data-transformations you can try. You can search for that tag on this site and you'll come up with many useful posts. (Note that "normalization" is defined differently from "data-transformation" and it is the latter that you want in order to create something closer to a bell curve.)

The first thing I would try would be to take the square root of each value. The result may still be skewed, but probably a lot less so. In other words, it would probably not "pass" a normality test (those are pretty unreliable anyway), but it would likely be a more workable variable for most purposes. As many on this site have reminded us, in multivariate analysis it is typically the residuals, rather than the univariate distributions, that must be close to normal in order to satisfy the normality assumption.

$\endgroup$
  • $\begingroup$ How are you interpreting these "shapes," rolando? To me, these graphs do not appear to relate rainfall or temperature to anything recognizable in any standard way, but perhaps you have seen something in them that I am missing. $\endgroup$ – whuber Nov 25 '11 at 15:17
  • $\begingroup$ @whuber: i interpreted the question to be, "these univariate distributions are not normal; how can they be made more so?" The earth's tilt is treated as merely a background factor affecting the temperature distributions; tilt is not plotted. $\endgroup$ – rolando2 Nov 25 '11 at 23:02
  • $\begingroup$ If there is some measurable variable that is believed to be related to the values of interest (such as rainfall or temperature) then we would want to re-express the values of interest to make their residuals with respect to the regression reasonably symmetric; it would not (in general) be helpful to make the values themselves more normally distributed. These distributions represent a combination of (a) locations chosen for sampling and (b) effects of latitude (and other variables such as elevation), so we would not expect them to be normal. $\endgroup$ – whuber Nov 26 '11 at 1:09

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.