Results_slp_sst_Adam

SLP, SST

**2. Your field 1 means and variances: space, time, spacetime**
 ==**2. Field1 means and variances: space, time, spacetime** == ==**3. Further decomposition of spacetime variations: Mean seasonal cycle and 'climate anomaly'** ==
 * 1) The GRAND (space+time) **mean** of my x and y: ** 1.0110e+003 hPa and** 27.5744 degC
 * 2) The GRAND (space+time) **variance** of my x and y: ** 2.3720 (hPa)^2 and ** 2.9326 (degC)^2
 * 3)   The GRAND standard deviations are : ** 1.5402 hPa and** 1.7125 degC
 * 4) The SPATIAL variance of my TIME MEAN longitude section is: 1.1617 (hPa)^2 and 1.8427 (degC)^2
 * 5) The TEMPORAL variance of my LONGITUDE MEAN time series is: 0.5488 (hPa)^2 and 0.3616 (degC)^2

1) Confirm that the 20-year mean of the anomalies as defined above is 0. Write math (on paper, for yourself) that proves it/ shows why. Yes. Mean(anomx)=0,

2) Is the spatial mean of the climate anomalies (as defined above) 0? Is it the same as the time series of the spatial mean of the raw data? Or is it a new object? Yes, it is zero. It is definitely not the same as the time series of the spatial mean of the raw data. For one thing, the mean of the anomalies is zero whereas the mean of the raw data is not.

3) My CLIMATOLOGICAL ANNUAL CYCLE has variance: _(units). var( climx12,1) print, var_n(climx12) <span style="color: blue; font-family: 'Arial','sans-serif'; font-size: 10pt; line-height: normal; margin: 0in 0in 10pt;">SLP: 1.8683 <span style="color: blue; font-family: 'Arial','sans-serif'; font-size: 10pt; line-height: normal; margin: 0in 0in 10pt;">SST: 2.4852

<span style="font-family: 'Arial','sans-serif'; font-size: 10pt; line-height: normal; margin: 0in 0in 10pt;">4) My INTERANNUAL ANOMALY ARRAYS has variance: _(units). var( anomx,1) print, var_n(anomx) <span style="color: blue; font-family: 'Arial','sans-serif'; font-size: 10pt; line-height: normal; margin: 0in 0in 10pt;">SLP: 0.5037 <span style="color: blue; font-family: 'Arial','sans-serif'; font-size: 10pt; line-height: normal; margin: 0in 0in 10pt;">SST: 0.4473

<span style="font-family: 'Arial','sans-serif'; font-size: 10pt; line-height: normal; margin: 0in 0in 10pt;">5) Fill out a variance decomposition table for field 1: feel free to add columns if you can define other parts. It may help you to look at an example:

<span style="font-family: 'Arial','sans-serif'; font-size: 10pt; line-height: normal; margin: 0in 0in 0pt;">A=G+H ie it checks out.
 * # || SLP || Uwnd (m/s) ||
 * a) total variance of x || 2.3720 || 2.9326 ||
 * b) purely spatial (variance of TIME mean at each lon) || 1.1617 || 1.8427 ||
 * c) temporal anomalies (x minus its TIME mean at each lon) || 1.2103 || 1.0899 ||
 * d) purely temporal (variance of LON mean at each time) || 0.5488 || 0.3616 ||
 * e) spatial anomalies (x minus its LON mean at each time) || 1.8232 || 2.571 ||
 * f) remove both means (space-time variability) || 2.3720 || 2.9326 ||
 * g) mean seasonal cycle || 1.8683 || 2.4852 ||
 * h) deseasonalized anomalies || 0.5037 || 0.4473 ||
 * i) variance of longitudinal mean of part h) || 0.16 || 1.025 ||
 * j) h minus i (anomalies from both space and monthly-climatological means) || 0.34 || .3448 ||

<span style="font-size: 1.3em; margin: 0px; padding-bottom: 0px; padding-left: 0px; padding-right: 0px; padding-top: 5px;">**4. Further decomposition of anomx by scale (using rebinning).**
<span style="font-family: 'Arial','sans-serif'; font-size: 10pt; line-height: normal; margin: 0in 0in 0pt;">variance_by_scalefactor =

<span style="font-family: 'Arial','sans-serif'; font-size: 10pt; line-height: normal; margin: 0in 0in 0pt;">0.5037 0.5022 0.4980 0.4884 0.4613 0.2811 <span style="font-family: 'Arial','sans-serif'; font-size: 10pt; line-height: normal; margin: 0in 0in 0pt;">0.3709 0.3698 0.3666 0.3592 0.3375 0.1861 <span style="font-family: 'Arial','sans-serif'; font-size: 10pt; line-height: normal; margin: 0in 0in 0pt;">0.2997 0.2988 0.2964 0.2903 0.2721 0.1388 <span style="font-family: 'Arial','sans-serif'; font-size: 10pt; line-height: normal; margin: 0in 0in 0pt;">0.2295 0.2287 0.2268 0.2221 0.2075 0.0951 <span style="font-family: 'Arial','sans-serif'; font-size: 10pt; line-height: normal; margin: 0in 0in 0pt;">0.1421 0.1416 0.1403 0.1371 0.1271 0.0536 <span style="font-family: 'Arial','sans-serif'; font-size: 10pt; line-height: normal; margin: 0in 0in 0pt;">0.0486 0.0484 0.0479 0.0464 0.0415 0.0142

There is a small gradient in the variance since there are small scale factors in space and time. For such a small scale factor fixed in time, increasing the scale of the longitude does not change significantly change the gradient.

Now consider the covariance and correlation of the two subset arrays entering your scatterplot.
 * 5. Scatter plot, correlation and covariance, regression-explained variance**Based on your data fields (which you've seen pictures of), make subsets of your 2 variables x and y and make a scatter plot of these showing the strongest (positive or negative) correlation of one field with the other you can find. The subset might simply be all (x,t) values if your fields are very similar (olr, precip), or maybe the 240 time values at one longitude, or 144 longitudinal values in the time mean, or time series at different longitudes if some variability is offset in your two fields (like pressure and wind).

rho= 1.0000 -0.7840 -0.7840 1.0000 standard deviation of slp at 190 longitude: 2.0855 standard deviation of sst at 190 longitude: 2.0085 2) What fraction of the variance of y can be 'explained' by linear regression on x (y = mx + b)? How does this relate to rho? How much y variance is explained? (variance: with units of y squared) What is m? //Use the value from rho squared*total variance to get explained variance:// The total variance of SST at 190 longitude is 4.0342. The explained variance is 2.4533, which is 61% of the total variance. The total variance of SLP at 190 longitude is 4.3292. The explained variance is 2.6460, which is 60% of the total variance. This exercise is well explained on this page: [] Since I use matlab, polyfit(x,y,1) gives the y=mx+b formula. I got m=-0.7551 y=-0.7551*x+792 3) Now add uncorrelated (random) noise with variance 1 to one of your variables. This might be like observation error. The variance of SST at 190 longitude with the noise is 5.3544, larger than the variance without noise (4.0342). The value of rho is smaller, with a value of -0.6637. The linear fit becomes y=-0.7364*x + 773 The total variance of SLP at 190 longitude with noise is 5.3544. The explained variance is 2.3586, which is 44% of the total variance. Predictably, adding noise results in less explained variance.
 * 6. Lagged correlation, covariance, and cross-covariance: hey let's compute all vs. all**