Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

Okay, here is the information that I have.

closingValues = 
  FinancialData["^GSPC", "Close", {"Jan 1., 2000", "Jan 1., 2012"}, 
   "Value"];
logr = Differences[Log[closingValues]];

x = Tally[Sort[Round[logr, .001]]];

What I want to do is this:

(logr >= x) / Length[logr]

I want to count all the values in logr that are greater than or equal to x (there are about 110 or so values in x).

I also need help on plotting a graph where x is on the x-axis, and the resulting number above is on the y-axis.

Any help or advice, or leading me to any direction of solving my problem is greatly appreciated. Thank you.

share|improve this question
    
What elements of "logr" and "x" are you trying to compare? The list x already consists of elements in logr, just rounded to the nearest 0.001. Are you trying to do an element by element compare? –  cartonn Dec 5 '12 at 1:02
    
What I'm trying to do is find typical values of logr, which i denoted x. And then count which values of logr are greater than x. I didn't know how to find "typical values" so I tallied all of the rounded values and am going to use those. –  user4341 Dec 5 '12 at 1:09
    
If you are looking for values in logr that are greater than the max element in x (hence all elements in x), all you will find are the ones that were rounded down to the max value in x. Am I right? –  cartonn Dec 5 '12 at 1:12
    
Right. So let's say the max value was .05, then I want to find count all values in logr that are greater than .05, and then divide how ever many numbers I found that were greater than .05 by Length[logr]. But instead of using just the max value, I want to do it for all listed values in x. –  user4341 Dec 5 '12 at 1:18
add comment

2 Answers 2

up vote 7 down vote accepted

Perhaps use UnitStep and Mean. This should be pretty fast.

f[logr_, x_] := Mean[UnitStep[logr - x]]

f[logr, 0]

(*1597/3018*)

Now to plot it.

Plot[f[logr, t], {t, -.1, .1}, Exclusions -> None]

enter image description here

share|improve this answer
    
Andy, I don't think this is what I'm trying to do, but I greatly appreciate you trying to help. What I wanted to do is find all values in logr that is greater than the first value in x (so I essentially count however many values are larger than x), and then divide by the length of logr (like a probability). And then repeat with the next value in x. –  user4341 Dec 5 '12 at 1:43
    
This counts all values in logr that are greater than a given x value and divides by the number of elements in logr. Is that not what you wanted? –  Andy Ross Dec 5 '12 at 1:49
    
Actually, this does what you want, much more easily than what I was doing. Hats off! –  cartonn Dec 5 '12 at 1:54
    
Ooh, my confusion lies in the := Mean[UnitStep[logr - x]] part. Could you explain what it is doing for me? –  user4341 Dec 5 '12 at 1:58
1  
UnitStep is listable and logr-x gives a list of numbers. If they are negative UnitStep gives 0 otherwise it gives 1. Mean of a list of zero's and 1's is the same as counting the non-negative values and dividing by the number of elements. –  Andy Ross Dec 5 '12 at 2:00
show 6 more comments

(Cannot connect to FinancialData, so I used fakeReturns instead of logr.)

 fakeReturns = RandomVariate[NormalDistribution[], 600];

 Plot[SurvivalFunction[HistogramDistribution[fakeReturns], x], {x, -3, 3},
   Exclusions -> None]

enter image description here

Or, Histogram with Round[logr,.001] ("typical values") as bin delimiters and "SF" (i.e., SurvivalFunction) as the height function gives:

Histogram[fakeReturns, {Union@Round[fakeReturns, 0.1]}, "SF", PlotRange -> Full]

enter image description here

or, with a different specification of "typical" values:

 Histogram[fakeReturns, {Union@Round[fakeReturns, 0.01]}, "SF", PlotRange -> Full]

enter image description here

share|improve this answer
    
OOH! That seems simple also! Now what if I wanted to take the log of the resulting values and plot it with the log of the x values? How would I do that? –  user4341 Dec 5 '12 at 3:21
    
@user4341, fakeReturns data has negative values as it is intended to represent your logr (Differences[Log[closingPrices]], that is, daily returns). With positive-valued data, you can use Histogram option ScalingFunctions -> {"Log", "Log"} get both axes in log scale. –  kguler Dec 5 '12 at 3:46
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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