Compute formula with sum of log and two loops

[SarahDew]

I want to apply the following formula to an input of the numbers 1 to 9 (=d) and n=2 (so k = 0 to 9): 

I am almost there, but I figure I need to add a second step somewhere to loop over the k. I currently use two do loops, for k and d, but I don't see how to get one loop within the sum, and one loop applied to the sum. I get the expected for each combination of digits, but I still need the sum of  the 'expected' where D remains the same. If I could also add a conversion from n to k, meaning I do "%let n=2" and then SAS computes the range of k, that would be even better.

my attempt:

data EXPECTED_D2;
 FORMAT EXPECTED 8.3;
 DO D = 0 TO 9;
 DO K = 1 TO 9;
 EXPECTED=sum(LOG10(1+(1/((10*K)+D))));
 OUTPUT;
 END;
 END;
run;

desired result:

data EXPECTED_D;
 input D EXPECTED;
 datalines;
 0 11.97
 1 11.39
 2 10.88
 3 10.43
 4 10.03
 5 9.67
 6 9.34
 7 9.04
 8 8.76
 9 8.50
;
run; 

[PaigeMiller · Accepted solution]

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@SarahDew wrote:

I want to apply the following formula to an input of the numbers 1 to 9 (=d) and n=2 (so k = 0 to 9): 

I am almost there, but I figure I need to add a second step somewhere to loop over the k. I currently use two do loops, for k and d, but I don't see how to get one loop within the sum, and one loop applied to the sum. I get the expected for each combination of digits, but I still need the sum of  the 'expected' where D remains the same. If I could also add a conversion from n to k, meaning I do "%let n=2" and then SAS computes the range of k, that would be even better.

---

Stay away from macro variables here. They are of no help in this situation.

Try this:

data EXPECTED_D2;
 FORMAT EXPECTED 8.3;
 DO D = 0 TO 9;
 expected=0;
 DO K = 1 TO 9;
 EXPECTED=expected+LOG10(1+(1/((10+K)+D)));
 END;
  OUTPUT;
 END;
 drop k;
run;

This doesn't give the same answers you provided, I don't know why, since it uses the exact same formula for EXPECTED that you are using.

[SarahDew]

That does work, only I mistakenly did 10+k instead of 10*K, maybe we should both edit this. And I forgot to mention I multiply the expected probabilities by 100.

[FreelanceReinh]

Hi @SarahDew,

10k is 10*k, not 10+k. So, very similar to @PaigeMiller's code:

data expected_d2(drop=k);
do d = 0 to 9;
  expected=0;
  do k = 1 to 9;
    expected+log10(1+(1/(10*k+d)));
  end;
  output;
end;
format expected percent8.2;
run;