Calculation equivalence in matlab

Question

I am calculating a log return series for a study on financial data.

By the laws of logarithms log(p_t/p_t-1) is the same as log(p_t) - log(p_t-1).

I calculate two different series that should be the same...

r = log(price(2:end)) - log(price(1:(end - 1)));

r1 = log(price(2:end))./price(1:(end - 1));

Out of curiosity I decided to check:

r1 == r

and I get a TON of 0's in the result array.

Is there some reason for this? These two should be absolutely equivalent. I wouldn't think weird floating point things would be a problem here, and if they were a problem they'd effect both vectors equally.

What is going on here?

You're probably seeing the limitations of numerical precision. — Oliver Charlesworth, Nov 18 '17 at 19:10
Possible duplicate of [Why is 24.0000 not equal to 24.0000 in MATLAB?](https://stackoverflow.com/questions/686439/why-is-24-0000-not-equal-to-24-0000-in-matlab) — Sardar Usama, Nov 18 '17 at 19:11
For floating point values, equality check must implement a tolerance: eq = abs(a - b) <= eps(max(abs(a), abs(b))); — Tommaso Belluzzo, Nov 18 '17 at 19:13
@TommasoBelluzzo Ah I see, isn't it strange that it wouldn't truncate/round all the numbers the same? — , Nov 18 '17 at 19:17
@rec: using TommasoBelluzzo suggestion you should have all ones (to be on the safe side I would even check for a bigger epsilon). — pacta_sunt_servanda, Nov 18 '17 at 19:52
@rec It isn't strange because the two numbers are computed in different ways meaning different rounding errors are introduced during intermediate computation. — jodag, Nov 18 '17 at 20:06

Alex · Accepted Answer · 2017-11-19T12:31:46.893

0

Your parentheses are wrong in the second equation, but you will still get floating point differences.

r = log(price(2:end)) - log(price(1:(end - 1)));

r1 = log(price(2:end)./price(1:(end - 1)));

edited Nov 19 '17 at 12:31

answered Nov 19 '17 at 12:26

Alex

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Calculation equivalence in matlab

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