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Returns used to compute the alpha and the beta

3 messages · Benoit.Schmid at unige.ch, Adams, Zeno, Patrick Burns

#
Good morning Julien,

Thanks for having taken the time to answer.

Quoting julien cuisinier <j_cuisinier at hotmail.com>:
For me a daily return is
1. Net Return = V(i)/V(i-1) - 1
2. Log return = log(V(i)/V(i-1))

Geometric or arithmetic are not a sample daily return they
are average return calculated in two different ways.

What is nice with log return, is that the arithmetic mean
of the daily log return is equal the log return
of the geometric mean of the net return.
You also need to take the log of the daily ratios
in the case of a log return before computing arithmetic
or geometric average.
Arithmetic with net return is useless as you said.

But arithmetic with logr is usefull because:
exp(arithmetic average of daily log return) - 1 = geometric average of  
net returns
Therefore arithmetic with the logr give the same as of geometric with
net return.

What is new for me is that quants also use geometric annualization
with logr if I understand well.
I thought that quants were only using arithmetic annualization with logr
and geometric with net returns.

If you have good urls on the motivations on applying geometric
on logr, please do not hesitate to share them.
I am interested.

If you have a simple R code that shows your computations of
annualization of logr and show geometric on logr is
a better aggregation, please do not hesitate to post it.
I think it would help me to be sure I understood
everything you wanted to explain.

Thanks for your answer.
#
<< Geometric or arithmetic are not a sample daily return they
 << are average return calculated in two different ways.

 << What is nice with log return, is that the arithmetic mean
 << of the daily log return is equal the log return
 << of the geometric mean of the net return.

 Just one remark: the arithmetic mean of log returns is only
approximately equal to the geometric mean of net returns (see example
below). I point this out because I have read this claim frequently and
was puzzled when I didn't get exactly the same results.



x <- numeric(100)
x[1] <- 100

set.seed(123)
for (i in 2:100) {
x[i] <- x[i-1] + rnorm(1,1,1) - 0.009*i*rnorm(1,1,1)^2
}

x1 <- c(x[-1],0)
ret <- x1/x - 1

#arithmetic mean of net returns:
amean <- mean(ret[-100])*25000 ; amean

#arithmetic mean of log returns:
amean2 <- mean(diff(log(x)))*25000 ; amean2

#geometric mean of net returns:
gmean<- ((prod(1+ret[-100]))^(1/length(ret[-100]))-1)*25000; gmean




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#
Adams, Zeno wrote:
'amean' is pretty much nonsense. 'amean2' and 'gmean'
should be equivalent except one is a log return and the
other is a simple return.  If you transform one return into
the other form, then you should get the same number.

r = log return
R = simple return

r = log(R + 1)
R = exp(r) - 1



Patrick Burns
patrick at burns-stat.com
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and "A Guide for the Unwilling S User")