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(peeping carefully from my cubicle....)
Just a note, we are neither calculating a simple nor an exponential
moving average with
> > x1:=Exp(Sum(Log(y),n)/n);
> > x2:=Exp(Mov(Log(y),n,S));
They should yield the same result, I suspect the difference between
the two are due to rounding.
What we are calculating here is the *geometric* average. How to
interpret this? It is especially easy if you are working with return
or interest rates. Say Jose invests 100 AUD in one of our systems and
we do not charge any fees for that (insert chuckle here somewhere).
Suppose the returns he gets from this system are as follows in the
next 10 bars:
1%, 2%, 3%, 4%, 5%, 6%, 7%, 8%, 9%, 10%
Now, to get to the final amount he would have in the account, you can
calculate it as follows:
100 x 1.01 x 1.02 x 1.03 x 1.04 x 1.05 x 1.06 x 1.07 x 1.08 x 1.09
x 1.10 = 170.2
So at the end he has 170.2 in the account. What was the average
return that he got per bar? We could calculate the average of 1, 2,
... 10 to get 5.5%. However, if we apply 5.5% per bar, we get an
overstated return of
100 x 1.055 x 1.055 x 1.055 x ... x 1.055 = 170.81
This is a small difference in this example, but next time a consultant
shows you 'linear' return figures, know he is overstating things! We
once had a guy trying to sell us some investment where he used this
method to get returns in line with what the market offered on some 5
year plan. If you calculated it correctly, as shown next, the return
was quite a bit less.....
How do we get the actual return? We calculate the geometric average
as discussed before, as follows:
( 1.01 x 1.02 x 1.03 ... x 1.10 ) ^ ( 1 / 10 )
or, using the method discussed before,
exp [ ( ln 1.01 + ln 1.02 + ln 1.03 + ... + ln 1.10 ) / 10 ]
which yields 1.054609 or 5.46% per bar - quite a bit less than the
salesman's 5.5%! Applying 5.46% for ten bars also correctly yields
the actual return of 170.2 (roughly, depending on how many digits you
ignored after the decimal....should be accurate if you use them all).
How can we use this in the market? Say you have a YTM series (bond
rates, for those who do not know what YTM stands for). If you
calculate the geometric mean of the yield, say
n := Input( "Period length", 1, 1000, 21 );
YTM := 1 + C / 100; {To convert 5.43 to 1.0543}
GM := Exp(Mov(Log(YTM),n,S)); {Geometric mean}
100 * ( GM - 1 ) {Convert 1.0543 back to 5.43}
you get the 'average' return that the bond yielded during the past n
days. If you had the bond the whole time, this is the return (pa)
that you received per day and you can compare it to say the federal
funds rate or whatever during this time.
To do the same thing for prices is actually very easy, since you
simply compare the final period price to the starting one and
calculate how much it has risen on average, percentage wise, per day
to get to the final. In math we have
yn = y0 x [ ( 1 + r / 100 ) ^ n ]
and we are looking for r. This reduces to
r = 100 x [ ( yn / y0 ) ^ ( 1 / n ) - 1 ]
which gives the average percentage increase. Using log and exp and
OTC MSFL (Off The Cuff MSFL) this then simply becomes
n := Input( "Period length", 1, 1000, 21 );
AVRET := 100 * ( Exp( Log( C / Ref( C, -n ) ) / n ) - 1 );
AVRET
and it plots the average percentage returns, per bar, that a price
yielded over the past n bars.
(Whoooop - I am back inside.....)
Regards
MG Ferreira
TsaTsa EOD Programmer and trading model builder
http://www.ferra4models.com
http://fun.ferra4models.com
--- In equismetastock@xxxxxxxxxxxxxxx, "Jose Silva" <josesilva22@xxxx>
wrote:
>
> > Comparing that to a 12EMA...
> > From my perspective, X1 and X2 actually appear smoother.
>
> So they should, Preston - both x1 & x2 are SMAs (Simple Moving
> Averages). Same data/period SMAs are always smoother than EMAs.
>
> There is little or no benefit in using Exp/Log over Mov(DataArray,
> Periods,S).
>
>
> jose '-)
>
>
> --- In equismetastock@xxxxxxxxxxxxxxx, pumrysh <no_reply@xxxx> wrote:
> >
> > MG,Jose,
> >
> > Not so quick to the cubicle!
> > There was a coding error. The fix is below:
> > y:=C;
> > n:=12;
> > x1:=Exp(Sum(Log(y),n)/n);
> > x2:=Exp(Mov(Log(y),n,S));
> > x1;x2
> >
> > That being said, there is a slight difference between x1 and x2 when
> > compared. Comparing that to a 12EMA of the close there is still
> > another difference. From my perspective, X1 and X2 actually appear
> > smoother. So we have a new way of writing an exponential moving
> > average.
> >
> > Finally, MG I've used the sum method to derive a SMA before.
> > I always enjoy new methods of writing moving averages.
> >
> > Thanks to both of you for the mental stimulus.
> >
> > Preston
> >
> >
> > --- In equismetastock@xxxxxxxxxxxxxxx, "Jose Silva"
> > <josesilva22@xxxx> wrote:
> > >
> > > > 1000 * exp( Cum( log( y / 1000 ) ) )
> > >
> > > In *theory*, this would be quite useful in accumulating large
> > values
> > > (such as Volume), but unfortunately MetaStock cannot handle large
> > > values for the Exp() function, so there is no advantage here.
> > >
> > > Here is an example:
> > > Exp(Cum(Log(V/1000)))*1000
> > >
> > >
> > > > Exp( Sum( Log( y ), n ) / n )
> > > > or simply
> > > > Exp( Mov( Log( y, n, S ) ) )
> > >
> > > Neither of these is an improvement on Mov(y,n,S).
> > >
> > >
> > > MG, back to your modeling cubicle with you...
> > >
> > >
> > > jose '-)
> > > http://www.metastocktools.com
> > >
> > >
> > >
> > > --- In equismetastock@xxxxxxxxxxxxxxx, "MG Ferreira" <quant@xxxx>
> > > wrote:
> > > >
> > > > OK, here is a quick, off the cuff example. The Cum function
> > > >
> > > > Cum(y)
> > > >
> > > > gives
> > > >
> > > > y1 + y2 + y3 + ...
> > > >
> > > > What if you want
> > > >
> > > > y1 * y2 * y3 * ...
> > > >
> > > > Well, use
> > > >
> > > > exp( Cum( log( y ) )
> > > >
> > > > This will generally give an overflow quickly, unless you use
> > > > smallish values, so you may need to rescale it to get it to
> work,
> > > > ie calculate
> > > >
> > > > 1000 * exp( Cum( log( y / 1000 ) ) )
> > > >
> > > > If you think the Cum function is useful, then this must appeal
> to
> > > > you as well!
> > > >
> > > > Another example, the geometric average. The simple moving
> average
> > > > is defined as
> > > >
> > > > ( y1 + y2 + y3 + ... + yn ) / n
> > > >
> > > > Of course, in MSFL we all use
> > > >
> > > > Mov(y,n,S)
> > > >
> > > > to calculate this, but we could also use
> > > >
> > > > Sum(y,n) / n
> > > >
> > > > to get the answer the brute-force way.
> > > >
> > > > The geometric moving average, which may actually be more
> > applicable
> > > > to markets due to the exponential growth often seen in prices,
> > > > is defined as
> > > >
> > > > ( y1 x y2 x y3 x ... x yn ) ^ ( 1 / n )
> > > >
> > > > To do this in MSFL, use
> > > >
> > > > Exp( Sum( Log( y ), n ) / n )
> > > >
> > > > or simply
> > > >
> > > > Exp( Mov( Log( y, n, S ) ) )
> > > >
> > > > Regards
> > > > MG Ferreira
> > > > TsaTsa EOD Programmer and trading model builder
> > > > http://www.ferra4models.com
> > > > http://fun.ferra4models.com
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