Comparing Ebbinghaus Forgetting Curve to the standard exponential half-life decay curve
Jennifer Murphy
http://en.wikipedia.org/wiki/Forgetting_curve
Hermann Ebbinghaus discovered that memory decays exponentially. The
formula he came up with for modeling forgetting curve was
R = e^(-t/S)
Where R is memory retention, S is the relative strength of the memory,
and t is time.
I have a couple of questions about this formula.
1. What are the units for S?
This formula is a variation of the standard exponential half-life
decay function
y = y0 * e^(-(ln(2)/h)t)
where
S = h/ln(2)
In that formula, h is in the same units as t (time), so I assume that
S is also in the same units as t (time).
Is that correct?
2. What are the units for R?
In the half-life formula, y and y0 are usually mass units (grams).
Since there is no R0, I am assuming that it is "1".
R = 1 * e^(-t/S)
This means that R goes from 1 to 0, which suggests that it's a
probability. Is it the probability of recall?
mike fee
JenM...@jm.invalid says...
> In 1885, according to this article in Wikipedia,
>
> http://en.wikipedia.org/wiki/Forgetting_curve
>
> Hermann Ebbinghaus discovered that memory decays exponentially. The
> formula he came up with for modeling forgetting curve was
>
> R = e^(-t/S)
>
> Where R is memory retention, S is the relative strength of the memory,
> and t is time.
Note that this formula is only a qualitative approximation to the decay
of memories. It provides little (if any) insight to the mechanism by
which memories may 'decay' and I doubt that
a) it is applicable over a wide range of t,
b) matches statistical data more than very approximately at best.
Also, it is a little self-fulfilling as there appears to be no way to
independantly estimate S other than by recording the R returned by
fitting the equation to some specific set of data. What I am saying here
is that while t is clearly 'time', and R is clearly some statistical
measure of how much 'memory is retained' after time t, it is not clear
that S represents 'strength of memory' in any real sense of the meaning
of S. Might be better to call S the 'memory decay constant' or 'memory
half-life'.
>
> I have a couple of questions about this formula.
>
>
>
> 1. What are the units for S?
>
> This formula is a variation of the standard exponential half-life
> decay function
>
> y = y0 * e^(-(ln(2)/h)t)
>
> where
>
> S = h/ln(2)
>
> In that formula, h is in the same units as t (time), so I assume that
> S is also in the same units as t (time).
>
> Is that correct?
Yes the units of S will be whatever time units are used to measure t.
>
>
>
> 2. What are the units for R?
The units of R will be defined by the method used to define memory
loss/retention.
> In the half-life formula, y and y0 are usually mass units (grams).
> Since there is no R0, I am assuming that it is "1".
>
> R = 1 * e^(-t/S)
>
> This means that R goes from 1 to 0, which suggests that it's a
> probability. Is it the probability of recall?
The probability of recall would be a reasonable guess, but without
access to Ebbinghaus' data it wouild be impossible to say for sure.
Mike
Jennifer Murphy
wrote:
>In article <vvek17h243th2t2g3...@4ax.com>,
>JenM...@jm.invalid says...
>> In 1885, according to this article in Wikipedia,
>>
>> http://en.wikipedia.org/wiki/Forgetting_curve
>>
>> Hermann Ebbinghaus discovered that memory decays exponentially. The
>> formula he came up with for modeling forgetting curve was
>>
>> R = e^(-t/S)
>>
>> Where R is memory retention, S is the relative strength of the memory,
>> and t is time.
>
>Note that this formula is only a qualitative approximation to the decay
>of memories. It provides little (if any) insight to the mechanism by
>which memories may 'decay' and I doubt that
>a) it is applicable over a wide range of t,
>b) matches statistical data more than very approximately at best.
While I'm no expert, it is my impression that the formula is a fairly
good approximation, at least after new material has been learned.
Ebbinghaus did his research using nonsense words, but I believe others
have extended that to more general data.
There are those who disagree.
>Also, it is a little self-fulfilling as there appears to be no way to
>independantly estimate S other than by recording the R returned by
>fitting the equation to some specific set of data. What I am saying here
>is that while t is clearly 'time', and R is clearly some statistical
>measure of how much 'memory is retained' after time t, it is not clear
>that S represents 'strength of memory' in any real sense of the meaning
>of S. Might be better to call S the 'memory decay constant' or 'memory
>half-life'.
For my purposes, I don't really care how memory really works or what
we call the parameters as long as the model is reasonably accurate. I
just wanted to get the units right.
>> I have a couple of questions about this formula.
>>
>>
>>
>> 1. What are the units for S?
>>
>> This formula is a variation of the standard exponential half-life
>> decay function
>>
>> y = y0 * e^(-(ln(2)/h)t)
>>
>> where
>>
>> S = h/ln(2)
>>
>> In that formula, h is in the same units as t (time), so I assume that
>> S is also in the same units as t (time).
>>
>> Is that correct?
>Yes the units of S will be whatever time units are used to measure t.
That's what I thought it had to be, thanks.
>> 2. What are the units for R?
>The units of R will be defined by the method used to define memory
>loss/retention.
>> In the half-life formula, y and y0 are usually mass units (grams).
>> Since there is no R0, I am assuming that it is "1".
>>
>> R = 1 * e^(-t/S)
>>
>> This means that R goes from 1 to 0, which suggests that it's a
>> probability. Is it the probability of recall?
>The probability of recall would be a reasonable guess, but without
>access to Ebbinghaus' data it wouild be impossible to say for sure.
Of course. I was hoping that someone could infer it from the formula
or might even be familiar with the research.
mike fee
JenM...@jm.invalid says...
> On Mon, 11 Jul 2011 13:26:18 +1200, mike fee <m....@irl.nospam.cri.nz>
> wrote:
>
> >In article <vvek17h243th2t2g3...@4ax.com>,
> >JenM...@jm.invalid says...
> >> In 1885, according to this article in Wikipedia,
> >>
> >> http://en.wikipedia.org/wiki/Forgetting_curve
> >>
> >> Hermann Ebbinghaus discovered that memory decays exponentially. The
> >> formula he came up with for modeling forgetting curve was
> >>
> >> R = e^(-t/S)
> >>
> >> Where R is memory retention, S is the relative strength of the memory,
> >> and t is time.
> >
> or might even be familiar with the research.
>
translation of Ebbinghaus' work. There should be enough information
there to answer your questions.
Mike
Jennifer Murphy
wrote:
>In article <4ank17t85mceu8cuj...@4ax.com>,
>JenM...@jm.invalid says...
>> On Mon, 11 Jul 2011 13:26:18 +1200, mike fee <m....@irl.nospam.cri.nz>
>> wrote:
>>
>> >In article <vvek17h243th2t2g3...@4ax.com>,
>> >JenM...@jm.invalid says...
>> >> In 1885, according to this article in Wikipedia,
>> >>
>> >> http://en.wikipedia.org/wiki/Forgetting_curve
>> >>
>> >> Hermann Ebbinghaus discovered that memory decays exponentially. The
>> >> formula he came up with for modeling forgetting curve was
>> >>
>> >> R = e^(-t/S)
>> >>
>> >> Where R is memory retention, S is the relative strength of the memory,
>> >> and t is time.
>> >
>...
>> Of course. I was hoping that someone could infer it from the formula
>> or might even be familiar with the research.
>>
>If you click on the first reference in the wiki article - you get a 1913
>translation of Ebbinghaus' work. There should be enough information
>there to answer your questions.
Well, maybe for someone with better statistics skills... One problem
is that most of the charts are poorly labeled or completely unlabeled.
It was often not clear to be what the independent and dependant
variables were.
Finally, the formula above attrobuted to Ebbinghaus is not in this
paper at all!
In Section 29: Discussion of results, the discussion seems to focus on
a binary recall/norecall (what I think he calls "production"). I am
assuming that the actual measure is the point at which the memory
slips below the recall strength.
In short, I believe that R *is* recall probabiity. If you disagree,
please point me to the relevant section of the paper.
Thanks for the help.
mike fee
JenM...@jm.invalid says...
the wikipedia article. The table just above the heading 'Section 29', in
chapter 7 appears to summarise his data - note that I haven't checked
that it is a valid summary but, assuming it is, it predicts a very
different behaviour from the wiki article.
The wiki claims
a) R ~= exp(-t/s),
whereas a simple fit to the data suggests
b) R ~= Ro - log(t/s), or R ~= Ao*t^(-b)
for appropriately chosen Ro,s,Ao,b, over a significant range of t (and
Ebberhaus comes to a similar conclusion about 20 lines down from the
table).
The two behaviours are very different.
a) claims that for any set of learned data, if (for example) you forget
10% on the first day, then you will forget 10% of what is left on the
second day, 10% of what is still left on the third day, another 10% on
the fourth day, etc...
b) claims that for the same set of data, if (for example) you forget
10% on the first day, you might forget another 10% of the total in the
next 2 days, another 10% of the total in the next 4 days, 10% in the
next 8 days, etc..
My guess is that someone, either the writer of the qwiki or the
reference they used to write it, misunderstood or misinterpreted the
papoer and data, or didnt really understand the nature of exponentials
and logarithms. Note that if formula b) is approximately correct then it
is also approximately true that exp(R) ~= k*t (for some constant k), so
there _is_ an exponential relationship between memory and time, but just
not of the nature shown by the wiki.
As a physicist (and I have to admit I have no background in psychology
or teh theory of brain function), I always like to check a proposed
formula by looking at what it suggests in real life. If you look at the
comment near the bottom of the wiki that states "In a typical schoolbook
application (e.g. learning word pairs), most students remember only 10%
after 3?6 days (depending on the material)", the (presumably incorrect
?) wiki formula suggests that on the same test the students would only
remember 1% in 6-12 days, and 0.1% in 9-18 days, which might be a little
far-fetched. Whereas formula b) (Ebberhaus' and mine from fitting the
data) might suggest that teh 1% retention might be after 30-60 days (for
example) and the 0.1% retention after a year or two - this 'feels' a bit
more reasonable when compared with my personal experience.
Mike
Jennifer Murphy
wrote:
>In article <j7cn17t6agi482hnu...@4ax.com>,
>JenM...@jm.invalid says...
>> On Tue, 12 Jul 2011 09:07:14 +1200, mike fee <m....@irl.nospam.cri.nz>
>> wrote:
>>
...snip...
I'm glad to hear this, because I couldn't reconcile the article with
the paper. I figured it was because my math skills are not good
enough.
>As a physicist (and I have to admit I have no background in psychology
>or teh theory of brain function), I always like to check a proposed
>formula by looking at what it suggests in real life. If you look at the
>comment near the bottom of the wiki that states "In a typical schoolbook
>application (e.g. learning word pairs), most students remember only 10%
>after 3?6 days (depending on the material)", the (presumably incorrect
>?) wiki formula suggests that on the same test the students would only
>remember 1% in 6-12 days, and 0.1% in 9-18 days, which might be a little
>far-fetched. Whereas formula b) (Ebberhaus' and mine from fitting the
>data) might suggest that teh 1% retention might be after 30-60 days (for
>example) and the 0.1% retention after a year or two - this 'feels' a bit
>more reasonable when compared with my personal experience.
What you say makes intuitive sense. Thanks for taking the time to
check it out.
Do you have any interest is doing some consulting on this and related
topics? I have some work I need to do and my math skills are not up to
the task.
mike fee
JenM...@jm.invalid says...
Could be interested if there is some small compensation for my time. You
could contact me privately by deleting the 'nospam' (and one of the '.')
from my e-mail address.
Cheers,
Mike
Jennifer Murphy
...snip...
>I had a closer look at the paper and believe that the error could be in
>the wikipedia article. The table just above the heading 'Section 29', in
>chapter 7 appears to summarise his data - note that I haven't checked
>that it is a valid summary but, assuming it is, it predicts a very
>different behaviour from the wiki article.
>
>The wiki claims
> a) R ~= exp(-t/s),
>whereas a simple fit to the data suggests
> b) R ~= Ro - log(t/s), or R ~= Ao*t^(-b)
>for appropriately chosen Ro,s,Ao,b, over a significant range of t (and
>Ebberhaus comes to a similar conclusion about 20 lines down from the
>table).
NOTE: I set the line length to 90 so keep the table from wrapping.
I did a little analysis on the Ebbinghaus data.
In the table at the end of Section 28, t is in hours (col 1).
I converted it to days to make the graph easier to read (col 2).
Col 3 contains his y-data.
Cols 4-6 are the results of a curve-fitting program for a power function.
The function is y = 0.3189725x^(-0.125822)
Cols 7-9 are the results of a curve-fitting program for a log function.
That function is y = -0.0455926 * LN(x) + 0.333989
Col 10 shows, as you said, that this data does not fit an exponential
decay function. With h = 0.0178 (days or 1.1 minutes), the first data
point is a fit, but then it decays much too rapidly.
Col 11 shows the same result for the Wikipedia formula, R = e^(-t/S).
Exp Wiki
Power Function Fit Log Function Fit Fit Fit
---------------------- ---------------------- h= S=
Hours Days %MS Value Error Squared Value Error Squared 0.0178 0.0257
0.000 0.000000 1.000 n/a n/a n/a n/a n/a n/a 1.0000 1.0000
0.333 0.013889 0.582 0.5463 0.0613 0.0038 0.5290 0.0911 0.0083 0.5825 0.5825
1.000 0.041667 0.442 0.4758 -0.0764 0.0058 0.4789 -0.0834 0.0070 0.1976 0.1976
8.800 0.366667 0.358 0.3619 -0.0109 0.0001 0.3797 -0.0607 0.0037 0.0000 0.0000
24.000 1.000000 0.337 0.3190 0.0535 0.0029 0.3340 0.0089 0.0001 0.0000 0.0000
48.000 2.000000 0.278 0.2923 -0.0516 0.0027 0.3024 -0.0877 0.0077 0.0000 0.0000
144.000 6.000000 0.254 0.2546 -0.0023 0.0000 0.2523 0.0067 0.0000 0.0000 0.0000
744.000 31.000000 0.211 0.2071 0.0186 0.0003 0.1774 0.1591 0.0253 0.0000 0.0000
Least Squares 0.0156 Least Squares 0.0521
If I did this right, always an iffy proposition, the data is slightly
better fit using a power function.
This next table is from the data at the end of Section 29.
In this table, he switched the units of t to minutes.
I again converted to days and compared his calculated values
to the pwoer function fit. They are very close, but the
power function is slightly better.
Ebbinghaus Log Fit Power Function Fit
--------------------- ---------------------
Mins Days %MS Value Error Squared Value Error Squared
0 0.000000 1.000 n/a n/a n/a n/a n/a n/a
20 0.013889 0.582 0.570 0.0206 0.0004 0.5467 0.0606 0.0037
64 0.044444 0.442 0.467 -0.0566 0.0032 0.4717 -0.0672 0.0045
526 0.365278 0.358 0.345 0.0363 0.0013 0.3611 -0.0086 0.0001
1440 1.000000 0.337 0.304 0.0979 0.0096 0.3178 0.0570 0.0033
2880 2.000000 0.278 0.281 -0.0108 0.0001 0.2910 -0.0469 0.0022
8640 6.000000 0.254 0.249 0.0197 0.0004 0.2532 0.0033 0.0000
44640 31.000000 0.211 0.212 -0.0047 0.0000 0.2056 0.0258 0.0007
Least Squares 0.0151 Least Squares 0.0144