David Yokum's Profile

altruism

We would expect people to be able to recognize altruism in
kinfolk, but not to recognize it in strangers.

Similarly, under the cheater perspective, we would
expect cheater detection of actions by strangers.

Putting these together, we expect, based on evolutionary theory, an interaction between
cheating/altruism and kin status. Detection will be high in all cases except where strangers are
altruistic to kin or to other strangers.

21.2 years

The first factor was whether or not the scenario involved
cheating or generous (altruistic) behavior on the part of the main actor in the scenario. The
second factor was whether the subject was cued to see the principal actor as a relative or as a
neighbour. The third factor was whether subsidiary actor(s) in the scenario were cued to be
relatives or not

The Job

A relative's altruism to
both other relatives and to strangers is easily identified. Cheating by strangers is equally well
identified when aimed at other strangers or at your relatives

Wundt is well aware of the common criticism that self-observation
seems inescapably to involve the paradoxical identity

The psychologist is
not then interested in the psychophysical connections between
the somatic or nervous sense-mechanisms and the elicited
“inner” phenomena, but solely in describing, “and
where possible measuring,” the psychological
regularities that such experiments can reveal, viz., regular causal
links within the domain of the psychic alone

Wundt sees WL as simply
a mathematical description of the more general experience that
“we possess in our consciousness no absolute, but merely a
relative measure of the intensity of the conditions
[Zustände] obtaining in it, and that we therefore
measure in each case one condition against another, with which we are
obliged in the first place to compare it” (PP I:
393). For this reason Wundt's “psychological
interpretation” makes WL into a special case of a more general
law of consciousness, viz. “of the relation or
relativity of our inner conditions
[Zustände]” (PP I: 393). WL is therefore
not a law of sensation so much as of apperception.

theory of actuality
(Aktualitätstheorie) (1911a: 145). Representations are
representational acts, never the “objects with constant
properties” propounded by adherents of a so-called theory of
substantiality

Representations may be either perceptions (Wahrnehmungen) or
intuitions (Anschauungen): the same representation is called a
“perception” if considered as the presentation of objective
reality, and an “intuition” if considered in terms of the
accompanying conscious, subjective activity

The chance that a patient with a "positive" result has breast cancer is
then the proportion of group A within the combined group A + C, or P*M
/
[P*M + (1 - P)*M], which, cancelling the common factor M from the
numerator and denominator, is P / [P + (1 - P)] or P / 1 or just
P.

Which is common sense.  Take, for example, the "test" of flipping
a coin; if the coin comes up heads, does it tell you anything about
whether a patient has breast cancer?  No; the coin has a 50%
chance
of coming up heads if the patient has breast cancer, and also a 50%
chance of coming up heads if the patient does not have breast
cancer.  Therefore there is no reason to call either heads or
tails
a "positive" result.  It's not the probability being "50/50" that
makes the coin a bad test; it's that the two probabilities, for "cancer
patient turns up heads" and "healthy patient turns up heads", are the
same.

prior probability

if the two
conditional probabilities are equal, the posterior probability equals
the prior probability

revised probability or the posterior probability

One thing that's confusing
about this notation is that the order of implication is read
right-to-left, as in Hebrew or Arabic

Reading from left to right, "|" means "given"; reading from
right to left, "|" means
"implies" or "leads to".

Looking at this applet, it's easier to see why the final answer depends
on all three probabilities; it's the differential
pressure between the two conditional probabilities,  p(blue|pearl) and p(blue|~pearl), that slides the prior probability p(pearl) to the posterior
probability p(pearl|blue).

Even when the prior probability changes, the
differential pressure of the two conditional probabilities always
slides
the probability in the same direction. 
If you learn the egg is painted blue, the probability the egg contains
a pearl always goes up - but
it
goes up from the prior
probability, so you need to know the prior probability in order to
calculate the final answer.

A study by Gigerenzer and Hoffrage in 1995 showed that some ways of
phrasing story problems are much more evocative of correct Bayesian
reasoning.  The least
evocative phrasing used probabilities.  A slightly more evocative
phrasing used frequencies instead of probabilities

The most effective presentation found so far is what's known as natural frequencies

the information about the prior
probability
is included in presenting the conditional probabilities

In this case, you might as well just say that 30% of eggs are painted blue, since
the probability of an egg being painted blue is independent of whether
the egg contains a pearl. 

If the bottom bar were renormalized
to the same length as the top bar, it would look like the left sector
had expanded.  This is why the proportion of "women with breast
cancer" in the group "women with positive mammographies" is higher than
the proportion of "women with breast cancer" in the general population
-
although the proportion is still not very high.

The evidence of
the positive mammography slides the prior probability of 1% to the
posterior probability of 7.8%.

You might intuit that since the test could have returned positive for
health, but didn't, then the failure of the test to return positive
must
mean that the woman has a higher chance of having breast cancer

Law of
Conservation of Probability - not a standard term, but the conservation
rule is exact.  If you take the revised probability of breast
cancer after a positive result, times the probability of a positive result,
and add that to the revised probability of breast cancer after a
negative result, times the probability
of a negative result, then you must always arrive at the prior
probability. 

p(A&B) is the same as p(B&A), but p(A|B) is not the same thing as
p(B|A)

For
example, the two quantities p(cancer)
and p(~cancer) have 1
degree of freedom between them, because of the general law p(A) + p(~A) = 1. 

p(positive|cancer) and p(~positive|cancer) also have
only one degree of freedom between them; either a woman with breast
cancer gets a positive mammography or she doesn't.  On the other
hand, p(positive|cancer)
and p(positive|~cancer)
have two degrees of
freedom.  You can have a mammography test that returns positive
for
80% of cancerous patients and 9.6% of healthy patients, or that returns
positive for 70% of cancerous patients and 2% of healthy patients, or
even a health test that returns "positive" for 30% of cancerous
patients
and 92% of healthy patients. 

You should recognize this operation
from the graph; it's the projection of the top bar into the bottom
bar.  p(cancer) is
the
left sector of the top bar, and p(positive|cancer)
determines how much of that sector projects into the bottom bar, and
the
left sector of the bottom bar is p(positive&cancer).

Similarly, if we know the number of patients with breast cancer and
positive mammographies, and also the number of patients with breast
cancer, we can estimate the chance that a woman with breast cancer gets
a positive mammography by dividing: p(positive|cancer) =
p(positive&cancer) / p(cancer). 

What about p(positive&cancer), p(positive&~cancer), p(~positive&cancer), and p(~positive&~cancer)? 
You might at first be tempted to think that there are only two degrees
of freedom for these four quantities - that you can, for example, get p(positive&~cancer) by
multiplying p(positive) *
p(~cancer), and thus that all four quantities can be found given
only the two quantities p(positive)
and p(cancer).  This
is not the case!  p(positive&~cancer)
= p(positive) * p(~cancer) only if the two probabilities are statistically independent - if the
chance that a woman has breast cancer has no bearing on whether she has
a positive mammography. 

groups A, B, C, and D

it follows that the entire set of 16 probabilities
contains only three degrees of freedom.  Remember that in our
problems we always needed three
pieces of information - the prior probability and the two conditional
probabilities