Hi dime365!
Definitions of concavity and quasi-concavity can be found in pages 1
and 3 respectively of the following pdf document.
Quasi-concavity
http://www.colorado.edu/Economics/courses/rubinchik/econ7010/quasiconc.pdf
Defintion of concavity: a function is concave if
f( a*x + (1-a)*x' ) >= a*f(x) + (1-a)*f(x') (1)
for all x, x', and for a belonging to (0,1)
Definition of quasiconcavity: a function is quasi-concave if
f (a*x + (1-a)*x' ) >= min( f(x) , f(x') ) (2)
Now, let's first see the proof that a concave function is also
quasi-concave. Given that the function is concave, then it satisfies
equation (1). But then it's immediate that it's quasiconcave. To see
this, without loss of generality, assume that f(x)>=f(x'). Then,
a*f(x) + (1-a)*f(x') >= a*f(x') + (1-a)*f(x') = f(x')
Therefore, by concavity of f,
f( a*x + (1-a)*x' ) >= a*f(x) + (1-a)*f(x') >= f(x')
But we assumed that f(x)>=f(x'), so min(f(x),f(x'))=f(x'), so we have that
f( a*x + (1-a)*x' ) >= min(f(x),f(x'))
which is definition of quasiconcavity. QED
In order to see that the converse is not true, we must find a
counterexample: a function that satisfies quasiconcavity but doesn't
satisfy concavity. A possible counterexample if:
f(x) = exp(x)
(exp(x) means "e to the power of x").
It's easy to verify that this function is quasi-concave (actually, all
monotonically increasing functions are quasi-concave). However, it's
clear that exp(x) is a convex function. You can check this by looking
at the second derivative of exp(x). Since the second derivative is
also exp(x), which is greater than zero, then this function is convex.
So this is an example of a quasiconcave function that is convex,
providing a counterexample for the converse of the theorem above.
Google search strategy
"quasi-concavity" definition
://www.google.com.ar/search?q=%22quasi-concavity%22+definition&ie=UTF-8&oe=UTF-8&hl=es&btnG=B%C3%BAsqueda+en+Google&meta=
I hope this helps! If you have any doubts regarding my answer, please
don't hesitate to request a clarification before rating it. Otherwise,
I await your rating and final comments.
Best wishes!
elmarto |
Request for Answer Clarification by
dime365-ga
on
10 Nov 2003 17:20 PST
Hi elmarto!
Ive been going over it the past few days and the definition of
quasi-concavity that we use is different.
It's: The function f is quasiconcave if and only if for all x E S, all
x' E s and all t [0,1] we have
f(x) >= f(x')
then
f(tx' + (1-t)x) >= f(x')
(im not sure if the x and x' are consistent to use with the previous
answer though, this is the characterization i was given, i just copied
it letter for letter)
This is what I have so far:
Concavity:
f(tx' + (1-t)x) >= tf(x') + (1-t)f(x)
Quasi Concavity:
f(x) >= f(x')
Statement:
f(tx' + (1-t)x) >= f(x')
Proving, set:
tf(x') + (1-t)f(x) >= f(x')
Rearranging:
tf(x') + f(x) - tf(x) >= f(x')
f(x) - tf(x) >= f(x') - tf(x')
(1-t)f(x) >= (1-t)f(x')
f(x) >= f(x')
Since
f(tx' + (1-t)x) >= tf(x') + (1-t)f(x)
tf(x') + (1-t)f(x) >= f(x')
Then
f(tx' + (1-t)x) >= f(x')
And i need a counter example that proves that something can be
quasi-concave but not concave, i didnt quite understand your
counterexample
Thanks!
|
Request for Answer Clarification by
dime365-ga
on
10 Nov 2003 17:22 PST
i mean that i need proof that quasiconcavity does not imply concavity
|
Request for Answer Clarification by
dime365-ga
on
10 Nov 2003 17:33 PST
Is my definition for quasi concavity correct for R --> R ? or am i
looking at the wrong theorem?
Sorry for all the questions lol
|
Clarification of Answer by
elmarto-ga
on
11 Nov 2003 05:44 PST
Hi dime!
Your definition for quasi-concavity is esentially the same as the one
I stated in my answer. Your definition is that:
f(x)>=f(x') implies that f(tx + (1-t)x')>=f(x')
But, if f(x)>=f(x'), then min(f(x),f(x')) is equal to f(x'), so we go
back to my definition! If you don't understand the proof I gave using
"my" definition (which is equivalent to yours) please let me know.
Regarding the proof that quasi-concavity doesn't imply concavity; in
order to show this it suffices to come up with a counterexample. The
function
exp(x)
constitutes a counterexample. Why is this so? Since exp(x) is an
increasing function, then it's also quasi-concave. Let's see why. We
have that, given x>x', then
t*x' + (1-t)*x > x'
but then, since exp(x) is increasing, we have that
exp(t*x' + (1-t)*x) > exp(x')
and also, since exp(x) is increasing, we have that exp(x)>exp(x'), so
min(exp(x),exp(x'))=exp(x'). So,
exp(t*x' + (1-t)*x) > min(exp(x),exp(x'))
Therefore, exp(x) is quasiconcave. Now we have to show that it's not
concave. Let's take x=0 and x'=1 and t=1/2. Then,
exp( (1/2)*0 + (1/2)*1 ) = exp(1/2) = 1.648
but
(1/2)*exp(0) + (1/2)*exp(1) = 1.859
so we have that, for this x, x' and t
exp(tx + (1-t)x') < t*exp(x) + (1-t)*exp(x')
so exp(x) is not concave.
If you still have any doubts, please request further clarification.
Best wishes!
elmarto
|
Request for Answer Clarification by
dime365-ga
on
11 Nov 2003 06:55 PST
ah i see!
but would the way i proved concavity implies quasiconcavity be correct aswell?
is one way "more" correct than the other? or are both valid for what
needed to be proved?
thanks =)
|
Clarification of Answer by
elmarto-ga
on
11 Nov 2003 10:28 PST
Hi dime!
I don't quite understand your proof. In particular, you state:
"Proving, set:
tf(x') + (1-t)f(x) >= f(x')
Rearranging:
tf(x') + f(x) - tf(x) >= f(x')
f(x) - tf(x) >= f(x') - tf(x')
(1-t)f(x) >= (1-t)f(x')
f(x) >= f(x')"
But why do you start by assuming tf(x') + (1-t)f(x) >= f(x')?
In any case, no proof can be "more" correct than a correct proof.
There might be different ways of proving statements, none of which
could be deemed "better". Short, clear proofs are usually preferred,
but a correct long one can't be rejected.
Best regards,
elmarto
|
Request for Answer Clarification by
dime365-ga
on
11 Nov 2003 10:37 PST
f(x) >= f(x')
then
f(tx' + (1-t)x) >= f(x')
and
f(tx' + (1-t)x) >= tf(x') + (1-t)f(x)
so if the following holds true: tf(x') + (1-t)f(x) >= f(x')
then: f(x) >= f(x')
does that work?
|
Request for Answer Clarification by
dime365-ga
on
11 Nov 2003 10:39 PST
oops, clicked submit too early, add this on
meaning: f(tx' + (1-t)x) >= f(x')
|
Clarification of Answer by
elmarto-ga
on
11 Nov 2003 11:08 PST
Hi dime!
OK, I get it now, and it's mostly correct. I think the only correction
you'd have to make is the following: instead of saying
if tf(x') + (1-t)f(x) >= f(x') then f(x) >= f(x')
I would say
if f(x) >= f(x') then tf(x') + (1-t)f(x) >= f(x')
(which is also true)
I would state it this way because the definition of quasi-concavity
starts by saying: if f(x) >= f(x'), then... So you would start with
the same hypothesis (f(x)>=f(x')) and arrive to the conclusion (f(tx'
+ (1-t)x) >= f(x')). This reasoning would thus imply quasi-concavity.
Best regards,
elmarto
|
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