Re: Fuzzy integral definition
- From: un student <un.student@xxxxxxxxx>
- Date: Tue, 1 Sep 2009 01:00:21 -0700 (PDT)
On Aug 28, 2:41 pm, "Dmitry A. Kazakov" <mail...@xxxxxxxxxxxxxxxxx>
wrote:
On Fri, 28 Aug 2009 00:41:06 -0700 (PDT), un student wrote:<..>
On Aug 27, 9:13 pm, "Dmitry A. Kazakov" <mail...@xxxxxxxxxxxxxxxxx>
I guess if it was meant to be
<math>
A_\alpha =
\{ \int_a^b g(t) dt |
g(t) \leq f_\alpha(t) \forall t \in [a,b] \}
</math>
I .e. the alpha cut of the fuzzy integral is a set of plain integrals over
[a,b] computed for each function q dominated by the alpha-cut of f.
f_\alpha(t) is a fuzzy set for any t. You take all real-valued g's such
that for any t the outcome of g is in the support set of f_\alpha(t), i.e.
<math>
f_\alpha(t)(g(t)) \ge 0
</math>
In other words, g may take any value from _\alpha(t). All these g's get
integrated on [a,b]. The results comprise some crisp set of "possible"
integrals. (It should be shown that with alpha increasing, the set get
narrower) Then the integral is proclaimed to have in x the truth value of
maximum of the alpha over sets of possible integrals containing x. (Zadeh's
extension principle).
I guess my reaction is "mmmkay...". Lets say I have the earlier
defined function:
f(t)(x) =
1+t-x, t\leq x \leq t+1
1-t+x, t-1\leq x \leq t \forall t \in R
0 elsewhere
And I would like to calculate \int_0^1 f(t) dt. Intuitively speking
this is a triangle-like fuzzy number (or set, for that matter) moving
along x-axis. Lets start with
A_0 = \int_0^1 f_0(t) dt = \{ \int_0^1 g(t) dt | g(t)\leq f_0(t)
\forall t \in [0,1]\}.
Now f_a(t) = [t-(1-a), t+(1-a)] and f_0(t) = [t-1, t+1], hence (I'm
cutting details out)
A_0 = \{ \int_0^1 g(t) dt | g(t)\leq f_0(t) \forall t \in [0,1] \}.
Now for any t < -1 we get min \int_0^1 g(t) dt = -1 (let g(t) = f_0
(t)) and max \int g(t) dt = 0 (g(t) == 0). Respectively for t > -1 we
get max being 1 and hence
A_0 = [-1,1].
As a now increases the area between g(t) and x-axis gets smaller, i.e.
absolute value of the integral decreases. From f_a(t) = [t-(1-a), t+(1-
a)] we can conclude that the width of the "triangle" (g(t), that is)
is t+(1-a)-(t-(1-a)) = t+1-a- t+1-a = 2*(1-a) for and height 1-a
giving the area (1-a)^2, hence A_a = [-(1-a)^2, (1-a)^2 ] which gives
F(x) =
1-sqrt(x), 0 \leq x \leq 1
1-sqrt(-x), -1\leq x < 0
0 otherwise
But I have doubts on this result. I'll double check my algebra later.
At least this gets clearer now. Thanks for your help!
.
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