- #1
dollyprane
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Hi everybody,
I am currently writing a master's thesis in probabilistic insurance, for which i am using prospect theory (derived by Kahneman and Tversky in 1979). .
Just as a preamble: this is not a probability problem, rather the study of a function.
I have an intuition but I cannot prove the result, and i have been working the whole day on it!
Let me explain simply:
Let [TEX]\pi[/TEX] be a probabililty weighting function such that, for all [TEX]p \in [0,1][/TEX] we have:
*****
[TEX]\pi(0) = 0[/TEX] and [TEX]\pi(1)=1[/TEX]
***** Overestimation of small probabilities and underestimation of large probabilities
[TEX]p <\bar{p} \ , \quad \pi(p) > p[/TEX]
[TEX]p >\bar{p} \ , \quad \pi(p) < p[/TEX]
where [TEX]\bar{p}[/TEX] is a fix point of the function [TEX]\pi[/TEX]***** Subadditivity of small probabilities
[TEX]p < \bar{p}, \ \pi(rp) > r \pi(p)[/TEX] for any [TEX]0<r<1[/TEX]***** Subcertainty
[TEX]\pi(p) + \pi(1-p) < 1[/TEX] for all [TEX]0<p<1[/TEX]***** Subproportionality
[TEX]\frac{\pi(pq)}{\pi(p)}\leq\frac{\pi(pqr)}{\pi(pr)}[/TEX] for all [TEX]0<p,q,r \leq 1[/TEX]A generic graphical representation of this function is to be found below:View attachment 3272My question:
Thanks to functions plotting softwares i have the impression that:
for all [TEX]p_1, p_2 < \bar{p}[/TEX] :
[TEX]\pi(p_1) \pi(p_2) < \pi(p_1 p_2) \left[ \pi(p_1) \pi(p_2) + \pi(p_1) \pi(1-p_2) + \pi(1-p_1) \right][/TEX]
But i cannot manage to prove it...
We have clearly: [TEX]\pi(p_1) \ \pi(p_2)< \pi(p_1 \ p_2)[/TEX] thanks to the subproportionality criterium, and we have that [TEX] \left[ \pi(p_1) \pi(p_2) + \pi(p_1) \pi(1-p_2) + \pi(1-p_1) \right] < 1 [/TEX]. But even when I try to group terms or use any criteria above, I am stuck...
I know this result is not valid on [0,1], but i think it is the case for small values of p ([tex] p < \bar{p} [/tex]).
I would be so grateful if you could help me out to confirm or invalidate my intuition. any help is appreciated!
thanks in advance.
Guillaume
I am currently writing a master's thesis in probabilistic insurance, for which i am using prospect theory (derived by Kahneman and Tversky in 1979). .
Just as a preamble: this is not a probability problem, rather the study of a function.
I have an intuition but I cannot prove the result, and i have been working the whole day on it!
Let me explain simply:
Let [TEX]\pi[/TEX] be a probabililty weighting function such that, for all [TEX]p \in [0,1][/TEX] we have:
*****
[TEX]\pi(0) = 0[/TEX] and [TEX]\pi(1)=1[/TEX]
***** Overestimation of small probabilities and underestimation of large probabilities
[TEX]p <\bar{p} \ , \quad \pi(p) > p[/TEX]
[TEX]p >\bar{p} \ , \quad \pi(p) < p[/TEX]
where [TEX]\bar{p}[/TEX] is a fix point of the function [TEX]\pi[/TEX]***** Subadditivity of small probabilities
[TEX]p < \bar{p}, \ \pi(rp) > r \pi(p)[/TEX] for any [TEX]0<r<1[/TEX]***** Subcertainty
[TEX]\pi(p) + \pi(1-p) < 1[/TEX] for all [TEX]0<p<1[/TEX]***** Subproportionality
[TEX]\frac{\pi(pq)}{\pi(p)}\leq\frac{\pi(pqr)}{\pi(pr)}[/TEX] for all [TEX]0<p,q,r \leq 1[/TEX]A generic graphical representation of this function is to be found below:View attachment 3272My question:
Thanks to functions plotting softwares i have the impression that:
for all [TEX]p_1, p_2 < \bar{p}[/TEX] :
[TEX]\pi(p_1) \pi(p_2) < \pi(p_1 p_2) \left[ \pi(p_1) \pi(p_2) + \pi(p_1) \pi(1-p_2) + \pi(1-p_1) \right][/TEX]
But i cannot manage to prove it...
We have clearly: [TEX]\pi(p_1) \ \pi(p_2)< \pi(p_1 \ p_2)[/TEX] thanks to the subproportionality criterium, and we have that [TEX] \left[ \pi(p_1) \pi(p_2) + \pi(p_1) \pi(1-p_2) + \pi(1-p_1) \right] < 1 [/TEX]. But even when I try to group terms or use any criteria above, I am stuck...
I know this result is not valid on [0,1], but i think it is the case for small values of p ([tex] p < \bar{p} [/tex]).
I would be so grateful if you could help me out to confirm or invalidate my intuition. any help is appreciated!
thanks in advance.
Guillaume