Solving exponential utility function with risk taking attitude

6 次查看（过去 30 天）

Salina Maharjan 2022-9-9

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1801630-solving-exponential-utility-function-with-risk-taking-attitude

评论： Torsten 2022-9-9

Hi,

It would be really helpful if someone can respond on how to solve the Exponential utility function equation in matlab. Here CE can vary from High to Low.

U(x) = A – B*EXP(–x/RT).

where,

A = EXP (–Low/RT) / [EXP (–Low/RT) – EXP (–High/RT)]

B = 1 / [EXP (–Low/RT) – EXP (–High/RT)]

and CE = –RT*LN[(A–EU)/B],

3 个评论
显示 1更早的评论隐藏 1更早的评论

Salina Maharjan 2022-9-9

the first equation is called the Utility Function. An iterative approach to all the three equations needs to be applied but I am not sure how its done.

The unknowns are A, B, RT

But CE can be a value between Hight and Low (variable).

U(x) = A – B*EXP(–x/RT).

where,

A = EXP (–Min(x) / RT) / [EXP (–Min(x) / RT) – EXP (–Max(x) / RT)]

B = 1 / [EXP (–Min(x) / RT) – EXP (–Max(x) / RT)]

and RT= –CE / LN[{ A– (0.5 * U (Max (x))-0.5 * U(Min (x))} / B ]

Torsten 2022-9-9

So you have a vector of values for x and U(x) and you try to determine A, B and RT such that norm(U(x)-(A-B*exp(-x/RT))) is minimized for a given value of CE ?

请先登录，再进行评论。

请先登录，再回答此问题。