Solving exponential utility function with risk taking attitude

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],

[Torsten]

What do you mean by "solve the Exponential utility function equation" ?

And where do CE and EU come into play ?

[Salina Maharjan]

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]

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 ?