Solving exponential utility function with risk taking attitude

Salina Maharjan

2022 9 9

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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],

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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 ?

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