Monday, February 07, 2011

Normally distributed asset returns and CARA utility make the mean variance approximation exact.

My correspondence

Dear Pro.Waldmann

I'm a [university name redacted] PhD student who is interested in your paper "Noise Trader Risk in Financial Markets" in 1990 (The journal of Political Economy,Vol98, No.4, P703-738). It's really a good early paper which theoreticaly brings noise traders' activities into the cause of price changes. However I was stopped at a simple beginning step when I tried to understand every detail of this paper. In page 708 there is a explanation for the utility maximization. "With normally distrituted returns to holding a unit of the risky asset, maximizing the expected value of (2) is equivalent to maximizing (3)".
I was stopped here, because I tried to prove this equivalence algebraically but failed to do so. I was wondering if there is a mathmetical method to prove that maximizing the exponential utility function equals to maximizing the function (3)?

Maybe it is a stupid question which is very simple to answer, but I will really appreciate your guidance. Thanks for your time and I'm sorry to trouble you.

Best wishes
Yours sincerely
[name redacted ]


I reply below. This is how I reply to correspondence about math. I don't know if that is a threat or a promise. Look below to decide for yourselves.


It is an intelligent question. I don't know a reference for the proof.
The mathematical result is that if x is normally distributed with mean m and variance s^2 ( s squared) then
E(-exp(lambda*x)) = -exp[-lambda*(m - (lambda/2)s^2)]

(* means times, exp() meand e to the () ^ means to the so ^2 means squared )

the function f(a) = -exp(-lambda*a) is monotonically increasing. Its derivative is lambda*exp(lambda*a) > 0 for any a. So maximizing
-exp[-lambda*(m - (lambda/2)s^2)] is just the same thing as maximizing m - (lambda/2)s^2.

The key step is
E(-exp(lambda*x)) = -exp[-lambda*(m - (lambda/2)s^2)]

This is just an integral (hard to type with plain ascii)

if x is normally distributed the expected value is
the integral from - infinity to infinity of
- exp(lambda*x) (1/(s*(2*pi)^0.5)) exp(-(x-m)^2/s^2)=

the integral from - infinity to infinity of
-(1/(s*(2*pi)^0.5)) exp(-lambda*x -(x-m)^2/(2s^2)) =


the integral from - infinity to infinity of -(1/(s*(2*pi)^0.5))exp(-lambda*x -(x^2-2mx + m^2)/(2s^2)) =

the integral from - infinity to infinity of -(1/(s*(2*pi)^0.5))exp(-lambda*x -(x^2-2mx + m^2)/s^2) =

the integral from - infinity to infinity of -(1/(s*(2*pi)^0.5))exp(-lambda*x +(-x^2 + 2mx - m^2)/(2s^2)) =

the integral from - infinity to infinity of -(1/(s*(2*pi)^0.5))exp((-x^2 + 2mx-2lambda*(s^2)x - m^2)/(2s^2))

OK now complete the square so =

the integral from - infinity to infinity of -(1/(s*(2*pi)^0.5))exp((-(x - (m-lambda*(s^2)))^2 + m^2 + 2m*lambda*s^2 - lambda^2(s^4) - m^2)/(2s^2))

Sorry about that but it is just algebra -- completing a square. If I typed correctly it works fine.

this equals

the integral from - infinity to infinity of (1/(s*(2*pi)^0.5))exp(-(x - (m-lambda*(s^2)))^2/(2s^2)) *
(-exp(-(2m*lambda*s^2 - lambda^2(s^4) )/(2s^2)))

OK the term exp(-(2m*lambda*s^2 - lambda^2(s^4) )/(2s^2)) is a constant and can be taken out of the integral so this =

-exp(-(2m*lambda*s^2 - lambda^2(s^4) )/(2s^2)) *
the integral from - infinity to infinity of (1/(s*(2*pi)^0.5))exp(-(x - (m-lambda*(s^2)))^2/(2s^2))

The integral is just the integral from - infinity to infinity of a normal density with mean m - lambda*(s^2) and variance s^2 so it equals one so the whole mess is eqaul to

-exp(-(2m*lambda*s^2 - lambda^2(s^4) )/(2s^2))

or dividing
-exp(-(lambda*m - (lambda^2/2)s^2)
=

-exp(-lambda(*m - (lambda/2)s^2))

as asserted.

Ok a lot of algebra but basically just an integral, complete the square, take a constant out of the integration -- hey that integral is the probability that a randome variable is greater than - infinity and less than infinity so it's one.

I had fun typing this.
I hope it is useful.

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