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Arrow-Pratt Intuition

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asked May 18 in BUS 4028F - Financial Economics by Daniel (490 points)

The ActEd notes maintain that the absolute risk aversion function:

A(w) = -U''(w)/U'(w)

Can be ascribed to the fact that "The absolute value of the certainty equivalent of a fair gamble is proportional to -U''(w)/U'(w)".

Attempting to verify this for the function U(w)=log(w) does not appear to yield that conclusion. Consider the following where w is the initial level of wealth:

A fair additive gamble with payoff x s.t. 

E(Gamble) = 1/2x + 1/2(-x) = 0

Now, we have that the certainty equivalent (CE) is given by:

log(CE) = 1/2 log(w+x) + 1/2 log (w-x)

log(CE) = 1/2 log( (w+x)(w-x) )

log(CE) = log( [(w+x)(w-x)] ^ 1/2)

|c_x| = [(w+x)(w-x)] ^ 1/2 - w


-U''(w)/U'(w) = 1/w

Thus c_x does not appear to be proportional to the absolute risk aversion measure A(w).

1 Answer

+1 vote
answered May 19 by ErichMaritz (740 points)
edited May 19 by ErichMaritz

How would you assess the behaviour of  [(w+x)(w-x)] ^ 1/2 - w for small x? Try Taylor:

[(w+x)(w-x)] ^ 1/2 is equal to w [(1+x/w)(1-x/w)] ^ 1/2, which, for small x, is approximately equal to [(1+0.5*x/w-1/8*x^2/w^2)(1-0.5*x/w-1/8*x^2/w^2)] = w (1-1/2x^2/w^2) ignoring terms x^3 and higher.

See if you can take it from here. Best to consider c_x to be a function of x and consider its behaviour as x becomes small.

commented May 20 by Daniel (490 points)

Hi Ehrich, thanks for your help. Been wrestling with the question again and trying to wrap my head around it as per your instruction, but still a little bit stuck.

Could you please explain how to expand 1+x/w to 1+0.5*x/w-1/8*x^2/w^2?

Our formula sheet would state the Taylor expansion as follows:

f(w+h)=f(w) + h*f'(w) + h^2/2! f''(w) + ...

I'm assuming that here we take x to be h. So I need to find some way to express 1+x/w as function of (w+x). 

What I attempted was to use the function f(r)=1/r since this would give me an approximation for f(w+x) = 1/(w+x).

Multiplying the resulting expression by w and then raising it to the power of -1, would give me an approximation for (w+x)/w = 1+x/w as required.

commented May 21 by ErichMaritz (740 points)
I think you forgot about the exponent 1/2. So we expand (1+x/w)^0.5.