Jason (jcreed) wrote,

Here's a cute little decision-theoretic puzzle that I fished out of a book:

Suppose you have a starting blob of money, say X dollars. Each day, you choose how to divide your money between consumption and investment. If you consume Y dollars, then you get ln(Y) utility out of that act of consumption. If you invest Z dollars, (really, Z = X - Y; any money you don't spend automatically gets "invested") then you get back R*Z dollars on the next day, where R is a number given to you by the universe, indicating the return you can manage to get on investments. Finally, we make the assumption that you're impatient: the actual value of you today of a util earned far in the future, say on day T, is really only D^T, where D is some "discounting factor", a number less than (but perhaps quite close to) one.

Given that, what strategy optimizes the sum of the present values of all future rewards?

So the book said it was to always consume (1-D)X and to invest DX, which surprised me, because it was totally independent of R!

Suppose the optimal strategy is at least of the above form; always to consume some fixed fraction k of my current money. The answer is ostensibly that k = 1 - D.

Let's abstract away from my logarithmic utility function for a moment and say that spending Y dollars gives me f(Y) utils. Then my total utility function is

f(kX) + D f(kXR(1-k)) + D^2 f(kX(R(1-k))^2) + ... + D^n f(kX(R(1-k))^n)

because immediately I consume kX, but I keep (1-k)X around, which grows to XR(1-k), and then I consume k of that, giving me f(kXR(1-k)) utility, but that all happens tomorrow, so it only counts as D f(kXR(1-k)) to me today. Then on the next day, the XR(1-k)^2 that I still invested becomes XR^2(1-k)^2, and I consume k of it, giving me f(kX(R(1-k))^2) utils, but that's two days into the future, so it really only counts as D^2 f(kX(R(1-k))^2). And so on.

To optimize, I take the k-derivative and set it to zero. Let me just rewrite the above compactly as
sum_n D^n f(kX(R(1-k))^n)
and then what I want to solve for is
0 = sum_n D^n R^n X f'(kX(R(1-k))^n) ((1-k)^n - nk(1-k)^(n-1))
Now if f is ln, then the derivative is just the reciprocal, and things cancel out a lot:
0 = sum_n D^n (1 - k - nk) / (k(1-k))
and we can multiply both sides by k(1-k) to get
0 = sum_n D^n (1 - k - nk)
Now I'm going to use the fact that sum_n D_n = 1/(1-D) and sum_n n D_n = D/(1-D)^2 to get
0 = (1 - k)/(1 - D) - kD/(1-D)^2
0 = (1 - k)(1-D) - kD
and so
k = 1 - D

There you have it! I guess both the logarithmic utility and the exponential discounting of utility over time are pretty sketchy assumptions, but it's interesting it works out this way. It's still completely weird to me that it doesn't depend on R.
Tags: decisions, math

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