I am not sure to solve the following investment problem:

I have an investor which receives an income $I_n\ge 0$ at the start of year $n$. The investor chooses a proportion $p_n\in[0,1]$ of this in the stock market and consumes the remainder. At the start of the next year his income is of course $I_{n+1}=I_n+p_nI_nX_n$, where $X_n>0$ is the retrn on the stock market in year $n$ and is IID with mean $\mu$.

The investor now chooses $p_n$ to maximise $\mathbb E[\sum_{n=0}^{N-1}(1-p_n)I_n+I_N]$

This equation obviously represents the consumption over $N$ years, which the investor wants to maximise.

My solution: Let $F_s(i)$ be the maximal reward obtainable in state $i$ when there is time $s=N-t$ to go, then the dynamic equation is $F_s(i)=\max_p[(1-p)i+F_{s-1}(i+piX)]$ with $F_0(i)=0$ because at time $N$ there is no consumption.

Now $F_1(i)=i, F_2(i)=\max[(1-p)i+i+piX]=\max[2i,i+piX]=i\max[2,1+pX]=i\rho_2$

There I get by induction $F_s(i)=i\rho_s$

How should the investor now choose his strategy? I am almost done but I do not see thr last crucial step.


1 Answer 1


For the last two years you want to maximise the expectation of $$I_N+I_{N-1}(1-p_{N-1}) = I_{N-1}(2+p_{N-1}(X_{N-1}-1))$$ and the expectation is maximised: if $\mu=E[X_{N-1}] \gt 1$ when $p_{N-1}=1$; and if $\mu=E[X_{N-1}] \lt 1$ when $p_{N-1}=0$.

If you invest everything in the penultimate year then it seems obvious you will also invest everything in all previous years.

So let's look at the $\mu=E[X_{N-1}] \lt 1$ so $p_{N-1}=0$ and $I_N+I_{N-1}(1-p_{N-1}) = 2I_{N-1}$: then the last three years are about maximising the expectation of $$I_N+I_{N-1}(1-p_{N-1}) + I_{N-2}(1-p_{N-2}) = I_{N-2}[3+ p_{N-2}(2X_{N-2} -1)]$$ and whether optimal $p_{N-2}$ is $1$ or $0$ depends on whether $\mu \gt \frac12$ or $\mu \lt \frac12$.

Taking this back in a similar manner, the overall optimal strategy becomes clear:

  • let $p_n=1$ (i.e. invest everything) when $\mu \gt \frac{1}{N-n}$, i.e. when $n \lt N - \frac{1}{\mu}$

  • let $p_n=0$ (i.e. invest nothing) when $\mu \lt \frac{1}{N-n}$, i.e. when $n \gt N - \frac{1}{\mu}$.

  • You can choose any $p_n\in [0,1]$ (i.e. invest anything) when $\mu = \frac{1}{N-n}$, i.e. when $n = N - \frac{1}{\mu}$

I would not recommend this as a personal investment strategy. Risk should have an impact too.

  • $\begingroup$ Thanks for your answer. Do you also have an idea how to use my approach using the dynamic programming equation? When you are saying "Taking this back in a similar manner" it is a more intuitive approach, I think using the Bellman Equation would be satosfactory in this case. $\endgroup$
    – Alkibiades
    Nov 22, 2013 at 1:51
  • $\begingroup$ When I say "Taking this back in a similar manner", I am saying that more generally given $I_{N-k+1} = I_{N-k}+p_{N-k}I_{N-k}X_{N-k}$ and if the sum of the last $k$ terms is $kI_{N-k+1}$ then the sum of the last $k+1$ terms is $k(I_{N-k}+ p_{N-k}I_{N-k}X_{N-k})+(1-p_{N-k})I_{N-k}$ which simplifies to $I_{N-k}[k+1 + p_{N-k}(kX_{N-k}-1)]$ which will be maximised by having $p_{N-k}=1 \text{ or }0$ depending on $E[kX_{N-k}-1]$, and if $E[kX_{N-k}-1] \lt 0$ then $p_{N-k}=0$ and the sum of the last $k+1$ terms is $(k+1)I_{N-k}$, thereby completing the induction. $\endgroup$
    – Henry
    Nov 22, 2013 at 7:37

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