How much to invest to reach a target?

Question

Your current wealth is $W$. Each day you can invest some of it; there's a probability $p$ that you will win as much as you invested, $1-p$ that you will lose it. You want to reach a target wealth $W_T$ within $n$ days. Each day, you can choose the fraction $f$ of your wealth to invest. How do you choose $f$ to maximise the chance to hit your target in time?

If it helps, assume $p > 0.5$, $n \gg 1$.

This is essentially a pure maths problem but I thought it would be interesting for quants. I have seen discussions of similar problems (e.g. "Can you do better than Kelly in the short run?", Browne (2000)), but they assume a continuous outcome and a few other things. I'd also be happy with a way to find $f$ via simulations, an analytical formula is not essential.

[Edit: you cannot bet more than you currently have. I should have specified this earlier.]

Answer 1

Let $w^*$ be your target wealth and $w_0$ be your initial amount. One very effective strategy one could apply is the following,

Day 1: Bet $w^* - w_0$; if the bet resolves in your favour then you have reached your target wealth and so stop; Else

Day 2: Bet $2(w^* -w_0)$; if the bet resolves in your favour then you have reached your target wealth and so stop; Else

Day 3: Bet $4(w^* -w_0)$; if the bet resolves in your favour then you have reached your target wealth and so stop; Else

... Day N: Bet $2^{N-1}(w^* -w_0)$ if the bet resolves in your favour then you have reached your target wealth and so stop; Else you have insufficient funds to reach your target wealth; i.e. $2w(t) < w^*$ :(

The number of bets you can place is then $N+1$ where $N$ is the largest integer such that,

$$ 2 \Big ( w_0 - (w^* - w_0) \sum_{k=0}^N 2^k \Big ) \geq w^*$$

That is,

The probability of the strategy being successful is then the complement of the strategy failing in each trial,

$$ 1 - (1-p)^{N+1} $$

For example, if probability of success is $p=0.6$, initial amount is $w(0)=w_0 = 100$, and target wealth is $w^* = 105$ then you can place $4$ bets each with the possibility of achieving the target wealth. The probability of success is the complement of the probability of failure which is, $(1-0.6)^4 = 0.0256$ and so probability of reaching your target wealth is $1 - 0.0256 = 0.9744 $ which is pretty good. You can visualize this with a tree diagram and it helps to explain the reasoning.

All in all, I am not sure if this is optimal but it seems very effective.

I realized also the policy, $f = f(w)$ can be expressed as,

$$ w(t+1) = w(t) + \max\{0,w^* - w(t)\}\mathcal{X}_{t+1}$$

for $t \leq T$ with $T$ chosen such that it is the largest integer for which $2w(T-1) \geq w^*$ in all cases, in particular where you lose in every bet. $\mathcal{X}_{t+1}$ is a random variable which takes the values $1$ and $-1$ with appropriate probabilities.

Answer 2

Trying to get this started I'll go for what I think is the easiest part, in a highly handwaving way if you'll excuse:

With number of days large enough that we will likely not end the last day with wealth $W$ between $0$ and target wealth $W_T$, the time distance to the last day loses significance and the problem could be stated in time invariant form with $W$ developing between infinite borders of bankruptcy and target. For $p > 0.5$ we will become risk adverse as excpected gains are positive and we want to bet cautiously to stay inside the interval to reap those gains with the added benefit of diversification between many bets. If instead $p < 0.5$ we become risk loving as we don't want to let negative expected returns grind down our wealth from repeated playing.

We would then stake it all in one go to expose us to negative expected returns only once, but not more than it would take to reach $W_T$; this trivially spells out as:

for $ n>>1 $ and $p<0.5$:

$$f(W) = min(1,W_T/W-1)$$

Update: By the same argument for $ n>>1 $ and $p>0.5$:$$W>0: f(W) -> 0; f(0) = 1$$ Now, this is still some way from the final answer as I guess the question is asking either for the exact expression for $f$ but preferably for some approximation given a specific value of $ n>>1 $?

Update2: To get optimal $f(w)$ for a given $n$ and $p>0.5$; write out the transition tree backwards starting with $n = 0$ for a couple of $n$:s. For nodes in that tree, use the lowest level of $W$ in each interval of $W$:s that have the same probability of winning. For $n=1$ for instance we thus have $W/W_t = 0.5$ as the one node inside the interval. In order to travel through as many nodes as possible towards a win or lose to exploit $p>0.5$ we have to only stake what we need to get to nearest nodes in the next timestep. We see in the tree that this amount is sometimes $0$, and sometimes $1/2^n$. Looking closer it is further seen that:

for $p>0.5; W>=W_T/2^n$:

if $int((W/W_T)/2^n)$ is an odd number: $$f(W) = (1/2^n)*W_T/W$$ otherwise: $$f(W) =0$$

Equally optimal but smoother you could use a sawtooth form for $f$ varying linearly between $0$ at even values for $int((W/W_T)/2^n)$ and $(1/2^n)*W_T/W$ at odd values.

Update3: For $p=0.5$ far away from the last day the game is fair and the outcome does not depend on how we play it. The only thing we have to ensure is to bet boldly enough not to end up with $0<W<W_T$ when the game is finished and of course not to reach for gains that would put us above $W_T$. So every choice: $$0<f<=min(1,W_T/W-1)$$ is equally good away from the endgame. Using the higher value for $f$ yields by symmetry and induction $$P(win) = W/W_T$$ Now you can use that and stationarity of $P(win)$ to show that the lower values for $f$ also works with these $P(win)$.

[Edited Update 2]