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Say that I observe a predictor $w_t \sim N(0,\sigma_1)$ for the returns in a single asset over the next time interval:

$$ r_t = \alpha w_{t-1} + z_t $$ where $z_t \sim N(0,\sigma_2)$ is unobserved and independent of $w_{t-1}$.

I want to find the function $f$ that maps $w_t$ to a position in the asset which maximizes the expected Sharpe Ratio of the next $n$ trading returns: $$\max_f \mathrm{E}\left[\frac{\sqrt{n}\sum_{i=1}^n p_i}{n \sqrt{\sum_{i=1}^n \left(p_i - \overline{p_i}\right)^2}} \right]$$ where the profit at time step $t$ is $$p_t = r_t f(w_{t-1}) $$

Is there an easy way using something like stochastic control to find the optimal $f$ without any constraints on its structure?

EDIT: As pointed out by Chris, I had oversimplified the problem by only looking at a single-step. I've modified the description to make it n-step.

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  • $\begingroup$ There is no such thing as a "one asset position". One possibility is the investor has wealth $W$ and allocates it between the risk free asset and the single risky asset according to whether the 1-step predicted return is greater or less than $R_f$. $\endgroup$
    – nbbo2
    Commented Apr 11, 2016 at 16:44
  • $\begingroup$ That's just semantics. Obviously in this example the fraction of the investor's wealth currently not invested in the risky asset would be in something like cash earning nothing. You could call it a 2nd asset if you like. $\endgroup$
    – user2303
    Commented Apr 11, 2016 at 22:32
  • $\begingroup$ Did you have any luck on this topic? $\endgroup$
    – mojovski
    Commented Dec 18, 2020 at 21:10

1 Answer 1

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As you have defined the Sharpe ratio, it is independent of your position. You have $$ \mathrm{E}[rf(z_1)] = \alpha z_1 f(z_1) $$ and $$ \mathrm{Var}[rf(z_1)] = \sigma_2^2f(z_1)^2 $$ and hence $$ \frac{\mathrm{E}[rf(z_1)]}{\sqrt{\mathrm{Var}[rf(z_1)]}} = \frac{\alpha z_1}{\sigma_2} $$ independent of the function $f(z_1)$.

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  • $\begingroup$ Thanks, you're right. I oversimplified the problem by only looking at one time step. I've edited the question to make it n-step. $\endgroup$
    – user2303
    Commented Apr 11, 2016 at 14:55

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