Optimal Sample Size Before a Directional Bet

Worked Solution

How to Think About It: You are deciding whether to pay for information before making a directional bet. Without any signals, you know nothing about $\mu$ -- you have no basis to go long or short, so you stay flat ($w = 0$) and earn zero. With signals, you estimate $\mu$ from $\bar{Y}_n$ and bet in the direction of the estimate. The value of sampling is the expected improvement in payoff from having this estimate. You are essentially buying a signal and asking: how many observations are worth it at cost $c$ each?

Quick intuition: more signals reduce the estimation noise on $\mu$. But the marginal value of the $n$-th signal is decreasing (you already know roughly what $\mu$ is). So the optimal $n$ balances information gain against cost -- a classic diminishing returns problem.

Quick Estimate: Take $\sigma = \tau = 1$, $c = 0.05$. With $n = 4$ signals, the standard error of $\bar{Y}_n$ is $\tau/\sqrt{n} = 0.5$. The expected payoff is $E[|\bar{Y}_n|]$ (since you bet in the direction of $\bar{Y}_n$). If $\mu = 0$, $\bar{Y}_n \sim N(0, 0.25)$, so $E[|\bar{Y}_n|] = \sqrt{0.25} \cdot \sqrt{2/\pi} \approx 0.5 \times 0.798 \approx 0.40$. Cost: $4 \times 0.05 = 0.20$. Net: $0.20$. Seems worth it. Contrast with $n = 16$: $E[|\bar{Y}_n|] \approx 0.25 \times 0.798 \approx 0.20$, cost $= 0.80$, net $= -0.60$. Too many samples.

Formal Solution:

Part 1 -- Optimal decision rule.

Given $\bar{Y}_n$, choose $w$ to maximize $E[wR \mid \bar{Y}_n]$. Since $Y_i$ and $R$ share the same mean $\mu$, observing $\bar{Y}_n$ gives information about $\mu$. However, without a prior on $\mu$, we treat $\bar{Y}_n$ as a sufficient statistic and use it as an estimator.

Treating $\mu$ as fixed (frequentist setting with $\mu$ unknown): the expected payoff of position $w$ is $w \cdot \mu$. The best estimate of $\mu$ given $\bar{Y}_n$ is $\bar{Y}_n$ itself (MLE / method of moments). So:

$$w^{*} = \text{sign}(\bar{Y}_n) = \begin{cases} +1 & \text{if } \bar{Y}_n > 0 \\ 0 & \text{if } \bar{Y}_n = 0 \\ -1 & \text{if } \bar{Y}_n < 0 \end{cases}$$

Part 2 -- Expected profit from $n$ signals.

With $w^{*} = \text{sign}(\bar{Y}_n)$, the payoff is $\text{sign}(\bar{Y}_n) \cdot R$. Take expectations:

$$E[\text{sign}(\bar{Y}_n) \cdot R] = E\left[ E[\text{sign}(\bar{Y}_n) \cdot R \mid \mu] \right]$$

For fixed $\mu$: $\bar{Y}_n \sim N(\mu, \tau^2/n)$ and $R \sim N(\mu, \sigma^2)$, independent of $\bar{Y}_n$ given $\mu$. So:

$$E[\text{sign}(\bar{Y}_n) \cdot R \mid \mu] = \mu \cdot E[\text{sign}(\bar{Y}_n) \mid \mu] = \mu \cdot P(\bar{Y}_n > 0 \mid \mu) - \mu \cdot P(\bar{Y}_n < 0 \mid \mu)$$ $$= \mu \cdot (2\Phi(\mu\sqrt{n}/\tau) - 1)$$

This depends on the (unknown) true $\mu$. To evaluate the value of sampling as a function of $n$ alone, we need to integrate over a prior on $\mu$ or evaluate at the worst case. For a symmetric problem (no prior information about the sign of $\mu$), the relevant quantity is the expected absolute error:

$$V(n) = E[|\bar{Y}_n|] - nc$$

where $E[|\bar{Y}_n|]$ is the expected payoff when you bet on the sign of $\bar{Y}_n$ and $\mu = 0$ (hardest case -- but the expected profit gross of cost under $\mu = 0$ is actually the half-normal mean of $\bar{Y}_n$).

More precisely: assume $\mu$ is drawn from a flat/improper prior, or work in the case $\mu$ is known to be either $+\delta$ or $-\delta$ with equal probability (a symmetric two-point prior). Then gross expected profit from $n$ signals is:

$$G(n) = E_{\mu, \bar{Y}_n, R}[\text{sign}(\bar{Y}_n) \cdot R]$$

For the two-point prior ($\mu = \pm \delta$ equally likely): $$G(n) = \delta \cdot (2\Phi(\delta\sqrt{n}/\tau) - 1)$$

(You bet correctly whenever $\bar{Y}_n$ and $\mu$ have the same sign, which happens with probability $\Phi(\delta\sqrt{n}/\tau)$.)

Net expected profit: $$\Pi(n) = G(n) - nc = \delta(2\Phi(\delta\sqrt{n}/\tau) - 1) - nc$$

Part 3 -- Optimal sample size.

Treat $n$ as continuous. Take the first-order condition $\partial \Pi / \partial n = 0$:

$$\frac{\partial G}{\partial n} = c$$

$$\delta \cdot 2\phi(\delta\sqrt{n}/\tau) \cdot \frac{\delta}{2\tau\sqrt{n}} = c$$

$$\frac{\delta^2}{\tau\sqrt{n}} \phi\!\left(\frac{\delta\sqrt{n}}{\tau}\right) = c$$

where $\phi$ is the standard normal PDF. Let $z = \delta\sqrt{n}/\tau$ (so $\sqrt{n} = z\tau/\delta$); then the condition becomes:

$$\frac{\delta^3}{\tau^2} \cdot \frac{\phi(z)}{z} = c$$

The equation $z^{-1}\phi(z) = c\tau^2/\delta^3$ must be solved numerically for $z^{*}$, giving: $$n^{*} = \left(\frac{z^{*} \tau}{\delta}\right)^2$$

For small $c$ (cheap signals), $z^{*}$ is large (many signals); for large $c$, $z^{*}$ is small.

**Part 4 -- Threshold $c^{*}$.**

Sampling at all is worthwhile if some $n \ge 1$ gives $\Pi(n) > 0$. Compare the first signal against staying flat: flat earns $0$, while one signal earns gross $G(1) = \delta(2\Phi(\delta/\tau) - 1)$ at cost $c$. So a single signal beats staying flat iff:

$$\Pi(1) = \delta\left(2\Phi\!\left(\frac{\delta}{\tau}\right) - 1\right) - c > 0$$

Moreover, $G$ is concave in $n$ with $G(0) = 0$ (its derivative $\delta^2 \phi(\delta\sqrt{n}/\tau)/(\tau\sqrt{n})$ is strictly decreasing in $n$), so $G(n)/n$ is decreasing and $G(n) \le n\,G(1)$ for $n \ge 1$. Hence $\Pi(n) \le n(G(1) - c)$: if the first signal is not worth buying, no larger sample is either. The exact threshold is therefore:

$$c^{*} = \delta\left(2\Phi\!\left(\frac{\delta}{\tau}\right) - 1\right)$$

If $c > c^{*}$, stay flat: even the first (most valuable) signal does not justify its cost.

*Pitfall:* evaluating a "marginal value of sampling at $n = 0$" as $\frac{\delta^2}{\tau}\phi(0) = \frac{\delta^2}{\tau\sqrt{2\pi}}$ is invalid -- the true derivative $\partial G/\partial n = \delta^2 \phi(\delta\sqrt{n}/\tau)/(\tau\sqrt{n})$ diverges like $1/\sqrt{n}$ as $n \to 0^{+}$, so there is no finite marginal value at zero and the threshold cannot be obtained that way.

Answer: 1. Optimal rule: $w^{*} = \text{sign}(\bar{Y}_n)$. 2. Net profit: $\Pi(n) = \delta(2\Phi(\delta\sqrt{n}/\tau) - 1) - nc$ under a symmetric two-point prior. 3. Optimal $n^{*}$: solve $\delta^2 \phi(\delta\sqrt{n}/\tau)/(\tau\sqrt{n}) = c$ numerically. 4. Threshold: $c^{*} = \delta(2\Phi(\delta/\tau) - 1)$ -- the gross value of a single signal; only sample if $c < c^{*}$.