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 $\delta = \tau = 1$ and $c = 0.05$. With $n = 4$ signals the standard error of $\bar{Y}_n$ is $\tau/\sqrt{n} = 0.5$, so you call the sign correctly with probability $\Phi(2) \approx 0.977$: gross value $\approx 1 \times (2(0.977) - 1) = 0.95$ against a cost of $4 \times 0.05 = 0.20$, netting about $0.75$. Push to $n = 16$ and the gross value only creeps up to $\approx 1.00$ while the cost quadruples to $0.80$, netting $0.20$. The gross value saturates at $\delta$ almost immediately while the cost grows linearly, so the optimum is small -- a handful of signals, not dozens.

Formal Solution:

Part 1 -- Optimal decision rule.

Given $\bar{Y}_n$, choose $w$ to maximize $E[wR \mid \bar{Y}_n] = w \, E[\mu \mid \bar{Y}_n]$, using that $R$ is independent of the signals once $\mu$ is fixed. This is linear in $w$, so the optimum sits at an endpoint of $\{-1, 0, +1\}$ -- there is never a reason to stay flat once you have looked.

Under the symmetric two-point prior, the posterior odds on $\mu = +\delta$ against $\mu = -\delta$ are $\exp(2n\delta\bar{Y}_n/\tau^2)$, which exceed 1 exactly when $\bar{Y}_n > 0$. So $E[\mu \mid \bar{Y}_n]$ carries the sign of $\bar{Y}_n$, and:

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

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:

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

Now average over the two branches of the prior. At $\mu = +\delta$ this is $\delta(2\Phi(\delta\sqrt{n}/\tau) - 1)$; at $\mu = -\delta$ it is $-\delta(2\Phi(-\delta\sqrt{n}/\tau) - 1)$, which equals the same thing. The two branches agree, so the gross expected profit is

$$G(n) = \delta \cdot (2\Phi(\delta\sqrt{n}/\tau) - 1)$$

Read it directly: you make $+\delta$ whenever $\bar{Y}_n$ has the same sign as $\mu$, which happens with probability $\Phi(\delta\sqrt{n}/\tau)$, and $-\delta$ otherwise.

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 $c < c^{*}$.