Profit-Aware Classification Threshold

Worked Solution

How to Think About It: This is a decision theory problem dressed up as trading. The standard classification threshold of $0.5$ minimizes misclassification error, but that is not what we care about -- we care about expected profit. The asymmetry comes from the cost $c$: every time you trade, you pay $c$ regardless of whether the return is positive or negative. So you need the expected return conditional on trading to exceed $c$, not just be positive. This pushes the optimal threshold above $0.5$.

Quick Estimate: If $c = 0$, you should trade whenever $\hat{p}_t > 0.5$ (any positive expected return is worth taking for free). If $c$ is large, you should be more selective -- only trade when you are very confident the return is positive. So $\theta^{*}$ increases with $c$. In the symmetric case where up and down moves have equal magnitude $\mu$, the break-even condition is roughly $\theta^{*} \cdot \mu - c = 0$, giving $\theta^{*} \approx c / \mu + 0.5$.

Approach: Write the expected profit as a function of $\theta$, take the derivative with respect to $\theta$, and find the optimum.

Formal Solution:

**(i) Deriving $\theta^{*}$:**

The expected profit per period is: $\Pi(\theta) = E\left[(r_{t+1} - c) \cdot \mathbf{1}(\hat{p}_t > \theta)\right]$

Using the law of iterated expectations, condition on $\hat{p}_t$: $\Pi(\theta) = E\left[E[r_{t+1} - c \mid \hat{p}_t] \cdot \mathbf{1}(\hat{p}_t > \theta)\right]$

Let $\mu(p) = E[r_{t+1} \mid \hat{p}_t = p]$ denote the expected return conditional on the model's predicted probability. Then: $\Pi(\theta) = \int_{\theta}^{1} (\mu(p) - c) \, f(p) \, dp$

where $f(p)$ is the density of $\hat{p}_t$.

Differentiating with respect to $\theta$ using Leibniz's rule: $\frac{d\Pi}{d\theta} = -(\mu(\theta) - c) \cdot f(\theta)$

Setting this to zero (and assuming $f(\theta) > 0$): $\mu(\theta^{*}) = c$

The optimal threshold $\theta^{*}$ is the value of $\hat{p}_t$ at which the conditional expected return equals the trading cost.

To make this more explicit, if we assume that the return conditional on $\hat{p}_t = p$ can be decomposed as: $E[r_{t+1} \mid \hat{p}_t = p] = p \cdot E[r_{t+1} \mid r_{t+1} > 0, \hat{p}_t = p] + (1-p) \cdot E[r_{t+1} \mid r_{t+1} \leq 0, \hat{p}_t = p]$

Denote the average up-move as $\mu^+$ and the average down-move magnitude as $\mu^-$ (both positive). In the simplest case where these are constant across $p$: $\mu(p) = p \cdot \mu^+ - (1-p) \cdot \mu^-$

Setting $\mu(\theta^{*}) = c$: $\theta^{*} \cdot \mu^+ - (1 - \theta^{*}) \cdot \mu^- = c$ $\theta^{*}(\mu^+ + \mu^-) = c + \mu^-$ $\theta^{*} = \frac{c + \mu^-}{\mu^+ + \mu^-}$

When $c = 0$ and $\mu^+ = \mu^-$: $\theta^{*} = 0.5$. When $c > 0$: $\theta^{*} > 0.5$.

**(ii) Why $\theta^{*} \neq 0.5$:**

The standard 0.5 threshold minimizes classification error -- it treats false positives and false negatives symmetrically. But in a trading context, the costs are asymmetric:

• False positive (predict up, market goes down): you lose the negative return AND pay cost $c$. Total loss: $|r_{t+1}| + c$.
• False negative (predict up but don't trade): you miss a gain, but you lose nothing. Total loss: $0$ (opportunity cost only).

Since false positives are more costly than false negatives (you pay $c$ every time you trade), you should demand a higher probability of being right before trading. This pushes $\theta^{*}$ above $0.5$.

Additionally, if down moves are larger than up moves on average ($\mu^- > \mu^+$, which is typical for equities -- "stairs up, elevator down"), the threshold shifts further above $0.5$ even without costs.

Answer: The optimal threshold is $\theta^{*} = \frac{c + \mu^-}{\mu^+ + \mu^-}$, where $\mu^+$ is the average up-move and $\mu^-$ is the average down-move magnitude. More generally, $\theta^{*}$ is the value where $E[r_{t+1} \mid \hat{p}_t = \theta^{*}] = c$. The threshold exceeds $0.5$ because trading costs make false positives more expensive than false negatives, so you need higher confidence before acting.