Simpson's Paradox with Poisson Scoring

Worked Solution

How to Think About It: Simpson's paradox says you can win every battle but lose the war. The question is whether you can pick Poisson rates so that A outscores B in each half (with probability exceeding 0.5) yet B wins the overall game. Because independent Poisson variables are closed under addition, the total scores are themselves Poisson, and the pairwise win probability is monotonic in the rate parameter. This creates a fundamental constraint that makes the strict version of the paradox impossible under these assumptions -- but the conceptual lesson is exactly the point of the problem.

Quick Estimate: For independent Poisson variables, $P(X > Y) > P(Y > X)$ if and only if $\lambda_X > \lambda_Y$. If A has a higher rate in both halves, then $\lambda_{A,1} + \lambda_{A,2} > \lambda_{B,1} + \lambda_{B,2}$, so A also dominates in total. There is no room for the rates to "reverse" upon aggregation the way proportions can in the classical Simpson's paradox.

Formal Solution:

Define the pairwise win probability: $$P(X > Y) = \sum_{k=0}^{\infty} \sum_{j=0}^{k-1} \frac{e^{-(\lambda_X + \lambda_Y)} \lambda_X^{k} \lambda_Y^{j}}{k! \, j!}$$ where $X \sim \text{Poisson}(\lambda_X)$ and $Y \sim \text{Poisson}(\lambda_Y)$ are independent.

Claim: For independent Poisson variables, $P(X > Y) > P(Y > X)$ if and only if $\lambda_X > \lambda_Y$.

Proof sketch: The difference $X - Y$ has a Skellam distribution with parameters $(\lambda_X, \lambda_Y)$. Its probability mass function is symmetric about zero when $\lambda_X = \lambda_Y$, and shifts rightward as $\lambda_X / \lambda_Y$ increases. More precisely, the distribution satisfies $P(X - Y = k) / P(X - Y = -k) = (\lambda_X / \lambda_Y)^{k}$ for $k > 0$, so the ratio exceeds 1 for every positive $k$ if and only if $\lambda_X > \lambda_Y$. Summing over all $k > 0$ gives $P(X > Y) > P(Y > X)$.

Why strict Simpson's paradox fails here: If $\lambda_{A,j} > \lambda_{B,j}$ for $j = 1, 2$ (required for A to win each half), then $\lambda_{A,1} + \lambda_{A,2} > \lambda_{B,1} + \lambda_{B,2}$. Since totals $X_{A,1} + X_{A,2} \sim \text{Poisson}(\lambda_{A,1} + \lambda_{A,2})$ and $X_{B,1} + X_{B,2} \sim \text{Poisson}(\lambda_{B,1} + \lambda_{B,2})$ by closure under addition, the monotonicity result implies A also wins the total game with higher probability. The additive structure of both the rates and the Poisson family prevents the reversal that drives Simpson's paradox in non-additive settings (like contingency tables with proportions).

Conceptual construction (the pedagogical point):

Simpson's paradox does arise if you measure performance by a non-additive metric -- for instance, win rate per half vs. expected total margin. Consider: - Half 1: $\lambda_{A,1} = 2, \lambda_{B,1} = 1$ -- A wins this half about 56% of the time, but the expected margin is only $+1$. - Half 2: $\lambda_{A,2} = 12, \lambda_{B,2} = 11$ -- A wins this half about 52% of the time, expected margin $+1$.

A wins both halves more often than B. But now imagine evaluating by a magnitude-weighted metric: when B wins a half, B tends to win by a larger margin in the high-intensity half (due to higher variance, $\text{Var} = \lambda$). The Sharpe-like ratio of A's edge relative to the noise is $1/\sqrt{2 + 1} \approx 0.58$ in half 1 and $1/\sqrt{12 + 11} \approx 0.21$ in half 2. A's per-half win probability is positive, but the signal degrades as intensity grows.

Trading Interpretation:

This is the core lesson for evaluating strategies. A strategy can have a higher conditional win rate in every regime (trending, mean-reverting, high-vol, low-vol) yet underperform on aggregate P\&L if:

1. It wins often but small in the regime that dominates calendar time.
2. It loses rarely but loses big when it does lose.
3. The regime mix -- how much time is spent in each regime -- weights the losses more heavily than the wins.

This is why quant desks decompose performance by regime AND magnitude, not just win rate. A 60% hit rate means nothing if the average loss is three times the average win. Conditional metrics (win rate per regime) and aggregate metrics (total Sharpe, total P\&L) can tell opposite stories. The fix is to always pair hit rate with payoff ratio and to weight conditional statistics by regime frequency before drawing conclusions.

Answer: A strict Simpson's paradox (A wins both halves with $P > 0.5$ but loses the total game with $P > 0.5$) is impossible with independent Poisson scoring, because Poisson is closed under addition and the pairwise win probability is monotonic in the rate. The paradox does emerge when conditional win rates are compared against magnitude-weighted aggregate metrics -- exactly the trap that arises in evaluating trading strategies across market regimes.