The Lindeberg-Feller Central Limit Theorem

A trading desk's daily P&L is the sum of many positions: a large S&P future, a handful of single-name options, a small basis trade. The ordinary Central Limit Theorem promises that a sum of many identical, independent shocks looks normal. But these positions are nothing alike in size, one line can be ten times the risk of another. Does the sum still settle into a bell curve, or can one oversized position wreck the average for everyone else?

The analogy: one loud voice in a choir

Imagine averaging the volumes of a hundred singers to predict the chorus's overall loudness. If everyone sings at roughly the same volume, the average is stable and predictable, that's the ordinary CLT. Now put one singer with a bullhorn in the back row. However quiet the other ninety-nine are, the bullhorn dominates the total, and the sum no longer behaves like an average of "typical" contributors, it behaves like that one voice. Lindeberg-Feller is the precise rule for how loud any single singer is allowed to get, relative to the whole choir, before this happens.

Writing it down

Let $X_1, X_2, \ldots, X_n$ be independent (not necessarily identical) random variables, each with mean zero and variance $\sigma_i^2$. Write $s_n^2 = \sum_{i=1}^n \sigma_i^2$ for the total variance of the sum. The Lindeberg condition asks that for every $\epsilon > 0$,

In words: look at the fraction of total variance coming from pieces large enough to be a noticeable slice of the whole sum's spread, and check that fraction shrinks to nothing as you add more pieces. No single term (or small cluster of terms) is allowed to keep contributing a fixed share of the total variance forever. If this holds, then

In words: the properly rescaled sum converges to a standard normal, even though the pieces have different sizes and different distributions. The ordinary CLT is the special case where all $\sigma_i$ are equal, there, no single term can ever be a large share of the total, so the condition is automatic.

Worked example 1: a diversified book

Ten independent positions, each with variance $\sigma_i^2 = 4$ (so $s_n^2 = 40$ after ten). The largest single variance is 4, only $4/40 = 10\%$ of the total, and that share shrinks toward zero as you add more similar positions. Lindeberg's condition is satisfied in the limit, and the standardized P&L $S_n / \sqrt{40}$ is well approximated by $N(0,1)$. If each position also has bounded, roughly symmetric outcomes (no lottery-ticket payoffs), 40–50 such positions already give a serviceable normal approximation for a 95% VaR estimate.

Worked example 2: one dominant name

Now replace one small position with a concentrated bet: nine positions with $\sigma_i^2 = 1$ each (total 9) and one large position with $\sigma_i^2 = 91$, so $s_n^2 = 100$. That one term supplies $91/100 = 91\%$ of total variance, nowhere near shrinking to zero, no matter how many small positions you bolt on around it. Lindeberg fails. The rescaled sum $S_n / 10$ does not converge to $N(0,1)$; instead it inherits the shape of the dominant position's own distribution. If that position's return is skewed (say, a short-dated option near expiry with a fat left tail), the "portfolio P&L is roughly normal" assumption a risk model quietly relies on is simply false, no matter how many other diversifying trades sit around it.

Run several batches of paths with very different starting notionals. Notice how a single oversized path can visibly drag the average path away from the smooth, symmetric spread the other paths would produce on their own, a live picture of one term dominating the sum.

What this means in practice

• Portfolio risk aggregation. A VaR model that sums many position P&Ls and calls the total normal is implicitly assuming Lindeberg holds. Concentrated books, one massive futures line plus small satellite trades, are exactly where that assumption breaks.
• Position limits. Capping any single position's variance at, say, 10–15% of book variance is a practical way to keep Lindeberg's condition roughly true, and keep the normal approximation trustworthy.
• Stress testing. When Lindeberg fails, don't approximate the tail with a normal quantile, simulate, or model the dominant term's own distribution directly (see The Pareto Distribution and Power Laws for what a dominant fat-tailed term can do to the sum).