Ratio and Regression Estimators in Surveys

You want to estimate a portfolio's total current market value by sampling 200 of its 5,000 positions and marking them individually, because pricing every position by hand is too slow. A plain average-times-count estimate ignores something you already have for free: last quarter's exact value for every position, recorded in the books. If a position's value tends to move roughly in proportion to its old value, that fully-known old number is a powerful hint about the new one you haven't sampled. Ratio and regression estimators are the two standard ways to fold that free auxiliary information into your estimate, tightening it without collecting a single extra sample.

An analogy: guessing weight from a scale you trust

Suppose you want ten people's current weights, but your scale is broken and you can only weigh three directly. You do have last year's weigh-in for all ten. If this year's weights move roughly proportionally to last year's — someone heavy stays relatively heavy — you can scale everyone's known old weight by the average change observed in your three-person sample, and get a decent guess for all ten. That's the ratio estimator: use a known, correlated auxiliary quantity to extrapolate what you didn't directly measure.

The formulas, one piece at a time

Let $y$ be the quantity you want (current value), and $x$ be an auxiliary variable whose population total $X$ you already know exactly (last quarter's value, summed over all 5,000 positions). From your sample, compute $\bar y$ and $\bar x$, the sample averages of $y$ and $x$. The ratio estimator of the population total is

$$\hat Y_{\text{ratio}} = \frac{\bar y}{\bar x} \cdot X.$$

In plain English: estimate the average ratio of new value to old value from your sample, then apply that same ratio to the known, exact old total for the whole population. This works well when $y$ is roughly proportional to $x$ — the ratio $y/x$ stays fairly constant across positions — and it typically has much lower variance than simply averaging $y$ alone, because the noisy part of $y$ that's explained by $x$ gets cancelled out.

The regression estimator generalizes this by not forcing the relationship through the origin:

$$\hat Y_{\text{reg}} = \bar y + b(X/N - \bar x),$$

where $b$ is the sample regression slope of $y$ on $x$ and $N$ is the population size. In plain English: start from the sample average of $y$, then nudge it up or down based on how far your sample's average $x$ was from the population's true average $x$, scaled by how strongly $y$ and $x$ move together. When the true relationship between $y$ and $x$ has a nonzero intercept, the regression estimator is more efficient than the ratio estimator; when the relationship genuinely passes through zero (a position with no old value has no new value), the ratio estimator does about as well and is simpler.

Worked example 1: the ratio estimator

The known population total of last quarter's value is $X = 48{,}000{,}000$ ($48 million) across 5,000 positions. In a sample of 200 positions, the average current value is $\bar y = 11{,}200$ and the average old value is $\bar x = 9{,}600$ (both in dollars). The ratio estimator gives

$$\hat Y_{\text{ratio}} = \frac{11{,}200}{9{,}600} \times 48{,}000{,}000 \approx 1.1667 \times 48{,}000{,}000 \approx 56{,}000{,}000,$$

i.e. about $56 million. The 200 sampled positions grew about 16.7% on average, and applying that same growth rate to the fully-known $48 million old total gives a full-portfolio estimate without pricing all 5,000 positions.

Worked example 2: the regression estimator

Suppose the same sample also shows a regression slope of $y$ on $x$ of $b = 1.05$ (each extra dollar of old value predicts about $1.05 of new value, not a perfectly proportional relationship), the population average old value is $X/N = 48{,}000{,}000/5{,}000 = 9{,}600$ dollars, and it happens the sample's $\bar x = 9{,}600$ matches the population average exactly. Then

$$\hat Y_{\text{reg}} = \bar y + b(9{,}600 - 9{,}600) = \bar y = 11{,}200 \text{ per position, i.e. } 56{,}000{,}000 \text{ total},$$

or $11,200 per position and $56 million overall.

Here the correction term vanishes because the sample happened to be representative on $x$; if instead the 200-position sample had over-sampled larger old positions, say $\bar x = 10{,}000$ ($10,000), the regression estimator would pull the estimate down by $1.05 \times (9{,}600 - 10{,}000) = -420$, i.e. $420 less per position — a correction the plain sample average could never make, since it has no way to know the sample's $x$-values were unrepresentative.

Drag the points above: the tighter the fit, the closer $b$ is to the true slope, and the more precision the auxiliary variable buys you.

What this means in practice

These estimators cut sampling cost anywhere a cheap, fully-known auxiliary variable correlates with an expensive-to-measure target: valuing a large book from a marked subsample, auditing inventory from a fraction of SKUs, or projecting revenue from a partial sample scaled by known totals. The gain is real but conditional — it only helps to the extent $x$ and $y$ are actually correlated.