What's semi-partial correlation, in one sentence?

The semi-partial (or 'part') correlation r_Y(X-Z) is the Pearson r between the raw Y and the residual of X after regressing X on Z — algebraically r_Y(X-Z) = (r_XY − r_XZ·r_YZ) / sqrt(1 − r_XZ²), which shares the numerator with partial correlation but divides by only one sqrt(1 − r²) factor instead of two.

How does semi-partial differ from partial correlation?

Partial correlation removes Z from BOTH X and Y, then correlates the two residuals; semi-partial removes Z from X only and correlates that residual with the raw Y. Partial answers 'what's the correlation of X and Y within strata of Z?' (a within-strata reading). Semi-partial answers 'how much RAW Y-variance is explained by the part of X that's orthogonal to Z?' (an incremental-fit reading). The two agree in sign (same numerator) but differ in magnitude — semi-partial is always closer to zero when |r_YZ| > 0, because the denominator sqrt(1 − r_XZ²) is bigger than sqrt((1−r_XZ²)(1−r_YZ²)).

Why is semi-partial the right stat when you're asking 'does X help predict Y beyond what Z already explains?'

Because r_Y(X-Z)² equals ΔR² exactly. If R²(Y ~ Z) is your baseline and R²(Y ~ X + Z) is the full model, then R²(full) − R²(baseline) = r_Y(X-Z)². Partial correlation squared does NOT equal ΔR² — it equals a different quantity, sometimes called the coefficient of partial determination, which measures 'what fraction of RESIDUAL Y-variance is explained by the residual of X'. The semi-partial version is the one that maps to raw R² gains in a regression comparison — so it's the version that answers 'is this predictor pulling its weight in the full model?'

Walk me through today's worked example.

Sample: CAD Retail Sales m/m × CADJPY 15m, n=197. Encode buckets as ordinal Z: big_miss=−2, small_miss=−1, in_line=0, small_beat=+1, big_beat=+2. Compute three Pearson r's: r(X=surprise_z, Y=move_pips)=+0.2095; r(X,Z)=+0.6488; r(Y,Z)=+0.4195. Numerator: r_XY − r_XZ·r_YZ = 0.2095 − 0.6488·0.4195 = 0.2095 − 0.2722 = −0.0627. Semi-partial denominator: sqrt(1 − 0.6488²) = sqrt(0.5791) = 0.7610. r_Y(X-Z) = −0.0627 / 0.7610 = −0.0823. Squared: 0.00678, or 0.68 percent. Cross-check by running two regressions: R²(Y ~ Z_ord) = 0.1760; R²(Y ~ X + Z_ord) = 0.1827; ΔR² = 0.0067 — matches the semi-partial squared to 3 decimals. Raw surprise_z, when added to bucket_ord as a second predictor, buys you 0.68 percentage points of variance explanation. Not much.

Why is r(X, Z_ord) only 0.65 here, when Stats #23's version got 0.95?

One release does it: 2020-06-19, the Canada Retail Sales COVID Q2 print with actual −26.4% MoM vs consensus −15.0%, surprise_z = −20.42. It's a big_miss (bucket_ord = −2) with a z that's 10x larger than any other big_miss in the sample. Ordinal bucket encoding treats it as identical to a big_miss at z=−1.6, but the raw z carries 10x the magnitude. That single outlier drives r(X, Z_ord) from around 0.90 (what you'd expect for a normal 5-quantile sample) down to 0.65. Drop it and n=196 gives r(X, Z_ord) = 0.9026, matching Stats #23's USDCAD sample structure. The semi-partial collapses to +0.013 (0.02pp of ΔR²) when the outlier is dropped — the leverage of that single point was actually PROPPING UP a mild negative semi-partial in the full sample.

So which number should I quote — with or without the outlier?

Both, honestly labeled. The reported semi-partial with n=197 is −0.082 (0.68 pp of ΔR²); without the 2020-06-19 COVID Q2 print it's +0.013 (0.02 pp). Both are indistinguishable from zero at 95% (Fisher CIs cross). The pedagogic point is the same either way — surprise magnitude beyond bucket assignment adds essentially no unique R² to Y. The COVID outlier is a real data-generating-process observation, not a data-entry error, so the default reported number keeps it in. But the sensitivity check is important: on a leverage-heavy sample, a single outlier can flip the sign of the semi-partial while leaving the qualitative reading unchanged.

How does the categorical (4-dummy) semi-partial compare?

R²(Y ~ Z_cat) with in_line as reference = 0.1997 (a bit higher than the 0.1760 of Z_ord because the categorical fully absorbs the tail-bucket asymmetry that the linear ordinal misses). R²(Y ~ X + Z_cat) = 0.2022. ΔR² = 0.00252 → semi-partial² = 0.00252 → semi-partial = −0.0502. So under a full 4-dummy control, X adds 0.25 percentage points of R² on top of Z_cat. Even smaller than the ordinal-based 0.68 pp, and for the right reason: Z_cat is a stronger baseline (5 bucket means fitted vs 1 slope), leaving less unique variance for X to explain. As always, the categorical version is the more conservative choice for asking 'does X buy me anything beyond bucket?'.

What's the practical takeaway for trading?

Semi-partial nails down what Stats #23 hinted at — surprise magnitude within a bucket is not a tradable edge over just knowing the bucket. If you're building a model that already uses the 5-bucket assignment, adding raw surprise_z as a second predictor buys you well under 1% additional R² on this sample. That 1% is inside the standard error of the R² estimate itself. Trade the bucket, not the fine-grained z inside it. This mirrors and extends the Stats #23 partial-correlation finding on USDCAD (that partial r collapsed to −0.05) with a more direct ΔR² interpretation. The bucket boundaries ARE the granularity of this event's signal.

Can I reproduce this in the tool?

Not directly — Vantage's News Impact Explorer doesn't expose ΔR² or semi-partial columns. But you can recompute: pull /api/v1/news-impact/releases?event=FF:CAD_RETAIL_MOM&instrument=CADJPY&window=15m&limit=500 and compute the three Pearson r's, then plug into the semi-partial formula. All numbers cross-verified against Python numpy corrcoef and sklearn LinearRegression on 2026-08-25.

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Statistics25 August 2026· 6 min read

Statistics for Traders #25: Semi-partial (part) correlation on today's slot-1 CAD Retail × CADJPY sample — raw surprise_z adds only 0.7 percentage points of R² over the 5-bucket assignment (0.02 pp if the COVID outlier drops)

Semi-partial (or “part”) correlation is the sibling of Stats #23’s partial correlation: partial removes Z from BOTH X and Y, semi-partial removes Z from X ONLY. Its distinguishing property: r_Y(X-Z)² is exactly the ΔR² added by X over Z when both predict Y. On today’s slot-1 CAD Retail × CADJPY 15m sample (n=197), semi-partial r_Y(X-Z_ord) = -0.082, squared 0.68 percent. Bucket_ord already explains 17.6% of move-pips variance; adding raw surprise_z lifts the two-predictor R² by 0.68 percentage points.

Drop the 2020-06-19 COVID Q2 print (z = -20.42, +6.6p move) and the semi-partial collapses further to +0.013, ΔR² = 0.02pp. Buckets carry the signal on this event; surprise magnitude beyond bucket adds essentially nothing.

Horizontal-bar chart with Fisher 95% CI whiskers. Top row: 'r(X,Y) overall' = +0.209 (or +0.392 without outlier) — solid positive Pearson r drawn in dark. Middle rows: 'Partial r_XY|Z_ord (n=197)' = -0.091, CI [-0.228, +0.050]; 'Semi-partial r_Y(X-Z_ord) (n=197)' = -0.082, CI [-0.220, +0.058]; 'Partial r_XY|Z_ord (n=196, no COVID)' = +0.014; 'Semi-partial r_Y(X-Z_ord) (n=196, no COVID)' = +0.013 — all coral, all crossing the r=0 reference line. Bottom rows: ΔR² decomposition of Y ~ X + Z_ord: R²(Y~Z_ord) 17.60%, R²(Y~X) 4.39%, R²(Y~X+Z_ord) 18.27% — the ΔR² added by X over Z_ord is 0.68 percentage points (matches semi-partial squared exactly). Coral vertical line at r=0 makes the semi-partial-crosses-zero result visible at a glance. — r(X,Y) vs partial r vs semi-partial r on the CAD Retail × CADJPY sample — with and without the COVID Q2 outlier. All partial/semi-partial CIs cross zero.

The formula and how it differs from partial

Both partial and semi-partial correlations use the same numerator:

Numerator = r_XY − r_XZ · r_YZ

They differ in denominator only:

Statistic	Denominator	Reading
Partial r_XY\|Z	sqrt((1 − r_XZ²)(1 − r_YZ²))	correlation of residuals; within-strata reading
Semi-partial r_Y(X-Z)	sqrt(1 − r_XZ²)	correlation of X-residual with raw Y; ΔR² reading

Since sqrt(1 − r_XZ²) > sqrt((1 − r_XZ²)(1 − r_YZ²)) whenever |r_YZ| > 0, semi-partial is always closer to zero (smaller in magnitude) than partial. They agree in sign because the numerator is shared. Both collapse to r_XY when Z is uncorrelated with both X and Y.

The key property: semi-partial² is ΔR²

If you regress Y on Z alone and then on both X and Z:

r_Y(X-Z)² = R²(Y ~ X + Z) − R²(Y ~ Z)

That’s the reason semi-partial is the right statistic for the question “does adding X to my model actually help, given I already have Z?”. It maps directly to raw variance-explained. Partial correlation squared has a different interpretation — it’s the coefficient of partial determination, the fraction of RESIDUAL Y-variance (after subtracting Z’s contribution) explained by the residual of X. Semi-partial keeps the full Y-variance denominator; that’s what makes it the ΔR² sibling.

Worked example on today’s slot-1 sample

n	197
r(X = surprise_z, Y = move_pips)	+0.2095
r(X, Z_ord)	+0.6488
r(Y, Z_ord)	+0.4195
r_XY − r_XZ·r_YZ (numerator)	−0.0627
sqrt(1 − 0.6488²) (semi-partial denom)	0.7610
Semi-partial r_Y(X-Z_ord)	−0.0822
Semi-partial squared = ΔR²	0.00676 (0.68 pp)
Fisher 95% CI (dof = n − 1 − 2 = 194)	[−0.220, +0.058]

Cross-check via two regressions (sklearn LinearRegression):

R²(Y ~ X only)	0.0439
R²(Y ~ Z_ord only)	0.1760
R²(Y ~ X + Z_ord)	0.1827
ΔR² of X over Z_ord	0.00676

The ΔR² matches the semi-partial squared to 5 decimals. That’s the identity — semi-partial correlation IS the square root of ΔR² (sign preserved from the numerator).

The COVID Q2 outlier and its sensitivity

The r(X, Z_ord) = 0.6488here is meaningfully lower than the ~0.90 you’d expect for a 5-quantile-based bucket variable. The 2026-06-19 COVID Q2 print explains it: actual Canada Retail Sales −26.4% MoM vs consensus −15.0%, surprise_z = −20.42. That’s 10x the magnitude of any other big_miss in the sample. Ordinal bucket encoding assigns it Z = −2, identical to a big_miss at z = −1.6, but the raw z carries 10x the leverage. That single point drags r(X, Z_ord) down by 25 percentage points.

Drop it and redo:

Metric	n = 197 (with)	n = 196 (without)
r(X, Y)	+0.210	+0.392
r(X, Z_ord)	+0.649	+0.903
r(Y, Z_ord)	+0.420	+0.429
Semi-partial r_Y(X-Z_ord)	−0.082	+0.013
Semi-partial² = ΔR²	0.68 pp	0.02 pp

Two features of the sensitivity check: (1) the sign of the semi- partial flips from negative (with COVID) to positive (without), and (2) the magnitude collapses toward zero either way — the pedagogic point survives. On a leverage-heavy sample, a single outlier can flip semi-partial’s sign while leaving the qualitative reading (“X adds ~zero unique variance beyond Z”) unchanged. Both Fisher CIs cross zero.

The categorical (4-dummy) semi-partial

Replace Z_ord (a single ordinal) with Z_cat (four dummies for the five buckets, with in_line as reference), and the same machinery gives ΔR² over a stronger baseline:

R²(Y ~ Z_cat)	0.1997
R²(Y ~ X + Z_cat)	0.2022
ΔR² of X over Z_cat	0.00252 (0.25 pp)
Semi-partial r_Y(X-Z_cat)	-0.0502

Even smaller than the ordinal version (0.25 pp vs 0.68 pp), for the right reason: Z_cat is a stronger baseline (5 bucket means fitted vs 1 slope), leaving less unique variance for X to explain. As always, the categorical is the more conservative choice for asking “does X buy me anything beyond bucket?”.

Cross-links to Stats #23 and Stats #24

Stats #23 introduced partial correlation on the CAD CPI × USDCAD sample and got a partial r of −0.05 (Fisher CI crosses zero). The semi-partial on THAT same sample would be r_Y(X-Z_ord) = (−0.4256 − 0.9488·(−0.4327)) / sqrt(1 − 0.9488²) = −0.0151 / 0.3162 = −0.0478, squared 0.23 pp — telling the same story (X adds ~zero unique variance beyond Z) via a slightly different denominator. The two statistics are close in absolute value here because |r_YZ| = 0.43 makes the two denominators similar in size.

Stats #24 introduced one-way ANOVA and R² on the same USDCAD sample and got η² = R²(Y ~ Z_cat) = 0.1952. Adding X to that model would give ΔR² of exactly the categorical semi-partial squared for that sample. The three installments are three views of the same underlying decomposition: partial (Stats #23) answers “is there a within-strata correlation?”; ANOVA / η² (Stats #24) answers “what fraction of Y-variance is between vs within?”; today’s semi-partial answers “does the fine-grained predictor buy me anything the coarse one doesn’t?”.

Verification note

All r values computed in a scratch Python script against the 197-release sample from /api/v1/news-impact/releases?event=FF:CAD_RETAIL_MOM&instrument=CADJPY&window=15m&limit=500 on 2026-08-25. Cross-verified against sklearn.linear_model.LinearRegression for the ΔR² identity (semi-partial squared = ΔR² of X over Z), matching to 5 decimal places. Chart written via a one-off script reusing scripts/insights-charts/svg.ts and theme.ts primitives; not committed under scripts/ since the horizontal-bar-with-whiskers layout is single-use.

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