Intuition What this page is forThe parent note built the formula . This page exercises it in every direction — big sensor noise, tiny sensor noise, zero prior uncertainty, a perfect sensor, correlated vector states, a real navigation word problem, and an exam twist. If a scenario can happen in the Kalman Filter — Update Step , there is a cell for it below and a worked example that lands in that cell.
Before we start, one reminder of every symbol so nobody is lost:
Definition The seven quantities in one place
P − prior covariance : how uncertain our prediction is (bigger = less sure). A single number in the scalar case; a matrix when the state is a vector.H measurement map : turns "state" into "what the sensor would read." If the sensor reads the state directly, H = 1 R sensor-noise covariance : how blurry the measurement is (bigger = blurrier).v sensor noise : the random error the sensor adds each reading, so z = H x + v E [ v ] = 0 E [ v v ⊤ ] = R R v S = H P − H ⊤ + R innovation covariance : total spread of the surprise, prior-uncertainty-mapped-into-sensor-space plus sensor noise.K = P − H ⊤ S − 1 Kalman gain , the dial from "trust prediction" to "trust measurement."P + = ( I − K H ) P − posterior covariance at the optimum (how sure we are after fusing). I
Intuition The two update formulas, once, so every example can just cite them
Fused estimate: x ^ + = x ^ − + K ( z − H x ^ − ) nudge it by the gain times the surprise. K Posterior covariance: P + = ( I − K H ) P − removes the fraction K H ( I − K H ) any gain use the Joseph form, introduced and used in Ex 10.)
Every cell below is a kind of situation the Kalman gain must handle. Each row is covered by at least one worked example.
Cell
Situation
Distinguishing feature
Example
A Balanced scalar
P − R Ex 1
B Sensor much better
R ≪ P − Ex 2
C Sensor much worse
R ≫ P − Ex 3
D Degenerate prior
P − = 0 Ex 4
E Non-unit map
H = 1 Ex 5
F Perfect sensor
R = 0 Ex 6
G Vector state, diagonal
2 × 2 P − Ex 7
H Vector state, correlated
off-diagonal P −
Ex 8
I Real navigation word problem
altitude fusion
Ex 9
J Exam twist: gain vs. trace curve
prove the U-shape numerically
Ex 10
Intuition How to read the dial figure belowFigure s01 is a number line for the gain K 0 1 K 1 0 0 is the story of how the same formula answers four opposite situations.
The Kalman gain as a slider: each scenario cell drops its computed K 0 → 1
Worked example Ex 1 — comparable uncertaintiesPrediction of a temperature: x ^ − = 20 P − = 3 z = 26 R = 3 H = 1 K x ^ + P +
Forecast: the two sources are equally trustworthy (P − = R K = 0.5 = 23
Innovation covariance. S = H P − H ⊤ + R = 1 ⋅ 3 ⋅ 1 + 3 = 6 Why this step? S Gain. K = P − H ⊤ S − 1 = 3/6 = 0.5 Why this step? K Innovation (the surprise). z − H x ^ − = 26 − 20 = 6 Why this step? This is how far the sensor disagrees with the prediction.Fused estimate. x ^ + = 20 + 0.5 ⋅ 6 = 23 Why this step? We apply x ^ + = x ^ − + K ( surprise ) 0.5 6 Posterior covariance. P + = ( 1 − K ) P − = ( 1 − 0.5 ) ⋅ 3 = 1.5 Why this step? P + = ( I − K H ) P − K H = 0.5
Verify: P + = 1.5 below both P − = 3 R = 3
Worked example Ex 2 — trust the measurementP − = 10 R = 0.1 H = 1 x ^ − = 5 z = 8
Forecast: sensor far sharper, so K 1 z = 8
S = 10 + 0.1 = 10.1 Why: total spread dominated by the shaky prior.K = 10/10.1 ≈ 0.990099 Why: prior dwarfs sensor, so we lean almost fully on the sensor.Innovation = 8 − 5 = 3 Why: how far the sensor disagrees with the prediction.
x ^ + = 5 + 0.990099 ⋅ 3 ≈ 7.970297 Why: the update x ^ − + K ( surprise ) z = 8 P + = ( 1 − 0.990099 ) ⋅ 10 ≈ 0.099010 Why: ( 1 − K H )
Verify: K ≈ 0.99 ≈ 7.97 z = 8 P + ≈ 0.099 R
Worked example Ex 3 — trust the predictionP − = 0.1 R = 10 H = 1 x ^ − = 5 z = 8
Forecast: mirror image of Ex 2. K 0 5
S = 0.1 + 10 = 10.1 K = 0.1/10.1 ≈ 0.009901 Why: tiny prior, huge sensor noise ⇒ barely nudge.Innovation = 3 Why: the sensor–prediction disagreement.
x ^ + = 5 + 0.009901 ⋅ 3 ≈ 5.029703 Why: the update adds under 1% of the surprise, so the estimate barely leaves the prediction.P + = ( 1 − 0.009901 ) ⋅ 0.1 ≈ 0.099010 Why: ( 1 − K H )
Verify: K ≈ 0.0099 ≈ 5.03 P + ≈ 0.099 P − = 0.1
Worked example Ex 4 — already certainP − = 0 zero uncertainty (a boundary case — mathematically we already "know" the state). R = 2 H = 1 z = 9
Forecast: if we're already perfectly sure, no sensor should move us. Expect K = 0 P + = 0
S = 1 ⋅ 0 ⋅ 1 + 2 = 2 Why: S K = 0 ⋅ 1/2 = 0 Why: zero prior confidence-times-nothing ⇒ zero numerator.x ^ + = x ^ − + 0 ⋅ ( z − x ^ − ) = x ^ − Why: the update multiplies the surprise by K = 0 P + = ( 1 − 0 ) ⋅ 0 = 0 Why: ( 1 − K H ) P − = 1 ⋅ 0 = 0
Verify: K = 0 P + = 0 or corrupted by a sensor — the filter correctly ignores the measurement. This is the exact mirror of the perfect-sensor case in Ex 6. ✅
Worked example Ex 5 — the sensor scales the stateOur state is a distance in metres , but the sensor reports in a scaled unit so that z = 2 x + v H = 2 v R P − = 5 R = 4 x ^ − = 3 z = 8
Forecast: because H = 2 K < 1 z
S = H P − H ⊤ + R = 2 ⋅ 5 ⋅ 2 + 4 = 24 Why: mapping the prior through H H 2 = 4 R v K = P − H ⊤ S − 1 = 5 ⋅ 2/24 = 10/24 ≈ 0.416667 Why: the H Predicted measurement H x ^ − = 2 ⋅ 3 = 6 = 8 − 6 = 2 Why: we must compare z sensor would read, not against x ^ −
x ^ + = 3 + 0.416667 ⋅ 2 ≈ 3.833333 Why: the update x ^ − + K ( surprise ) K H − 1 P + = ( 1 − K H ) P − = ( 1 − 0.416667 ⋅ 2 ) ⋅ 5 = ( 1 − 0.833333 ) ⋅ 5 ≈ 0.833333 Why: here the removed fraction is K H = 0.833 K H
Verify: note K H = 0.8 3 < 1 ( 1 − K H ) P + > 0 P + ≈ 0.833 < P − = 5
z x ^ − H = 1 Why it feels right: both are "the value we care about." The fix: the sensor lives in measurement space ; you must map the estimate forward with H z − H x ^ − z − x ^ −
Worked example Ex 6 — the sensor is exactBoundary limit opposite to Ex 4: now the sensor is flawless, R = 0 P − = 4 H = 1 x ^ − = 10 z = 13
Forecast: a flawless sensor should be believed completely. Expect K = 1 = z = 13 P + = 0
S = 1 ⋅ 4 ⋅ 1 + 0 = 4 Why: with R = 0 K = P − H ⊤ S − 1 = 4/4 = 1 Why: a noiseless sensor earns full trust — the dial hits its maximum.Innovation = z − H x ^ − = 13 − 10 = 3
x ^ + = 10 + 1 ⋅ 3 = 13 Why: the update adds the entire surprise, so the estimate lands exactly on the measurement — the prediction is fully overwritten.P + = ( 1 − K H ) P − = ( 1 − 1 ) ⋅ 4 = 0 Why: ( 1 − K H ) = 0 all prior uncertainty — an exact reading collapses our spread to zero.
Verify: K = 1 x ^ + = z = 13 P + = 0 R → 0
Worked example Ex 7 — position–velocity, uncorrelatedState is [ position velocity ] P − = [ 4 0 0 1 ] , H = [ 1 0 ] , R = 2. position only . Find the gain vector K P +
Forecast: the sensor touches position directly, so it should shrink the position variance a lot and the velocity variance not at all (they're uncorrelated here).
H P − H ⊤ = [ 1 0 ] [ 4 0 0 1 ] [ 1 0 ] = 4 Why: pick out the position variance the sensor actually sees.S = 4 + 2 = 6 P − H ⊤ = [ 4 0 0 1 ] [ 1 0 ] = [ 4 0 ] K = P − H ⊤ S − 1 = [ 4 0 ] /6 = [ 2/3 0 ] Why: the zero second entry says "this measurement can't directly correct velocity."P + = ( I − K H ) P − K H = [ 2/3 0 ] [ 1 0 ] = [ 2/3 0 0 0 ] I − K H = [ 1/3 0 0 1 ] Why: ( I − K H ) K 0 P + = [ 1/3 0 0 1 ] [ 4 0 0 1 ] = [ 4/3 0 0 1 ] Why: posterior uncertainty; the diagonal ( I − K H ) P −
Verify: position variance dropped 4 → 4/3 ≈ 1.333 1
Worked example Ex 8 — off-diagonal terms let one sensor fix two statesSame position sensor H = [ 1 0 ] R = 2 correlated :
P − = [ 4 2 2 3 ] . 2 K P +
Forecast: because the states are linked, a position measurement should also sharpen velocity — the velocity variance should drop below 3
H P − H ⊤ = [ 1 0 ] P − [ 1 0 ] = 4 Why: still the top-left entry — the position variance the sensor reads.S = 4 + 2 = 6 Why: mapped prior variance plus sensor noise, a scalar since there is one sensor.P − H ⊤ = [ 4 2 2 3 ] [ 1 0 ] = [ 4 2 ] Why: the second entry is now 2 0 — the correlation makes velocity respond to a position reading.K = P − H ⊤ S − 1 = [ 4 2 ] /6 = [ 2/3 1/3 ] Why: a nonzero velocity gain — the filter will correct velocity via the shared uncertainty. Here S − 1 = 1/6 S 1 × 1 K H = [ 2/3 1/3 ] [ 1 0 ] = [ 2/3 1/3 0 0 ] I − K H = [ 1/3 − 1/3 0 1 ] Why: ( I − K H ) − 1/3 bleeds into velocity.P + = ( I − K H ) P − = [ 1/3 − 1/3 0 1 ] [ 4 2 2 3 ] = [ 4/3 2/3 2/3 7/3 ] Why: posterior covariance; the correlation channelled information from position into velocity, so both diagonal entries shrink.
Verify: velocity variance dropped 3 → 7/3 ≈ 2.333 strictly below the 3 two states. Position variance also dropped 4 → 4/3
Intuition The lesson of Ex 7 vs Ex 8With a diagonal prior a position sensor helps only position. With correlation , the shared uncertainty acts like a hidden wire: measuring one end tightens the other. This is why real GNC filters (Sensor Fusion in Navigation ) keep full covariance matrices, not just diagonals.
Worked example Ex 9 — fusing barometer altitude into a GNC estimateA drone's navigation filter predicts altitude x ^ − = 100 m P − = 9 m 2 ± 3 z = 106 m R = 16 m 2 ± 4 H = 1
Forecast: the prediction (± 3 ± 4 K < 0.5 100 106
S = 9 + 16 = 25 m 2 Why: add the two uncertainties in the same space.K = 9/25 = 0.36 Why: prediction contributes 9/25 Innovation = 106 − 100 = 6 m Why: the barometer–prediction disagreement.
x ^ + = 100 + 0.36 ⋅ 6 = 102.16 m Why: the update nudges the predicted altitude by 36% 6 P + = ( 1 − 0.36 ) ⋅ 9 = 5.76 m 2 ± 2.4 Why: ( 1 − K H ) 64%
Verify: units: K P + m 2 102.16 100 106 ± 2.4 both ± 3 ± 4
Intuition How to read the U-valley figure belowFigure s02 plots the posterior trace tr ( P + ) curve against the gain K every K K is the Kalman gain. Sample points at K = 0.6 K = 1.0
Definition The Joseph form (general posterior covariance)The formula P + = ( I − K H ) P − optimal gain. For an arbitrary gain K Joseph form
P + = ( I − K H ) P − ( I − K H ) ⊤ + K R K ⊤ . K R K ⊤ H = 1 P + = ( 1 − K ) 2 P − + K 2 R K
Worked example Ex 10 — is the Kalman gain really the minimum?With P − = 4 H = 1 R = 1 K ⋆ = 0.8 tr ( P + ) Joseph form at K = 0.6 K = 0.8 K = 1.0
Forecast: the trace is a parabola in K K = 0.8
Joseph form (valid for any K P + = ( 1 − K ) 2 P − + K 2 R = 4 ( 1 − K ) 2 + K 2 Why this step? below the optimum we cannot use P + = ( 1 − K ) P − K ⋆
K = 0.6 4 ( 0.4 ) 2 + ( 0.6 ) 2 = 4 ⋅ 0.16 + 0.36 = 0.64 + 0.36 = 1.00 Why: evaluate on the left wall of the valley.K = 0.8 4 ( 0.2 ) 2 + ( 0.8 ) 2 = 4 ⋅ 0.04 + 0.64 = 0.16 + 0.64 = 0.80 Why: evaluate at the predicted optimum K ⋆ = 0.8 K = 1.0 4 ( 0 ) 2 + ( 1 ) 2 = 0 + 1 = 1.00 Why: evaluate on the right wall (full trust in sensor).
Verify: values 1.00 , 0.80 , 1.00 K = 0.8 0.80 P + K ⋆ ( 1 − K ) P − = 0.2 ⋅ 4 = 0.8
The trace of P + K
Line the ten cells up as one sentence: balanced → sensor-wins → prediction-wins → already-sure → rescale → perfect-sensor → vectorize → correlate → fly it → check the valley. Same formula K = P − H ⊤ S − 1
Recall Quick self-testIn Ex 1, why did K 0.5 P − = R P − = 0 P − H ⊤ = 0 K = 0 v = z − H x ^ − = 8 − 2 ⋅ 3 = 2 mapped prediction; v R P + = 0 R = 0 K = 1 ( 1 − K H ) = 0 K P + = ( 1 − K ) P − K = 0.6 K ( 1 − K ) 2 P − + K 2 R
Intuition What the flowchart saysThe chain below is the reading order of this page: each box is one scenario cell, and the arrows mean "next harder case." It shows the deliberate escalation — start with a balanced scalar, push the dial to each extreme, hit the two degenerate boundaries, then climb into vectors (first uncorrelated, then correlated), apply it to a real flight, and finally prove the optimum. Following the arrows left to right walks you through every example on this page in the order it builds understanding.
