Kalman gain — minimizes trace of covariance: Exercises

Before we start, here is the toolbox, all in one place, so no symbol appears unexplained.

Here, $\overset{x}{^}^{-}$ is our prediction and $P^{-}$ its uncertainty (variance); $z$ is the sensor reading; $H$ turns a state into what the sensor should read; $R$ is the sensor noise variance; $K$ is the gain, the dial in $[0, 1]$ (in the scalar case) telling us how far to slide from prediction toward measurement.

The figure below makes the "slider" idea concrete using the numbers from Problem 2.2: a prediction at $10$ , a measurement at $15$ , and the fused estimate landing at $14$ because the gain is $K = 0.8$ . Look at how the orange diamond sits between the two sources, pulled toward the measurement.

Figure — Kalman gain — minimizes trace of covariance

L1 — Recognition

Problem 1.1

In the scalar update, which of these is the innovation, and which is the innovation covariance? (a) $z - H \overset{x}{^}^{-}$ (b) $H P^{-} H^{⊤} + R$ (c) $P^{-} H^{⊤} S^{- 1}$

Recall Solution 1.1

WHAT: match names to formulas.

(a) $z - H \overset{x}{^}^{-}$ is the innovation $y$ : measurement minus what we expected to measure.
(b) $H P^{-} H^{⊤} + R$ is the innovation covariance $S$ : the prior uncertainty pushed into measurement space, plus sensor noise.
(c) $P^{-} H^{⊤} S^{- 1}$ is the gain $K$ — not asked, but good to recognise. WHY it matters: $y$ tells you how surprised you are; $S$ tells you how surprising a surprise of that size is. You always divide the first by the second (via $K$ ).

Problem 1.2

A scalar filter has $P^{-} = 9$ , $H = 1$ , $R = 0$ (a perfect sensor). Without computing derivatives, what should $K$ be, and what does $\overset{x}{^}^{+}$ equal?

Recall Solution 1.2

WHAT: a perfect sensor ( $R = 0$ ) has zero noise, so we should trust it fully. $S = H P^{-} H^{⊤} + R = 1 \cdot 9 \cdot 1 + 0 = 9$ . Then $K = P^{-} H^{⊤} S^{- 1} = 9 \cdot 1 \cdot \frac{1}{9} = 1$ . With $K = 1$ : $\overset{x}{^}^{+} = \overset{x}{^}^{-} + 1 \cdot (z - \overset{x}{^}^{-}) = z$ . The estimate equals the measurement. ✅ WHY: $K = 1$ means the slider is pushed all the way to the measurement.

L2 — Application

Problem 2.1

Scalar: $P^{-} = 4$ , $H = 1$ , $R = 1$ . Compute $S$ , $K$ , and $P^{+}$ .

Recall Solution 2.1

Step 1 (build $S$ ): $S = 1 \cdot 4 \cdot 1 + 1 = 5$ . Step 2 (gain): $K = P^{-} H^{⊤} S^{- 1} = 4 \cdot 1 \cdot \frac{1}{5} = 0.8$ . Step 3 (posterior): at the optimum, $P^{+} = (1 - K H) P^{-} = (1 - 0.8) \cdot 4 = 0.8$ . WHY the sanity check: $P^{+} = 0.8$ is smaller than both $P^{-} = 4$ and $R = 1$ . Fusion beats either source alone — that is the entire point.

Problem 2.2

Same numbers as 2.1 ( $K = 0.8$ ). The prediction was $\overset{x}{^}^{-} = 10$ , the sensor reads $z = 15$ . Find the corrected estimate $\overset{x}{^}^{+}$ .

Recall Solution 2.2

Step 1 (innovation): $y = z - H \overset{x}{^}^{-} = 15 - 10 = 5$ . Step 2 (correct): $\overset{x}{^}^{+} = \overset{x}{^}^{-} + K y = 10 + 0.8 \cdot 5 = 14$ . WHY it lands where it does: the answer $14$ sits between prediction $10$ and measurement $15$ , closer to the sensor because $K = 0.8 > 0.5$ means we trust the sensor more here. This is exactly the slider figure above.

Problem 2.3

Verify the Joseph form gives the same $P^{+}$ as the shortcut for the optimal $K = 0.8$ in Problem 2.1.

Recall Solution 2.3

Joseph: $P^{+} = (1 - K H)^{2} P^{-} + K^{2} R = (0.2)^{2} \cdot 4 + (0.8)^{2} \cdot 1 = 0.16 + 0.64 = 0.8$ . ✅ Shortcut: $(1 - K H) P^{-} = 0.2 \cdot 4 = 0.8$ . ✅ They agree — because $K$ is optimal.

L3 — Analysis

Problem 3.1

For the scalar case with $P^{-} = 4$ , $R = 1$ , $H = 1$ , write $P^{+} (K) = (1 - K)^{2} \cdot 4 + K^{2} \cdot 1$ as a function of $K$ . Show it is a parabola, find its minimum by differentiating, and confirm you recover $K = 0.8$ .

Recall Solution 3.1

Step 1 (expand): $P^{+} (K) = 4 (1 - 2 K + K^{2}) + K^{2} = 4 - 8 K + 4 K^{2} + K^{2} = 5 K^{2} - 8 K + 4$ . WHAT this is: a parabola in $K$ opening upward (coefficient $5 > 0$ ), so it has a unique minimum — this is the "U-shaped valley" from the parent note. Step 2 (differentiate): $\frac{d P ^{+}}{d K} = 10 K - 8$ . Set to zero: $10 K = 8 \Rightarrow K = 0.8$ . ✅ Step 3 (value at minimum): $P^{+} (0.8) = 5 (0.64) - 8 (0.8) + 4 = 3.2 - 6.4 + 4 = 0.8$ . ✅ WHY differentiate here: minimizing the trace is setting the derivative of this parabola to zero — the general matrix derivation collapses to exactly this in the scalar case. The coefficient $10 = 2 S$ is the positive second derivative guaranteeing a minimum, not a maximum.

The figure below plots this exact parabola. Trace the violet curve: it bottoms out at the orange dot ( $K = 0.8$ , $P^{+} = 0.8$ ), rises to $P^{+} = 4$ at $K = 0$ (over-trusting the model, navy dot) and back up to $P^{+} = 1$ at $K = 1$ (over-trusting the sensor, magenta dot). One valley, one optimum.

Problem 3.2

Suppose an engineer hard-codes $K = 1$ (full trust in the sensor) even though $R = 1$ , $P^{-} = 4$ , $H = 1$ . Compute the actual resulting $P^{+}$ using the Joseph form and compare with the optimal $0.8$ .

Recall Solution 3.2

Must use Joseph ( $K = 1$ is not optimal): $P^{+} = (1 - 1)^{2} \cdot 4 + 1^{2} \cdot 1 = 0 + 1 = 1$ . Compare: optimal gave $0.8$ . So over-trusting the sensor gives $P^{+} = 1 > 0.8$ — worse, because we swallowed all the sensor noise $R = 1$ and threw away the prediction's help. WHY: at $K = 1$ the parabola $5 K^{2} - 8 K + 4$ gives $5 - 8 + 4 = 1$ , past the bottom of the valley (magenta dot in the figure above). Extra gain now adds noise faster than it removes prediction error.

L4 — Synthesis

Problem 4.1

Two sensors read the same scalar state, independently: $z_{1}$ with variance $R_{1} = 2$ and $z_{2}$ with variance $R_{2} = 6$ . Starting from a "flat" prior (infinite $P^{-}$ , meaning no useful prediction), fuse them by applying the Kalman update twice in sequence and find the final variance. Then compare to the classic inverse-variance-weighting result $P_{final} = (\frac{1}{R _{1}} + \frac{1}{R _{2}})^{- 1}$ .

Recall Solution 4.1

Step 1 — first measurement with an infinite prior. We must be careful: with $P^{-} \to \infty$ the shortcut $(1 - K H) P^{-}$ is the indeterminate form $0 \cdot \infty$ , so we use the Joseph form $P_{1} = (1 - K)^{2} P^{-} + K^{2} R_{1}$ (valid for any $K$ , no cancellation assumed). The optimal gain is $K = \frac{P ^{-}}{P ^{-} + R _{1}}$ . Substitute: $P_{1} = (\frac{R _{1}}{P ^{-} + R _{1}})^{2} P^{-} + (\frac{P ^{-}}{P ^{-} + R _{1}})^{2} R_{1} = \frac{R _{1}^{2} P ^{-} + P ^{- 2} R _{1}}{( P ^{-} + R _{1} ) ^{2}} = \frac{R _{1} P ^{-} ( R _{1} + P ^{-} )}{( P ^{-} + R _{1} ) ^{2}} = \frac{R _{1} P ^{-}}{P ^{-} + R _{1}} .$ Now take $P^{-} \to \infty$ : divide top and bottom by $P^{-}$ , giving $\frac{R _{1}}{1 + R _{1} / P ^{-}} \to R_{1}$ . So after step 1, $P_{1} = R_{1} = 2$ — the first measurement simply becomes the estimate, but the Joseph form got us there cleanly without any $0 \cdot \infty$ sleight of hand. Step 2 — fuse $z_{2}$ : now use $P^{-} = P_{1} = 2$ , $R = R_{2} = 6$ , $H = 1$ . $S = 2 + 6 = 8$ , $K = 2/8 = 0.25$ , $P_{2} = (1 - 0.25) \cdot 2 = 1.5$ . Cross-check with inverse-variance weighting: $(\frac{1}{2} + \frac{1}{6})^{- 1} = (\frac{3}{6} + \frac{1}{6})^{- 1} = (\frac{4}{6})^{- 1} = \frac{6}{4} = 1.5$ . ✅ WHY this is beautiful: the Kalman filter, applied recursively, reproduces optimal least-squares / sensor fusion weighting. Each measurement is folded in one at a time, yet the answer matches processing them all at once.

Problem 4.2

A 2-state filter tracks position and velocity. The prior covariance and measurement model are $P^{-} = [4001], H = [10], R = 2.$ Only position is measured (so $R$ is a single scalar here). Compute $S$ , the gain $K$ (a $2 \times 1$ vector), and the posterior $P^{+}$ .

Recall Solution 4.2

Step 1 ( $H P^{-}$ ): $H P^{-} = [10] [4001] = [40]$ . Step 2 ( $S$ ): $S = H P^{-} H^{⊤} + R = [40] [10] + 2 = 4 + 2 = 6$ . Step 3 ( $K = P^{-} H^{⊤} S^{- 1}$ ): $P^{-} H^{⊤} = [40]$ , so $K = \frac{1}{6} [40] = [0.667 0]$ . Step 4 ( $P^{+} = (I - K H) P^{-}$ ): here $I$ is the $2 \times 2$ identity. $K H = [0.667 0] [10] = [0.667 0 00]$ , so $I - K H = [0.333 0 01]$ . $P^{+} = [0.333 0 01] [4001] = [1.333 0 01]$ . WHY velocity is untouched: the second entry of $K$ is $0$ , so velocity variance stays at $1$ . A position sensor cannot directly improve velocity when the two are uncorrelated in $P^{-}$ (off-diagonals are zero). Position variance drops $4 \to 1.333$ . See Covariance Matrices and Uncertainty.

L5 — Mastery

Problem 5.1

Prove, for the scalar case, that the optimal $K = \frac{P ^{-}}{P ^{-} + R}$ (taking $H = 1$ ) always satisfies $0 \leq K < 1$ for finite positive $R$ , and that $P^{+} < min (P^{-}, R)$ . Interpret both facts.

Recall Solution 5.1

Bound on $K$ : with $H = 1$ , $K = \frac{P ^{-}}{P ^{-} + R}$ . Since $P^{-} > 0$ and $R > 0$ , the numerator is positive and strictly less than the denominator, so $0 < K < 1$ . As $R \to 0$ , $K \to 1$ ; as $R \to \infty$ , $K \to 0$ . The dial lives strictly inside $[0, 1]$ — never over-trusts either source. ✅ Posterior: $P^{+} = (1 - K) P^{-} = \frac{R}{P ^{-} + R} P^{-} = \frac{P ^{-} R}{P ^{-} + R}$ . This is the harmonic-style combination (product over sum). Because $\frac{P ^{-} R}{P ^{-} + R} < P^{-}$ (multiply top and bottom insight: $\frac{R}{P ^{-} + R} < 1$ ) and symmetrically $< R$ , we get $P^{+} < min (P^{-}, R)$ . ✅ Interpretation: fusing two independent estimates is always strictly better than the best single one — the certainty adds like parallel resistors: $\frac{1}{P ^{+}} = \frac{1}{P ^{-}} + \frac{1}{R}$ .

Problem 5.2

Using $P^{-} = 4$ , $R = 1$ , confirm the parallel-resistor identity $\frac{1}{P ^{+}} = \frac{1}{P ^{-}} + \frac{1}{R}$ numerically, and state what this means for stacking many sensors.

Recall Solution 5.2

$\frac{1}{P ^{-}} + \frac{1}{R} = \frac{1}{4} + \frac{1}{1} = \frac{5}{4} = 1.25$ . So $P^{+} = \frac{1}{1.25} = 0.8$ . ✅ Matches Problem 2.1's $P^{+} = 0.8$ . Meaning: each new independent measurement adds its precision $\frac{1}{R}$ (called information). Stack $n$ identical sensors of noise $R$ and the fused precision is $\frac{1}{P ^{-}} + \frac{n}{R}$ — uncertainty shrinks toward zero as $n$ grows. This "information adds" view underlies the EKF and all recursive fusion.

Recall Feynman recap of the whole ladder

L1: name the parts ( $y$ , $S$ , $K$ ) so no symbol is a stranger. L2: plug numbers in and watch $\overset{x}{^}^{+}$ land between the two sources, pulled by the size of the gain. L3: see the trace of $P^{+}$ is a U-shaped parabola with exactly one bottom — the optimum is a balance, not an extreme. L4: fold measurements in one at a time and recover the classic inverse-variance (least-squares) weighting. L5: prove fusion is always a win because precisions (information) add like parallel resistors, so $P^{+}$ can only shrink. Same slider, five depths — from "spot the pieces" to "prove it can never hurt."

Connections

Kalman gain — minimizes trace of covariance (parent)
Kalman Filter — Update Step
Kalman Filter — Predict Step
Covariance Matrices and Uncertainty
Least Squares Estimation
Innovation and Residuals
Joseph Form Covariance Update
Extended Kalman Filter (EKF) in GNC
Sensor Fusion in Navigation