Unscented Kalman Filter (UKF) — sigma points, better for nonlinear: Question bank

Definition The symbols this page will keep using

Before you can spot the traps you must have these firmly in mind.

$x$ — the state (what we estimate), dimension $n$ ; $\overset{x}{ˉ}$ its mean, $P$ its covariance.
$f$ — the state-transition function: it maps the state one step forward in time, $x_{k} = f (x_{k - 1}) + w$ . It may be nonlinear (that is the whole reason we bother with a UKF).
$h$ — the measurement function: it maps a state to what the sensor would read, $z_{k} = h (x_{k}) + v$ , also possibly nonlinear.
$w$ — random process noise with covariance $Q$ : models the fact that $f$ is never a perfect description of reality (unmodelled forces, disturbances).
$v$ — random measurement noise with covariance $R$ : models sensor imperfection. $Q$ and $R$ are the two "how-much-do-I-trust" knobs; they must be added back after the sigma-point sums because the points only carry the prior uncertainty.

Two words you must also have in mind:

sigma points ::: the $2 n + 1$ deterministically chosen sample points ( $n$ = state dimension) whose weighted mean and spread reproduce the input mean $\overset{x}{ˉ}$ and covariance $P$ exactly.
Unscented Transform (UT) ::: the procedure of pushing sigma points through the true nonlinear function ( $f$ or $h$ ) and reading off the output mean and covariance from the transformed points.

True or false — justify

The UKF approximates the nonlinear function

f

with a polynomial.

False. The UKF approximates the distribution (with sigma points) and applies the true, un-approximated

f

to each point — approximating a distribution is easier than approximating an arbitrary function.

The UKF requires computing a Jacobian matrix.

False. It never linearizes; it estimates how uncertainty propagates by transforming sigma points directly, so no derivatives of

f

h

are ever taken.

Sigma points are drawn randomly like Monte Carlo samples.

False. They are deterministic and minimal (

2 n + 1

), placed to match the first two moments exactly; randomness would only add sampling noise, not accuracy.

For a purely linear system, the UKF gives the same result as the ordinary Kalman Filter (linear).

True. When

f

and

h

are linear the sigma-point mean and covariance reproduce the exact linear propagation, so the UKF collapses to the standard Kalman recursion.

The mean weights

W_{i}^{(m)}

and covariance weights

W_{i}^{(c)}

are identical for every sigma point.

False. They differ only at the center point

X_{0}

W_{0}^{(c)} = W_{0}^{(m)} + (1 - α^{2} + β)

. The outer

2 n

points share the same weight for both.

The UKF is a Bayesian sampling method that improves with more particles, like a Particle Filter.

False. A particle filter uses many random particles and represents arbitrary (non-Gaussian) distributions; the UKF uses a fixed minimal set matching only two moments of an assumed Gaussian.

Weights in the UT must all be positive.

False.

W_{0}^{(m)} = λ / (n + λ)

can be negative when

λ < 0

(e.g. large

n

with

κ = 0

), yet the weighted moment conditions still hold — negative weights are legal here.

The UT is exact for the mean of any nonlinear function.

False. It is 3rd-order accurate for the mean under Gaussian inputs — exact only for polynomials up to that order (and for special cases like

y = x^{2}

), not for arbitrary

f

Spot the error

"After propagating sigma points through

f

, the predicted covariance is

P_{k}^{-} = \sum_{i} W_{i}^{(c)} (X_{i}^{-} - \overset{x}{^}_{k}^{-}) (X_{i}^{-} - \overset{x}{^}_{k}^{-})^{⊤}

Missing

+ Q

. The sigma points carry only the prior uncertainty; the additive process-noise covariance

Q

must be added, or the filter grows overconfident and can diverge.

"I'll use

W_{0}^{(m)}

to build the innovation covariance

P_{z z}

Wrong weight. Covariance sums use

W_{i}^{(c)}

; only the mean

\overset{z}{^}_{k}

uses

W_{i}^{(m)}

. Swapping them corrupts the Gaussian-tuning (

β

) correction that lives inside

W_{0}^{(c)}

"The measurement covariance is

P_{z z} = \sum_{i} W_{i}^{(c)} (Z_{i} - \overset{z}{^}_{k}) (Z_{i} - \overset{z}{^}_{k})^{⊤}

Missing

+ R

. The measurement-noise covariance

R

is injected additively into

P_{z z}

; without it the gain

K

trusts the sensor too much.

"The matrix square root of

P

can be any

S

with

S^{⊤} S = P

We need

S S^{⊤} = P

: the columns of

S

become the axis-directions along which sigma points are placed, so those columns must satisfy

\sum_{j} S_{j} S_{j}^{⊤} = P

. In practice the lower-triangular Cholesky Decomposition factor is used; picking

S^{⊤} S = P

instead transposes the geometry and mis-places the axes.

"Since

α = 1 0^{- 3}

makes sigma points sit almost on the mean, the covariance estimate becomes tiny."

No. The shrink is exactly cancelled by the

(1 - α^{2} + β)

term in

W_{0}^{(c)}

and the large

1/ (n + λ)

outer weights, so the recovered covariance stays correct — that is the point of the scaled UT.

λ = α^{2} (n + κ)

Missing the subtraction: the correct definition is

λ = α^{2} (n + κ) - n

. Forgetting the

- n

misplaces every sigma point and breaks the moment match.

"The Kalman gain in the UKF is

K = P H^{⊤} (H P H^{⊤} + R)^{- 1}

There is no

H

in a UKF. The gain is

K = P_{x z} P_{z z}^{- 1}

, using the sigma-point cross-covariance

P_{x z}

in place of

P H^{⊤}

— the whole point is that no Jacobian

H

exists.

Why questions

Why does the UKF beat the Extended Kalman Filter (EKF) on sharply curved

f

The EKF keeps only the linear term of the Taylor Series Expansion and discards curvature; the UKF evaluates the true

f

at spread-out points, so it captures 2nd/3rd-order effects the EKF throws away.

Why are there exactly

2 n + 1

sigma points and not more?

One center point fixes the mean and two points per dimension (plus/minus along each principal axis of

P

) fix the covariance — that is the minimal set that matches both moments exactly.

Why do we need the matrix square root

(n + λ) P

at all?

To spread the points along the shape and orientation of the uncertainty ellipsoid; the square root's columns are the principal-axis directions scaled by their standard deviations.

Why split into

W^{(m)}

and

W^{(c)}

instead of one weight set?

So the Gaussian-tuning term

(1 - α^{2} + β)

can be injected into the covariance (via

W_{0}^{(c)}

) without shifting the mean, since it touches only the center weight.

Why is

β = 2

said to be optimal for Gaussian inputs?

β

enters only through

W_{0}^{(c)}

, so it re-weights the center point's contribution to the 4th-moment (kurtosis) term of the covariance expansion; setting

β = 2

exactly cancels the leading kurtosis error for the Gaussian, whose kurtosis is known — hence "optimal for Gaussians."

Why does the UKF handle a measurement like

h (x) = ∥ x ∥

where the EKF struggles at the origin?

The EKF's Jacobian

x^{⊤} /∥ x ∥

is undefined at

x = 0

, but the UKF's

P_{x z}

is a finite weighted sum of evaluated points — derivative-free, so no singularity.

Why is "match the distribution's two moments" enough for a Kalman-style filter?

A Kalman filter only ever propagates a mean and a covariance; if the sigma cloud reproduces those two moments, it feeds the recursion everything it uses — higher moments are irrelevant to the update.

Edge cases

What happens when

f

is exactly linear?

The UT returns the exact transformed mean and covariance (no approximation error), so the UKF is identical to the linear Kalman Filter (linear) on that step.

For a scalar

x \sim N (0, σ^{2})

and

y = x^{2}

, what mean does the EKF predict versus the truth?

The EKF uses Jacobian

2 x ∣_{x = 0} = 0

and predicts

\overset{y}{ˉ} = 0

; the truth (and the UT) is

\overset{y}{ˉ} = σ^{2}

— the EKF fails completely at a symmetric extremum.

What if

λ

comes out negative and makes

W_{0}^{(m)}

negative?

Perfectly valid: the moment conditions still hold, though a negative center weight can occasionally yield a non-positive-definite

P

estimate — a known reason to prefer the scaled/square-root UKF forms.

What if you forget the additive

Q

in prediction over many steps?

The covariance shrinks without bound, the filter becomes overconfident, ignores new measurements, and eventually diverges from the true state.

Is the UKF exact when the input is non-Gaussian?

No. Sigma points and

β = 2

are tuned to the Gaussian; for heavy-tailed or multimodal inputs the two-moment match misses structure — that is where a Particle Filter is preferred.

What is the smallest state dimension where the UKF differs from just evaluating

f

at the mean?

Any

n \geq 1

: even for

n = 1

the two outer points at

\pm (n + λ) σ

sense curvature that a single center evaluation cannot — this is exactly the

y = x^{2}

case.

If the sigma points are placed with the wrong scaling factor (e.g.

P

instead of

(n + λ) P

), what breaks?

The recovered covariance is scaled incorrectly — the points are too close or too far, so

P_{y}

no longer matches the true propagated covariance even for linear

f