The Zakai Equation for Unnormalized Filtering

Why Use an Unnormalized Filter?

In a filtering problem, the hidden state evolves randomly while observations provide incomplete information about it. The conditional distribution of the state given the observations is the filter. Its evolution is generally nonlinear because the observation process has a state-dependent drift. The Zakai equation replaces this normalized distribution by a likelihood-weighted, unnormalized one. That change produces a linear equation, and dividing by its total mass recovers the desired conditional distribution.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$. Under the physical probability measure $\mathbb{P}$, suppose the signal $X_t$ and scalar observation $Y_t$ satisfy $dX_t=b(X_t)\,dt+a(X_t)\,dV_t$ and $dY_t=h(X_t)\,dt+dW_t$, where $V_t$ and $W_t$ are independent Brownian motions, and $X_0$ is independent of both. The observation information at time $t$ is $\mathcal{F}_t^Y=\sigma(Y_s:0\le s\le t)$. Assume $b,a,h$ are sufficiently regular for the signal equation to have a solution, $b$ and $a$ are globally Lipschitz and of linear growth, and $h$ is bounded. For a twice continuously differentiable test function $f$, define the signal generator by $\mathcal{L}f(x)=b(x)f'(x)+\tfrac12a(x)^2f''(x)$.

Definition: A test-function form of a measure-valued equation specifies the evolution of the measure through its integrals against suitable functions $f$. Here the unnormalized filter is $\rho_t(f)$, a random linear functional of $f$; when represented by a measure, it is $\rho_t(f)=\int f(x)\rho_t(dx)$. The normalized filter is $\pi_t(f)=\mathbb{E}^{\mathbb{P}}[f(X_t)\mid\mathcal{F}_t^Y]$.

Reference Measure and the Zakai Equation

To define $\rho_t$, use a reference probability measure $\mathbb{Q}$ under which the signal has the same dynamics and $Y_t$ is a Brownian motion independent of the signal noise $V_t$. The physical observation law is recovered by the likelihood $\Lambda_t=\exp\{\int_0^t h(X_s)\,dY_s-\tfrac12\int_0^t h(X_s)^2\,ds\}$, so $d\mathbb{P}/d\mathbb{Q}|_{\mathcal{F}_t}=\Lambda_t$. The integral against $Y$ here is an Ito integral. Its integrand $h(X_s)$ is progressively measurable and square-integrable on finite horizons; boundedness of $h$ also gives Novikov's condition $\mathbb{E}^{\mathbb{Q}}[\exp(\tfrac12\int_0^T h(X_s)^2ds)]<\infty$. Thus the exponential is a true density martingale on each finite horizon. This use of the exponential-martingale and change-of-measure result is standard.

Theorem (Zakai equation): Under the assumptions above, for a suitable test function $f$, define $\rho_t(f)=\mathbb{E}^{\mathbb{Q}}[\Lambda_t f(X_t)\mid\mathcal{F}_t^Y]$. Then its weak equation is $$ d\rho_t(f)=\rho_t(\mathcal{L}f)\,dt+\rho_t(fh)\,dY_t, \qquad \rho_0(f)=\mathbb{E}[f(X_0)]. $$ The integral with respect to $Y$ is an Ito integral under $\mathbb{Q}$, with progressively measurable integrand $\rho_t(fh)$. The equation is understood at least for $f\in C_b^2$ when the coefficients and integrability assumptions ensure the displayed terms are defined; extensions to unbounded test functions require corresponding moment bounds.

Proof. Under $\mathbb{Q}$, Ito's formula, with $f'(X_t)$ and $f''(X_t)$ as the derivatives, gives $df(X_t)=\mathcal{L}f(X_t)\,dt+a(X_t)f'(X_t)\,dV_t$. The likelihood satisfies $d\Lambda_t=\Lambda_t h(X_t)\,dY_t$. Since $V$ and $Y$ are independent Brownian motions under $\mathbb{Q}$, their covariation is zero. The product rule therefore yields $d(\Lambda_t f(X_t))=\Lambda_t\mathcal{L}f(X_t)\,dt+\Lambda_t f(X_t)h(X_t)\,dY_t+\Lambda_t a(X_t)f'(X_t)\,dV_t$. Taking conditional expectations given the observation history gives the two drift and $dY$ terms in the theorem. More explicitly, conditional projection sends the $dY$ integral with integrand $H_s=\Lambda_s f(X_s)h(X_s)$ to the $dY$ integral with integrand $\mathbb{E}^{\mathbb{Q}}[H_s\mid\mathcal{F}_s^Y]=\rho_s(fh)$. The projected $dV$ integral is zero: approximating its integrand by adapted step processes, each Brownian increment of $V$ has conditional mean zero given the observation path and the earlier signal information. These projection identities extend from step processes by the Ito isometry. The integrals are legitimate under the stated boundedness and moment assumptions; if needed, one first stops the processes and then removes the stopping by the corresponding integrability bounds. Finally $\Lambda_0=1$, giving the stated initial value. $\square$

The conventions $dY_t\,dY_t=dt$, $dY_t\,dt=0$, and $dt\,dt=0$ are Ito multiplication rules, not ordinary chain rules. The correction $\tfrac12a^2f''$ in $\mathcal{L}f$ comes from the signal's quadratic variation; there is no additional cross term between the signal noise and observation noise because those Brownian motions are independent under $\mathbb{Q}$.

Normalization Recovers the Conditional Distribution

Proposition: If $\rho_t(1)>0$, then for every integrable test function $f$, $$ \pi_t(f)=\frac{\rho_t(f)}{\rho_t(1)} =\mathbb{E}^{\mathbb{P}}[f(X_t)\mid\mathcal{F}_t^Y]. $$ In particular, the normalized measure is obtained by dividing the unnormalized measure by its total mass.

Proof. The change-of-measure formula for conditional expectations gives $\mathbb{E}^{\mathbb{P}}[f(X_t)\mid\mathcal{F}_t^Y]= \mathbb{E}^{\mathbb{Q}}[\Lambda_t f(X_t)\mid\mathcal{F}_t^Y]/ \mathbb{E}^{\mathbb{Q}}[\Lambda_t\mid\mathcal{F}_t^Y]$ whenever the denominator is positive and the numerator is integrable. The numerator is $\rho_t(f)$; taking $f=1$ in its definition shows that the denominator is $\rho_t(1)$. Since $\Lambda_t>0$ almost surely, its conditional expectation is positive almost surely: if it were zero on an observation event of positive probability, the nonnegative variable $\Lambda_t$ would have zero expectation on that event and hence be zero there almost surely, a contradiction. This proves the identity. $\square$

Normalization is nonlinear even though the Zakai equation is linear in $\rho$. The denominator has a direct interpretation: $\rho_t(1)$ is the observation likelihood accumulated up to time $t$, expressed relative to the reference model. No division is required before solving the unnormalized equation.

Worked Example: A Two-Point Hidden State

Let $X_t=X_0$ be constant, with $\mathbb{P}(X_0=1)=\mathbb{P}(X_0=-1)=\tfrac12$, and let the observation drift be $h(x)=x$. Under $\mathbb{Q}$, the observation is Brownian and independent of $X_0$. Find the normalized conditional mean from the Zakai likelihood.

For a fixed state $x$, the likelihood at time $t$ is $\exp(xY_t-\tfrac12x^2t)$. The generator is zero because the signal is constant. Thus, for any test function $f$, $\rho_t(f)=\tfrac12 f(1)e^{Y_t-t/2}+\tfrac12 f(-1)e^{-Y_t-t/2}$. Taking $f=1$ gives $\rho_t(1)=e^{-t/2}\cosh(Y_t)$, and taking $f(x)=x$ gives $\rho_t(x)=e^{-t/2}\sinh(Y_t)$. Therefore $\pi_t(x)=\tanh(Y_t)$.

Check: at $t=1$ and $Y_t=\ln 3$, the likelihood factors for states $1$ and $-1$ are $3e^{-1/2}$ and $\tfrac13e^{-1/2}$. The posterior probability of $1$ is $(3e^{-1/2})/(3e^{-1/2}+\tfrac13e^{-1/2})=3/(3+\tfrac13)=\tfrac9{10}$. The posterior mean is $\tfrac9{10}-\tfrac1{10}=\tfrac45$, which agrees with $\tanh(\ln3)=(3-\tfrac13)/(3+\tfrac13)=\tfrac45$.

Worked Example: A Signal with No Observation Information

Let $X_t=B_t$ be a Brownian signal with $X_0=0$, independent of the observation, and set $h=0$. Under the reference and physical measures the observation is Brownian and carries no information about the signal. Compute $\rho_t(x^2)$ and check its Zakai equation.

Here $b=0$, $a=1$, and $\mathcal{L}f=\tfrac12f''$. Since $h=0$, $\Lambda_t=1$, so $\rho_t(x^2)=\mathbb{E}[B_t^2\mid\mathcal{F}_t^Y]=t$; independence makes the conditional expectation equal to the unconditional one. For $f(x)=x^2$, $f'(x)=2x$, $f''(x)=2$, and $\mathcal{L}f=1$. The Zakai equation becomes $d\rho_t(x^2)=\rho_t(1)\,dt=dt$, with initial value zero. Integrating gives $\rho_t(x^2)=t$.

Check: substituting the answer into the equation gives $d(t)=dt$, and at $t=0$ it gives $0=\mathbb{E}[X_0^2]$. Since $\rho_t(1)=1$, normalization leaves the conditional second moment equal to $t$, as expected when the observation is uninformative.

Worked Example: Verifying the Zakai Equation for a Known Signal

Suppose $X_t=x_0$ is known and constant, and $h(x_0)=c$, where $c$ is a fixed real number. For a fixed test function $f$, show directly that the likelihood-weighted filter solves the Zakai equation.

The signal generator vanishes and the likelihood is $\Lambda_t=\exp(cY_t-\tfrac12c^2t)$. Hence $\rho_t(f)=f(x_0)\Lambda_t$, $\rho_t(\mathcal{L}f)=0$, and $\rho_t(fh)=c f(x_0)\Lambda_t$. Ito's formula applied to $g(t,y)=\exp(cy-\tfrac12c^2t)$ has $g_t=-\tfrac12c^2g$, $g_y=cg$, and $g_{yy}=c^2g$. Therefore $dg(t,Y_t)=(g_t+\tfrac12g_{yy})dt+g_y\,dY_t=cg\,dY_t$. Multiplying by $f(x_0)$ gives $d\rho_t(f)=\rho_t(fh)dY_t$, precisely the Zakai equation.

Check: the drift terms cancel because $-\tfrac12c^2g+\tfrac12c^2g=0$. Also $\rho_t(1)=\Lambda_t>0$, so $\rho_t(f)/\rho_t(1)=f(x_0)$, the conditional expectation of a known state.

Why the Test-Function Form Matters

The weak equation can be used without assuming that the filtering distribution has a density. If a density $q_t(x)$ exists and the coefficients are regular enough, the same evolution can sometimes be written as a stochastic partial differential equation. But the test-function statement is more general: it says exactly how the random measure acts on each admissible $f$, whether or not a density is available.

For a fixed $f$, the Zakai equation is linear in $\rho$: both $\rho(\mathcal{L}f)$ and $\rho(fh)$ depend linearly on the measure. This is useful for analysis and computation. The normalized filter, by contrast, is a ratio and is therefore nonlinear. For example, in the two-point case the unnormalized weights are two linear likelihood-weighted contributions, while the posterior probability divides one contribution by their sum.

Common Pitfall: Confusing the Two Measures

A frequent error is to define $\rho_t(f)$ as $\mathbb{E}^{\mathbb{P}}[f(X_t)\mid\mathcal{F}_t^Y]$ and then write the Zakai equation for it. That is the normalized filter, not the likelihood-weighted unnormalized filter. The Zakai equation above is stated under the reference measure $\mathbb{Q}$ through $\rho_t(f)=\mathbb{E}^{\mathbb{Q}}[\Lambda_t f(X_t)\mid\mathcal{F}_t^Y]$. Its denominator $\rho_t(1)$ is essential.

For a concrete wrong calculation, take the known-state example with $x_0=1$, $c=1$, and $f=1$. The incorrect substitution $\rho_t(1)=\mathbb{E}^{\mathbb{P}}[1\mid\mathcal{F}_t^Y]=1$ would make the claimed equation $d\rho_t(1)=\rho_t(h)dY_t$ read $0=dY_t$, which is false. The correction is to use the unnormalized quantity $\rho_t(1)=\exp(Y_t-\tfrac12t)$; Ito's formula gives $d\rho_t(1)=\rho_t(1)dY_t$, exactly as the Zakai equation requires. Dividing $\rho_t(f)$ by this positive mass then recovers the physical conditional expectation.

Check Your Understanding

Use the measure definitions and the test-function equation to justify each answer.

Write the likelihood process and the Zakai equation for a signal with generator $\mathcal{L}$ and observation function $h$.
Why is the cross-variation between the signal noise and the observation noise zero under the reference measure in the derivation?
For a constant known signal $X_t=x_0$, calculate $\rho_t(1)$ when $h(x_0)=c$.
Explain why dividing $\rho_t(f)$ by $\rho_t(1)$ gives a conditional expectation under $\mathbb{P}$, rather than under $\mathbb{Q}$.
In the example with no observation information and $X_t=B_t$, compute $\rho_t(x^4)$ and identify which test-function generator term determines its evolution.