Solved – Likelihood calculation in Particle Filtering

bayesianparticle filter

I have a doubt with likelihood calculation in particle filtering.

In my understanding the particle filter consists of the following steps

Generate particles from initial point
Propagate through system model ($X_p(k) = A*X_f(k-1) + Q$)
(I am generating Gaussian noise and adding to state equation based on $Q$ for each particle)
Weight update using likelihood calculation
For likelihood calculation I need measurement from sensor (with Gaussian noise) and predicted measurement

For the predicted measurements, I have to use the measurement equation
$$y = H*x + R$$ For each particle I have to calculate corresponding $y$ value.

Should I generate Gaussian noise based of R for every particle while calculating predicted measurement y?

In the Kalman filter we use $y = H*x$ ( since we are calculating mean), what should I do in particle filter…?

Best Answer

Just compute the likelihood of data given the particles. For 1-d

normpdf(y(k),x_particles,standard_dev_of_noise)

For multi-dimensional case

mvnpdf(y(k),H*x_particles,R)

These will be your weights. Then get the resampling indices according to your favorite resampling algorithm and resample your particles.

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Solved – Particle filter in Matlab – what is going wrong

The problem may stem from the fitness ratio you use in your code. The inverse of the distances may give very large differences in the resampling probabilities of the new generation of the samples. For example, let x stand for the rms values in your design and let them be

x = [0.1 1 1 3 10 50]

When you use the inverses, that leads to

y = (1./x) ./ norm(1./x)
y = 
    0.9896    0.0990    0.0990    0.0330    0.0099    0.0020

As you see, the first sample dominates the next generation. It may converge to wrong proposals very easily. Alternatively, you may use a sigmoid-like function. For example,

z = exp(-x) ./ norm(exp(-x))
z = 
    0.8659    0.3521    0.3521    0.0476    0.0000    0.0000

Now, the particles in the next generation will probably involve some samples similar to the second and third ones. So, the generation scheme will be more robust to erronous proposals.

Edit: Moreover, after dividing by the norms, the sum of the result is not 1. You might use sum instead.

y = (1./x) ./ sum(1./x)
y = 
    0.8030    0.0803    0.0803    0.0268    0.0080    0.0016

z = exp(-x) ./ sum(exp(-x))
z =
    0.5353    0.2176    0.2176    0.0295    0.0000    0.0000

Best Answer

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