Understanding one step in particle_filter_base.py

[wilhem]

Hi, I looked at the code and found the following step in compute_likelihood method:

            # Map difference true and expected distance measurement to probability
            p_z_given_x_distance = \
                np.exp(-(expected_distance-measurement[i][0]) * (expected_distance-measurement[i][0]) /
                       (2 * self.measurement_noise[0] * self.measurement_noise[0]))

It seem an implementetation of the very well known gauss normal distribution:

but to me the term:

1 / (sigma sqrt(2.0 pi))

is missing. Why?

PS: Thank you very much for the tutorial... you are helping thousand and thousand of people, who are train to understand particle filters

[jelfring]

Short answer: because multiplying all weights with the same constant will not change the estimates of the filter. The main reason therefore is computational efficiency.

Longer answer: As you can see, the term is constant, in other words, it is independent of the state of a particular particle. For ease of writing we can define: c = 1 / (sigma sqrt(2.0 pi)).

Now let's assume this term will be added. The particle weights will now change as follows: weight_1 = weight_1_without_term c weight_2 = weight_2_without_term c ... weight_n = weight_n_without_term * c where weight without term refers to the weight the way it is computed in the code. As you can see, the ratio between the non-normalized weights remains unchanged after adding c.

Now let's normalize the new weights, such that they sum up to one. The sum of all weights after adding the term c changes as follows: sum_all_weights = weight_1_without_term c + weight_2_without_term c + ... + weight_n_without_term c, which can be rewritten as: sum_all_weights = c (weight_1_without_term + weight_2_without_term + ... + weight_n_without_term)

If we now normalize the weights, the term c cancels out: normalized_w1 = (weight_1_without_term c) / (c (weight_1_without_term + weight_2_without_term + ... + weight_n_without_term)).

So, the normalized weights will be equal to the normalized weights without the term c.

I hope this helps.

[wilhem]

Many many thanks!!!!! Very clear and understandable!