Q5: Filtering – Fundamentals of Sensing, Tracking, and Autonomy 1

Question 1: Deterministic state flow¶

Let 𝑋 = ℝ be the state space. Let 𝑓 : 𝑋 → 𝑋 be the deterministic state flow defined as 𝑓(𝑥) = 2𝑥 + 1.

Let

X_{k+1}(X_k) = \{x_{k+1} \in X \mid x_k \in X_k \text{ and } x_{k+1} = f(x_k) \}. \\\

𝑋ₖ₊₁([0, 1]) = [1, 3]

𝑋ₖ₊₁([−1, 1]) = [−1, 1]

𝑓( 𝑓( 𝑓(1))) = 15

𝑋ₖ₊₁( 𝑋ₖ₊₁([−1, 1])) = [−1, 1]

𝑋ₖ₊₁(ℝ) = ℝ

𝑋ₖ₊₁(∅) = ∅

𝑋ₖ₊₁(∅) = {0}

𝑋ₖ₊₁( 𝑋ₖ₊₁([−1, 1])) = [−1, 7]

𝑋ₖ₊₁({−1} ) = {−1}

𝑓( 𝑓([0, 1])) = 𝑓( 𝑓( 𝑓( 𝑓([0, 1]))))

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Question 2: Nondeterministic state flow¶

Let 𝑋 = ℝ be the state space. Let 𝑓 : 𝑋 → pow(𝑋) be the nondeterministic state flow defined as:

f(x) = \{x + d \in X \;|\; d \in [-1,1] \}. \\\

Let

X_{k+1}(X_k) = \{x_{k+1} \in X \mid x_k \in X_k \text{ and } x_{k+1} \in f(x_k) \}. \\\

𝑓(0) = [−1, 1]

𝑓(1) = [−1, 1]

𝑋ₖ₊₁([−2, 2]) = [−4, 4]

𝑋ₖ₊₁([−1, 1]) = [−2, 2]

If 𝐴 ⊂ 𝐵, then 𝑓(𝐴) ⊂ 𝑓(𝐵)

If 𝐴 ⊂ 𝐵, then 𝑋ₖ₊₁(𝐴) ⊂ 𝑋ₖ₊₁(𝐵)

𝑋ₖ₊₁([0, ∞)) = ℝ

𝑋ₖ₊₁( 𝑋ₖ₊₁((−1, 1))) = [−4, 4]

𝑋ₖ₊₁({0}) = [0, 1]

𝑋ₖ₊₁([−1, 0]) = 𝑋ₖ₊₁([0, 1])

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Question 3: Nondeterministic temporal filtering¶

Consider the same setup as in the previous problem, but now include a sensor mapping ℎ : 𝑋 → 𝑌 defined by:

y = \lfloor x + \tfrac{1}{2} \rfloor, \\\

in which ⌊·⌋ denotes the floor function (equivalent to the floor function in Python).

The preimage of 0 yields ℎ⁻¹(0) = [−1/2, 1/2].

The preimage of 0 yields ℎ⁻¹(0) = (−1/2, 1/2).

The preimage of 0 yields ℎ⁻¹(0) = [−1/2, 1/2).

The preimage of 0 yields ℎ⁻¹(0) = (−1/2, 1/2).

If 𝑋ₖ = [0, 1] and the next observation is 𝑦ₖ₊₁ = 1, then 𝑋ₖ₊₁ = [1/2, 3/2).

If 𝑋ₖ = [0, 1] and the next observation is 𝑦ₖ₊₁ = 1, then 𝑋ₖ₊₁ = (1/2, 3/2).

If 𝑋ₖ = [0, 1/2] and the next observation is 𝑦ₖ₊₁ = 2, then 𝑋ₖ₊₁ = {3/2}.

If 𝑋ₖ = [0, 1/2] and the next observation is 𝑦ₖ₊₁ = 2, then 𝑋ₖ₊₁ = ∅.

If 𝑋ₖ = [0, 1] and the next observation is 𝑦ₖ₊₁ = 2, then 𝑋ₖ₊₁ = [5/2, 2].

If 𝑋ₖ = ℝ, then 𝑋ₖ₊₁ = ℎ⁻¹(𝑦ₖ₊₁).

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Question 4: Discrete probability statements¶

Select ALL correct statements:

Bayes’ rule states that 𝑃(𝑎 | 𝑏) 𝑃(𝑏) = 𝑃(𝑎) 𝑃(𝑏 | 𝑎).

For any 𝑎 and 𝑏, 𝑃(𝑎 | 𝑏) ≥ 𝑃(𝑎).

If 𝑎 and 𝑏 are independent, then 𝑃(𝑎, 𝑏) = 𝑃(𝑏) 𝑃(𝑎).

If 𝑎 and 𝑏 are conditionally independent given 𝑐, then 𝑃(𝑎 | 𝑏, 𝑐) = 𝑃(𝑎 | 𝑏).

If 𝑎 and 𝑏 are conditionally independent given 𝑐, then 𝑃(𝑎 | 𝑏, 𝑐) = 𝑃(𝑎 | 𝑐).

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Question 5: More discrete probability statements¶

Select ALL correct statements:

By marginalization, 𝑃(𝑎) is obtained by summing 𝑃(𝑎 | 𝑏) 𝑃(𝑏) over all values of 𝑏.

By marginalization, 𝑃(𝑎 | 𝑐) is obtained by summing 𝑃(𝑎 | 𝑏,𝑐) 𝑃(𝑏 | 𝑐) over all values of 𝑏.

𝑃(𝑎 | 𝑏) = 𝑃(𝑏 | 𝑎).

If 𝑎 and 𝑏 are independent, then 𝑃(𝑎 | 𝑏) = 𝑃(𝑎).

If 𝑎 and 𝑏 are independent, then 𝑃(𝑏 | 𝑎) = 𝑃(𝑏).

If 𝑃(𝑎 | 𝑐) = 𝑃(𝑎), then 𝑎 and 𝑐 are independent.

If 𝑃(𝑎, 𝑏) = 0, then at least one of 𝑃(𝑎) or 𝑃(𝑏) must be 0.

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Question 6: Gaussians and sampling¶

Let 𝑆 ⊂ ℝ be a collection of 10,000 distinct samples that have mean 𝜇 and variance 𝜎².

Make a set 𝑆₁ = { 𝑥 ∈ ℝ ∣ 𝑥 − 1 ∈ 𝑆 } by adding 1 to all of the samples. Make another set 𝑆₂ = { 𝑥 ∈ ℝ ∣ 𝑥 / 2 ∈ 𝑆 } by multiplying all of the samples by 2. Let 𝜇ᵢ and 𝜎ᵢ² denote the mean and variance, respectively, of 𝑆ᵢ for 𝑖 = 1, 2.

𝜇₁ = 𝜇 + 1

𝜇₂ = 2𝜇

𝜇₂ = 2𝜇₁ − 1

𝜇₁ = 𝜇₂

𝜎₁ = 𝜎 + 1

𝜎₁ = 𝜎

𝜎₂ = 2𝜎

𝜎₂ = 4𝜎

𝜇₁ = 𝜇 + 1 only if the samples are drawn from a Gaussian distribution.

𝜇₂ = 2𝜇 only if the samples are drawn from a Gaussian distribution.

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Question 7: Probabilistic filters¶

Select all correct statements:

𝑃(𝑥ₖ₊₁ | 𝑦̃ₖ) is the pdf of the state at stage 𝑘 + 1, given observations up to stage 𝑘.

𝑃(𝑥ₖ₊₁ | 𝑦̃ₖ) is the pdf of the state at stage 𝑘 + 1, given the observation at stage 𝑘.

𝑃(𝑥ₖ₊₁ | 𝑥ₖ, 𝑦̃ₖ) = 𝑃(𝑥ₖ₊₁ | 𝑥ₖ, 𝑦̃ₖ) because 𝑥ₖ₊₁ is assumed to be conditionally independent of 𝑦̃ₖ, given 𝑥ₖ.

𝑃(𝑥ₖ₊₁ | 𝑥ₖ, 𝑦̃ₖ) = 𝑃(𝑥ₖ₊₁ | 𝑥ₖ) because 𝑥ₖ₊₁ is assumed to be conditionally independent of 𝑥ₖ, given 𝑦̃ₖ.

When 𝑦ₖ₊₁ is taken into account, the probabilistic sensing model is 𝑃(𝑦ₖ₊₁ | 𝑥ₖ₊₁).

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Question 8: More probabilistic filters¶

Select all correct statements:

𝑃(𝑦ₖ₊₁ | 𝑥ₖ₊₁, 𝑦̃ₖ) = 𝑃(𝑦ₖ₊₁ | 𝑥ₖ₊₁) because 𝑦ₖ₊₁ is assumed to be conditionally independent of 𝑦̃ₖ, given 𝑥ₖ₊₁.

𝑃(𝑦ₖ₊₁ | 𝑥ₖ₊₁, 𝑦̃ₖ) = 𝑃(𝑦ₖ₊₁ | 𝑥ₖ₊₁) because 𝑦ₖ₊₁ is assumed to be conditionally independent of 𝑥ₖ₊₁, given 𝑦̃ₖ.

If 𝑋 = {1, 2, 3, 4, 5, 6, 7}, then the dimension of 𝑰ₚᵣₒᵦ is 7.

If 𝑋 = {1, 2, 3, 4, 5, 6, 7}, then the dimension of 𝑰ₚᵣₒᵦ is 6.

We must always have that 𝑃(𝑥ₖ | 𝑦̃ₖ) > 0 for all 𝑘, 𝑥ₖ, and 𝑦̃ₖ.

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Lovelace

QUIZ 5: Filtering¶

Question 1: Deterministic state flow¶

Question 2: Nondeterministic state flow¶

Question 3: Nondeterministic temporal filtering¶

Question 4: Discrete probability statements¶

Question 5: More discrete probability statements¶

Question 6: Gaussians and sampling¶

Question 7: Probabilistic filters¶

Question 8: More probabilistic filters¶

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