*biased*for any finite sample size—i.e., the expected value of the location computed by the algorithm differs from the true expected value—because of the way particle filtering works. In this question, you are asked to quantify this bias.

To simplify, consider a world with four possible robot locations: $X=\{x_1,x_2,x_3,x_4\}$. Initially, we draw $N\geq $ samples uniformly from among those locations. As usual, it is perfectly acceptable if more than one sample is generated for any of the locations $X$. Let $Z$ be a Boolean sensor variable characterized by the following conditional probabilities:

$$\begin{aligned} P(z | x_1) = 0.8 \qquad\qquad P(z | x_1) = 0.2 \\ P(z | x_2) = 0.4 \qquad\qquad P(z | x_2) = 0.6 \\ P(z | x_3) = 0.1 \qquad\qquad P(z | x_3) = 0.9 \\ P(z | x_4) = 0.1 \qquad\qquad P(z | x_4) = 0.9 \end{aligned}$$

MCL uses these probabilities to generate particle weights, which are subsequently normalized and used in the resampling process. For simplicity, let us assume we generate only one new sample in the resampling process, regardless of $N$. This sample might correspond to any of the four locations in $X$. Thus, the sampling process defines a probability distribution over $X$.

1. What is the resulting probability distribution over $X$ for this new sample? Answer this question separately for $N=1,\ldots,10$, and for $N=\infty$.

2. The difference between two probability distributions $P$ and $Q$ can be measured by the KL divergence, which is defined as $${KL}(P,Q) = \sum_i P(x_i)\log\frac{P(x_i)}{Q(x_i)}\ .$$ What are the KL divergences between the distributions in (a) and the true posterior?

3. What modification of the problem formulation (not the algorithm!) would guarantee that the specific estimator above is unbiased even for finite values of $N$? Provide at least two such modifications (each of which should be sufficient).

Monte Carlo localization is
*biased* for any finite sample size—i.e., the expected
value of the location computed by the algorithm differs from the true
expected value—because of the way particle filtering works. In this
question, you are asked to quantify this bias.

To simplify, consider a world with four possible robot locations:
$X=\{x_1,x_2,x_3,x_4\}$. Initially, we
draw $N\geq $ samples uniformly from among those locations. As
usual, it is perfectly acceptable if more than one sample is generated
for any of the locations $X$. Let $Z$ be a Boolean sensor variable
characterized by the following conditional probabilities:

$$\begin{aligned}
P(z | x_1) = 0.8 \qquad\qquad P(z | x_1) = 0.2 \\
P(z | x_2) = 0.4 \qquad\qquad P(z | x_2) = 0.6 \\
P(z | x_3) = 0.1 \qquad\qquad P(z | x_3) = 0.9 \\
P(z | x_4) = 0.1 \qquad\qquad P(z | x_4) = 0.9
\end{aligned}$$

MCL uses these probabilities to generate particle weights, which are
subsequently normalized and used in the resampling process. For
simplicity, let us assume we generate only one new sample in the
resampling process, regardless of $N$. This sample might correspond to
any of the four locations in $X$. Thus, the sampling process defines a
probability distribution over $X$.

1. What is the resulting probability distribution over $X$ for this new
sample? Answer this question separately for
$N=1,\ldots,10$, and for $N=\infty$.

2. The difference between two probability distributions $P$ and $Q$ can
be measured by the KL divergence, which is defined as
$${KL}(P,Q) = \sum_i P(x_i)\log\frac{P(x_i)}{Q(x_i)}\ .$$ What are
the KL divergences between the distributions in (a) and the true
posterior?

3. What modification of the problem formulation (not the algorithm!)
would guarantee that the specific estimator above is unbiased even
for finite values of $N$? Provide at least two such modifications
(each of which should be sufficient).