MHB [Cognitive Robotics] state estimation

AI Thread Summary
The discussion revolves around calculating probabilities using Bayes' rule for a robot's position and sensor measurements in a cognitive robotics context. The user seeks assistance in determining the likelihoods prob(x1|z1) and prob(z1|x1), specifically how to account for Gaussian noise in their calculations. Participants clarify the use of Gaussian distribution density functions and the implications of systematic errors in measurements, noting that the noise should have a mean of zero. The conversation emphasizes understanding the relationship between expected readings and actual measurements to derive the necessary probabilities. Overall, the thread provides insights into applying probabilistic models in robotics for state estimation.
RandomUserName
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Hey guys, I need some help. Here's my task:

View attachment 6189

Here's what I came up with: I need to calculate prob(x1|z1), prob(x1|z2) and so on and then just write down which has the highest probability. To do that, I would use Bayes rule, but for that, I need p(z1|x1) and p(z1). How do I get these?

I have no idea what to calculate for the case: robot is at position x1, what is prob to measure z1. What do I do with the Gaussian noise? :(
 

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RandomUserName said:
Hey guys, I need some help. Here's my task:

Here's what I came up with: I need to calculate prob(x1|z1), prob(x1|z2) and so on and then just write down which has the highest probability. To do that, I would use Bayes rule, but for that, I need p(z1|x1) and p(z1). How do I get these?

I have no idea what to calculate for the case: robot is at position x1, what is prob to measure z1. What do I do with the Gaussian noise? :(

Hi RandomUserName! Welcome to MHB! ;)

We have:
$$P(z_1\mid x_1) = P(z_1 \le Z < z_1 + dz \mid x_1) = f_1(z_1) dz$$
where $f_1$ is the gaussian distribution density for $x_1$.

And we have:
$$P(z_1) = P(z_1 \wedge (x_1 \vee x_2 \vee x_3 \vee x_4)) \\
= P((z_1 \wedge x_1) \vee (z_1 \wedge x_2) \vee (z_1 \wedge x_3) \vee (z_1 \wedge x_4))
= P(z_1 \wedge x_1) + P(z_1 \wedge x_2) + P(z_1 \wedge x_3) + P(z_1 \wedge x_4) \\
= P(z_1 \mid x_1)P(x_1) + P(z_1 \mid x_2)P(x_2) + P(z_1 \mid x_3)P(x_3) + P(z_1 \mid x_4)P(x_4)
$$
Since we assume an equally distributed prior, we have:
$$P(x_1)=P(x_2)=P(x_3)=P(x_4)=\frac 14$$

Can we find $P(x_1\mid z_1)$ now?
 
I like Serena said:
Hi RandomUserName! Welcome to MHB! ;)

We have:
$$P(z_1\mid x_1) = P(z_1 \le Z < z_1 + dz \mid x_1) = f_1(z_1) dz$$
where $f_1$ is the gaussian distribution density for $x_1$.

And we have:
$$P(z_1) = P(z_1 \wedge (x_1 \vee x_2 \vee x_3 \vee x_4)) \\
= P((z_1 \wedge x_1) \vee (z_1 \wedge x_2) \vee (z_1 \wedge x_3) \vee (z_1 \wedge x_4))
= P(z_1 \wedge x_1) + P(z_1 \wedge x_2) + P(z_1 \wedge x_3) + P(z_1 \wedge x_4) \\
= P(z_1 \mid x_1)P(x_1) + P(z_1 \mid x_2)P(x_2) + P(z_1 \mid x_3)P(x_3) + P(z_1 \mid x_4)P(x_4)
$$
Since we assume an equally distributed prior, we have:
$$P(x_1)=P(x_2)=P(x_3)=P(x_4)=\frac 14$$

Can we find $P(x_1\mid z_1)$ now?
Since you're asking like that, probably yeah :D.
But I'm not 100% sure how yet...

Can I just use this formula now?
prob(z1|x2) = this

prob(z2|x2) = this
 
RandomUserName said:
Since you're asking like that, probably yeah :D.
But I'm not 100% sure how yet...

Can I just use this formula now?
prob(z1|x2) = this

prob(z2|x2) = this

That's the formula yes, but if I'm not mistaken we should take $\mu_2=4$ instead of $\mu_2=5$.

To be honest, the problem statement is a bit confusing saying that the added Gaussian noise has $\mu=1$.
That just makes no sense - added Gaussian noise always has $\mu=0$.
I can only assume that they meant that the given Gaussian distribution was for $x_1$ with $\mu_1=1$, while the other Gaussian distributions would have the respective expectations for $x_i$ and $\sigma=2$. That is, $\mu_1=1,\mu_2=4,\mu_3=7,\mu_4=10$.
 
I like Serena said:
That's the formula yes, but if I'm not mistaken we should take $\mu_2=4$ instead of $\mu_2=5$.

To be honest, the problem statement is a bit confusing saying that the added Gaussian noise has $\mu=1$.
That just makes no sense - added Gaussian noise always has $\mu=0$.
I can only assume that they meant that the given Gaussian distribution was for $x_1$ with $\mu_1=1$, while the other Gaussian distributions would have the respective expectations for $x_i$ and $\sigma=2$. That is, $\mu_1=1,\mu_2=4,\mu_3=7,\mu_4=10$.
I asked about this, and this was the reply:
The noise has a shift of 1 (the mean), so if the real reading is 10, then the noisy reading (which will be the most probable) will be 11. Therefore, the difference between the expected reading at a certain x_i location (taking into account the noise effect ex: 11) and the real reading z_t should determine the probability p(z_t|x_i).​
 
RandomUserName said:
I asked about this, and this was the reply:
The noise has a shift of 1 (the mean), so if the real reading is 10, then the noisy reading (which will be the most probable) will be 11. Therefore, the difference between the expected reading at a certain x_i location (taking into account the noise effect ex: 11) and the real reading z_t should determine the probability p(z_t|x_i).​

Okay. So we have a systematic error in our measurements.
In that case I believe you're good to go.
 
I like Serena said:
Okay. So we have a systematic error in our measurements.
In that case I believe you're good to go.

Alright :)
Thanks for all your help!
 
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