# Thread: [Cognitive Robotics] state estimation

1. 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?

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?

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?

Originally Posted by I like Serena

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 .
But I'm not 100% sure how yet...

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

prob(z2|x2) =

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, $\mu_1=1,\mu_2=4,\mu_3=7,\mu_4=10$.

Originally Posted by I like Serena
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, $\mu_1=1,\mu_2=4,\mu_3=7,\mu_4=10$.
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).

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.

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