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  1. MHB Apprentice

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    #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?
    Last edited by MarkFL; November 13th, 2016 at 13:35. Reason: Attach Image Inline

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    #2
    Quote Originally Posted by RandomUserName View Post
    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?

  3. MHB Apprentice

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    Quote Originally Posted by I like Serena View Post
    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 .
    But I'm not 100% sure how yet...

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

    prob(z2|x2) =

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    #4
    Quote Originally Posted by RandomUserName View Post
    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, $\mu_1=1,\mu_2=4,\mu_3=7,\mu_4=10$.

  5. MHB Apprentice

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    Quote Originally Posted by I like Serena View Post
    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).

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    #6
    Quote Originally Posted by RandomUserName View Post
    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.

  7. MHB Apprentice

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    #7 Thread Author
    Quote Originally Posted by I like Serena View Post
    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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