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Math Help - Conditional density function?

  1. #1
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    Conditional density function?

    Hi, I'm not sure what the notation is supposed to mean in the highlighted box and how it relates to conditional probability:
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    The definition of conditional probability that I'm familiar with is P(A|B) = P(A, B)/P(B). How (if at all) does that relate to what the authors have written there? Some special properties of the normal distribution perhaps?

    PS. Theta is a vector of reals, as is x, so \theta^Tx^{(i)}=\theta\cdot x^{(i)}
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  2. #2
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    Firstly, check that you understand how this relates to the previous line. It is simply making the subsititution y_i = \epsilon_i + \theta^T x_i


    The notation appears to be a conditional pdf for Y, ie (very loosely*):

     p(y_i | x_i ; \theta) \approx \mathbb{P}(Y_i = y_i | x_i ; \theta)

    That is to say, in very loose terms* it is telling your the conditional probability of Y|X, if the parameter value is equal to theta. The parameter value has to be included because it made the substitution y_i = \epsilon_i + \theta^T x_i, which will be different for every possible value of \theta


    *in fact, pdfs do not give probabilities, but thats a seperate topic...
    Last edited by SpringFan25; June 29th 2010 at 07:27 AM. Reason: inaccuracies fixed
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  3. #3
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    Thanks a lot! But didn't they make the substitution eps = y - theta . x instead? Since the mean of the normal dist is theta . x ?
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  4. #4
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    yeh i just typoed the subsitituion formula


    Should have been

    y_i=\epsilon + \theta^T x_i which rearranges to what you said
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  5. #5
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    But then I suppose my question is, if they made that substitution for \epsilon, why did they change

    p(\epsilon) to p(y|x), and not p(y-\theta^Tx) or something like that (if that even makes sense)? I.e. Why did it become conditional all of a sudden?
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  6. #6
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    i think:
    Remember that once you condition on x, \theta^T x is just a number. This means you can make the substitution \epsilon = y_i - \theta^T x_i without worrying about how the randomness of x will affect the distribution of y.
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