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Thread: Conditional normal implying independence

  1. #1
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    Conditional normal implying independence

    So reading a book someone comment that e \lvert x \sim N(0,1) hence e is independent of x. Why is this the case?

    Trying to prove this to myself I note that the density of the normal distribution is fully specified ones variance and mean is specified and does not depend on x hence I can reason:

    f(e) = \int f(e,x) dx = \int f(e\lvert x) f(x) dx = f(e\lvert x) \int  f(x) dx =  f(e\lvert x)

    is that the way to go?

  2. #2
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    Re: Conditional normal implying independence

    In the second last equality you cannot simply pull the conditional pdf out of the integral in general, because it is possibly dependent on the integrating variable. The step is valid only when it is independent of the integrating variable, or in other words the random variables are actually independent. So overall the proof itself is ok because the condition given guaranteed this. You just need to note that the given condition is hold for all possible values of x inside its support.

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    JesperHP (01-06-2015)

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