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Definition of conditional probability density

  1. Oct 8, 2011 #1
    Hello, I'm somewhat confused by the expression [itex]f(X = x | Y = y) = \frac{f(X=x)}{f(Y=y)}[/itex] (which, if I'm right, is the definition of a conditional probability density? My course seems to state it as a theorem, without proof, but then again my course is a little bit vague; although I welcome replies on this part, this is not the essential of this topic)

    Anyway, the confusion is the following: let the s.v. Y be the s.v. X, then of course [itex]f(X = a | X=b)[/itex] should be zero if a is not equal to b (if the expression means what it is meant to mean), however it is equal to [itex]\frac{f(X = a)}{f(X=b)}[/itex] and there's no real reason why this should be zero.
     
  2. jcsd
  3. Oct 8, 2011 #2

    mathman

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    Your definition is incorrect. Let A and B be events, then P(A|B) = P(A and B)/P(B).
     
  4. Oct 8, 2011 #3
    Oh chucks that was stupid of me :) thank you...
     
  5. Oct 8, 2011 #4
    While we're at it, could you suggest me how to prove the following? (NOT a homework question)

    [itex]f(X_1 = x_1, \dots, X_n = x_n, X_{(n)} = a) =\left\{
    \begin{matrix}
    f(X_1 = x_1, \dots, X_n = x_n) & \quad \text{if max($x_1, \dots, x_n$)$=a$}\\
    0 & \quad \text{otherwise}\\
    \end{matrix} \right.[/itex]
     
  6. Oct 8, 2011 #5
    Could you explain the notation X(n)?
     
  7. Oct 8, 2011 #6
    My apologies, I should have:

    it is the n-th order statistic,
    i.e. [itex]X_{(n)} := \textrm{max}(X_1,\dots,X_n)[/itex]
     
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