Definition of conditional probability density

  • #1
1,434
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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.
 

Answers and Replies

  • #2
mathman
Science Advisor
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Your definition is incorrect. Let A and B be events, then P(A|B) = P(A and B)/P(B).
 
  • #3
1,434
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Oh chucks that was stupid of me :) thank you...
 
  • #4
1,434
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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]
 
  • #5
Could you explain the notation X(n)?
 
  • #6
1,434
2
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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