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mr. vodka
Oct8-11, 04:10 PM
Hello, I'm somewhat confused by the expression f(X = x | Y = y) = \frac{f(X=x)}{f(Y=y)} (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 f(X = a | X=b) 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 \frac{f(X = a)}{f(X=b)} and there's no real reason why this should be zero.

mathman
Oct8-11, 04:36 PM
Your definition is incorrect. Let A and B be events, then P(A|B) = P(A and B)/P(B).

mr. vodka
Oct8-11, 04:55 PM
Oh chucks that was stupid of me :) thank you...

mr. vodka
Oct8-11, 05:21 PM
While we're at it, could you suggest me how to prove the following? (NOT a homework question)

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.

Useful nucleus
Oct8-11, 07:31 PM
Could you explain the notation X(n)?

mr. vodka
Oct8-11, 07:49 PM
My apologies, I should have:

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