20.06.2012, 22:38 |
Gast11022013 |
Auf diesen Beitrag antworten » |
Beweis falsch - wieso?
Meine Frage:
Hallo, ich habe hier ein Theorem, das bewiesen wird - und danach wird beschrieben, warum dieser Beweis recht besehen nicht stimmt. Aber diesen Einwand verstehe ich nicht und daher meine Nachfrage.
Voraussetzungen:
Zitat: |
"A Note on substitution in conditional distribution", Perlman, Wichura
Let be a probabilty space. Let , and be measurable spaces and suppose that , and are , and measurable. Let be a probability measure on .
The following proposition is commonly used in multivariate distribution theory and elsewhere:
Proposition 1:
(1) is independent of , and for each the random object has distribution .
Then
(2) the random Object also has distribution and is independent of .
[...]
Sometimes, however, one needs to weaken the assumption (1) slightly, as in the following.
Proposition 2. Suppose that
(3) for each , the random object has distribution and is independent of .
Then (2) holds.
[...]
Proposition 2 has a deceptively easy "proof", which runs as follows:
has distribution and is independent of , for all
Conditional distribution of , given , is , for all and
Conditional distribution of , given , is , for all
Conditional distribution of , given , is , for all
has distribution , and is independent of .
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So, jetzt kommt der Teil, der mir nicht klar werden möchte:
Zitat: |
Quelle wie oben
On close inspection, however, this argument breaks down. [...]
The major flaw in the above "proof" occurs in the first step, wherein the correct conclusion is that for each the conditional distribution of given is for -almost all ; as the null sets here may depend on , the second step may not be permissable.
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Diese Begründung, wieso obiger "Beweis" nicht korrekt ist, verstehe ich einfach auch nach hundertmaligem Durchlesen nicht! Könnte mir das jemand vielleicht erklären?
Meine Ideen:
Ich habe leider keine! |