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.dsy:it. : Powered by vBulletin version 2.3.1 .dsy:it. > Didattica > Corsi A - F > Calcolo delle probabilità e statistica matematica > Esame 5 luglio
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elpampero
Aniversario

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quindi il domandone finale è:
anche se le mie monete sono truccate (per esempio ho che la probabilità di avere testa con un lancio sia 2/3) la mia probabilità condizionata in quel modo è sempre 1/2??
03-07-2007 11:06
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Striker
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Originally posted by Striker
Penso che sia cosi'

P(X1=1|X1diversoX2) =
P(X1=1 AND ((X1=1 AND X2=0) OR (X1=0 AND X2=1)) / P((X1=1 AND X2=0) OR (X1=0 AND X2=1))


Andando avanti da qua dovrebbe essere:

P[(X1=1 AND (X1=1 AND X2=0)) OR (X1=1 AND (X1=0 AND X2=1))] / P((X1=1 AND X2=0) OR (X1=0 AND X2=1))

= pq + 0 / 2pq = 1/2 ?

E' giusto? Lo 0 sta ad indicare che e' impossibile che la v. c. X1 assuma due valori differenti.

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03-07-2007 11:25
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elpampero
Aniversario

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Ho ragionato esattamente come te....
03-07-2007 13:08
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Striker
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Speriamo di non sbagliare in due :-D

In pratica
(X1=1 AND (X1=1 AND X2=0)) si riduce a (X1=1 AND X2=0) perche' include entrambi. E la probab. di questo e' pq

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03-07-2007 13:11
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elpampero
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Sì sì..ti ho capito perfettamente...secondo me non fa una piega...quindi confermi che se avessimo che il lancio di una moneta è sbilanciato (cioè testa e croce non siano ugual probabili), sarebbe ancora 1/2 perchè non dipende da alcun parametro?
03-07-2007 13:17
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Striker
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Boh sembrerebbe cosi'...anche se non sono completamente convinto...

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03-07-2007 13:30
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elpampero
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Esatto..sembrerebbe così ma mi puzza
03-07-2007 13:38
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elpampero
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ALTRO QUESITO:
ho media=0.5, varianza=0,0041. Voglio che la mia variabile disti in valore assoluto meno di 0.05 da 0,5

E' possibile che applicando Chebishev con questi valori salti fuori un numero negativo?!?!?!
03-07-2007 14:38
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khelidan
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Originally posted by elpampero
ALTRO QUESITO:
ho media=0.5, varianza=0,0041. Voglio che la mia variabile disti in valore assoluto meno di 0.05 da 0,5

E' possibile che applicando Chebishev con questi valori salti fuori un numero negativo?!?!?!


Non ho fatto i conti ma chebichev puo benissimo dare un risultato negativo ma a quel punto non ti dice niente di importante,dato che una probabilita e definata tra 0 e uno

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Khelidan
03-07-2007 15:08
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rora
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con chebicev a me la P(|X-m|<.05) viene maggiore o uguale a - .64 ; qualcuno mi corregge?
dove hai trovato l'es?

Last edited by rora on 03-07-2007 at 15:28
03-07-2007 15:11
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elpampero
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L'ho calcolato io..ho fatto:
P(|X-0,5|<0.05)>=1-VAR(X)/(0.05)^2
03-07-2007 15:18
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khelidan
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viene -0.64 è plausibile ma non ci dice niente su questa stima!

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Khelidan
03-07-2007 15:24
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rora
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sì anche usando l'altra formula viene minore di 1.66... come dire che non se ne sa niente, per gli stessi motivi di prima
03-07-2007 15:27
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elpampero
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A meno che la x non sia una media standardizzata e da lì otterremmo:
P(|X*|<(0.05-0.5)/rad Var(X)) e da qui risolviamo con il teorema centrale della statistica (per cui P(|X*|<k)=F(k)-F(-k))
03-07-2007 15:30
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rora
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scusa se insisto, ma questi esercizi da dove arrivano?
03-07-2007 15:32
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