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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 del 11/02/2009 De Falco
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middu
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3.3. La funzione generatrice dei momenti di una somma di variabili aleatorie indipendenti e identicamente distribuite è dato dal prodotto delle funzioni generatrici dei momenti delle singole variabili aleatorie bernulliane e quindi (q+pe^t) è la funzione generatrice dei momenti di una bernulliana e per ottenere la funzione generatrice dei momenti moltiplico per c volte tale funzione ottenendo ((q+pe^t))^c che rappresenta la funzione generatrice di una nota variabile aleatoria discreta che è la binomiale di parametri c, p
12-02-2009 15:19
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middu
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3.4
t= 1 c=1 p = 1/3 ms(1) = (0,7+0,3*2,7)^1 = 1,51
t= 1 c =10 p = 1/30 = ms(1) =(0,96 + 0,03*2,7)^10 = 1,49 = 1,51
12-02-2009 15:24
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middu
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c = 1000000000 p = 1/3000000000 ms(1) = 1,77
c = 1=^23 p = 10^-23/3 = ms(1) = 1,77
12-02-2009 15:34
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middu
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il parametro λ = c*p
12-02-2009 15:35
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middu
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c = 1000000000 p = 1/3000000000 ms(1) = 1,77
c = 1=^23 p = 10^-23/3 = ms(1) = 1,77
12-02-2009 15:39
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gq690051
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 domanda

qualcuno riesce a dimostrarmi in formule perchè: 1/k∑y sia uno stimatore consistente?

si cerca di dimostrare la consistenza semplice, giusto? o la consistenza in media quadratica?
12-02-2009 17:23
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middu
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io ho dimostrato la consistenza semplice.
12-02-2009 18:58
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darkman13
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La soluzione dell'esercizio 4 e 5 non li postate?!!
13-02-2009 08:47
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hyperion
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Originally posted by middu
1.2 lo stimatore non distorto, consistente e asintoticamente normale è ad esempio la media campionaria. Perchè direte ??? Si tratta di uno stimatore non distorto in quanto il suo valore atteso coincide con il parametro da stimare e quindi uguale alla var(Yi). é consistente in quanto limite per k->ad infinito è uguale a 0 e asintoticamente normale in quanto la distribuzione radice(k)/σ * [Tk - ϴ] è una normale di valore atteso pari a 0 e varianza pari a 1. Questo lo si capisce quindi calcolando il valore atteso e la varianza di quest'ultima distribuzione e ottengo proprio i valori di 0 e 1 dove zero corrisponde al valore atteso e il secondo alla varianza.


la dimostrazione della normalità asintotica dello stimatore media campionaria c'entra qualcosa con il teorema centrale della statistica?
13-02-2009 10:09
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darkman13
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La soluzione dell'esercizio 4 e 5 non li postate?!!
13-02-2009 12:36
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middu
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darkam13 hai capito come viene calcolata quella probabilità???
13-02-2009 12:39
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darkman13
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Quale prob. MIddu
13-02-2009 12:45
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middu
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ma l'orale come si prepara
13-02-2009 13:00
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middu
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ricercando le parole da estrapolare????
13-02-2009 13:01
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hyperion
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Originally posted by middu
ricercando le parole da estrapolare????


cosi ci ha fatto capire De Falco l'11...poi una domanda ce l'ha i già scritta nel testo...
13-02-2009 13:08
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