Flashcards
Sem título
Flashcards do conjunto (4)
- 1Chapter 3 - Families of Discrete Random Variables For the Bernoulli random variable X~Bern(p) state the PMF Pr(X=x), the PGF Gx(s), E(X) and Var(X) For the Binomial random variable X~Bin(n,p) state the PMF, PGF, E(X) and Var(X) For the Poisson random variable X~Po(lambda) states the PMF, PGF, E(X) and Var(X)X~Bern(p):
Pr(X=x) = p, x=1
1-p, x=0
Gx(s) = ps+1-p
E(X) = p Var(X) = p(1-p)
X~Bin(n,p):
Pr(X=x) = (nCx)(p^x)(1-p)^(n-x)
Gx(s) = (ps+1-p)^n
E(X) = np Var(X) = np(1-p)
X~Po(lambda):
Pr(X=x) = [(lambda)^x*e^(-lambda)]/x!
Gx(s) = e^(lambda(s-1))
E(X) = lambda Var(X) = lambda - 2Chapter 3 - Families of Discrete Random Variables For the Geometric random variable X~Geom(p) state the PMF, PGF, E(X) and Var(X) For the Negative Binomial random variable X~NBin(r,p) state the PMF, PGF, E(X) and Var(X) For the Discrete Uniform random variable X~U(1,2,...,n) state the PMF, PGF, E(X) and Var(X)X~Geom(p):
Pr(X=x) = (1-p)^(x-1)p
Gx(s) = (ps)/[1-(1-p)s]
E(X) = 1/p Var(X) = (1-p)/p^2
X~NBin(r,p:
Pr(X=x) = [(x-1)C(r-1)]*(p^r)*(1-p)^(n-r)
Gx(s) =[(ps)/1-(1-p)s]^r
E(X) = r/p Var(X) = [r(1-p)]/p^2
Pr(X=x) = 1/n
Gx(s) = [s(1-s^n)]/[n(1-s)]
E(X) = (n+1)/2 Var(X) = (n^2-1)/12 - 3Chapter 3 - Sums of Random Variables If X1, X2, ..., Xn are independent and Tn = SIGMA(Xi) then GTn(s)=? Prove the above statementGTn(s) = GX1(s)GX2(s)...GXn(s)
Answer to proof on a separate sheet - 4Chapter 3 - Random sums of independent random variables A random variable Sn is defined as? We then get what 3 properties from Sn?Sn = X1+X2+...+Xn, where the Xi's have PGF Gx(s) and n has PGF Gn(S)
1. GSn(s) = Gn(Gx(s))
2. E(Sn) = E(N)E(X)
3. Var(Sn) = E(N)Var(X) + Var(N)[E(X)]^2
Informações
- Descrição
- FlashCards sobre Sem título, criado por Cara Wales em 27-01-2016.
- Criado por
- Cara Wales
- Visualizações
- 2
- Avaliação