Hint EX - Information Management Systems & Serviceszuev/teaching/2013Spring/Quiz-6.pdfMath 408...
1
Math 408 • Section 3963xD • Mathematical Statistics Quiz # 6 Solution, April 16, 2013 Teaching Assistant: Greg Sokolov This was a 20 minute quiz worth 10 points. Let p be an unknown parameter, 0 <p< 1, and suppose that X is a discrete random variable with p.m.f. π X (k|p)= P p (X = k) = (1 - p) k p, k =0, 1, 2, ··· . Hint. EX = (1 - p)/p and VarX = (1 - p)/p 2 . You observe three values 6, 1, 2. 1.) (2 points) Find the method of moments estimator of p. Solution. X 1 + ··· + X n n = 1 - ˆ p MoM ˆ p MoM ⇒ ˆ p MoM = n n + X 1 + ··· + X n . In our case, ˆ p MoM =3/(3 + 6 + 1 + 2) = 1/4. 2.) (4 points) Find the maximum-likelihood estimator of p. Solution. L(p)= n Y i=1 (1 - p) Xi p = (1 - p) X1+···+Xn p n , log L(p)=(X 1 + ··· + X n ) log(1 - p)+ n log p, ∂ ∂p log L(p)= -(X 1 + ··· + X n ) 1 1 - p + n p . Therefore, ˆ p MLE = n n + X 1 + ··· + X n =ˆ p MoM . In our case, ˆ p MLE =1/4. 3.) (4 points) Find the Cramer-Rao bound for the variance of unbiased estimators of p. Solution. π(X|p) = (1 - p) X p, log π(X|p)= X log(1 - p) + log p, ∂ ∂p log π(X|p)= - X 1 - p + 1 p = - 1 1 - p X - 1 - p p , E ∂ ∂p log π(X|p) 2 = 1 (1 - p) 2 E X - 1 - p p 2 = 1 (1 - p) 2 VarX = 1 p 2 (1 - p) . Therefore, for any unbiased estimator ˆ p, Var(ˆ p) ≥ p 2 (1 - p) n .
Transcript of Hint EX - Information Management Systems & Serviceszuev/teaching/2013Spring/Quiz-6.pdfMath 408...
Math 408 • Section 3963xD • Mathematical StatisticsQuiz # 6 Solution, April 16, 2013 Teaching Assistant: Greg Sokolov
This was a 20 minute quiz worth 10 points.
Let p be an unknown parameter, 0 < p < 1, and suppose that X is a discrete random variable with p.m.f.
πX(k|p) = Pp(X = k) = (1− p)k p, k = 0, 1, 2, · · · .
Hint. EX = (1− p)/p and VarX = (1− p)/p2.You observe three values 6, 1, 2.
1.) (2 points) Find the method of moments estimator of p.
Solution.X1 + · · ·+Xn
n=
1− p̂MoM
p̂MoM⇒ p̂MoM =
n
n+X1 + · · ·+Xn.
In our case, p̂MoM = 3/(3 + 6 + 1 + 2) = 1/4.
2.) (4 points) Find the maximum-likelihood estimator of p.
Solution.
L(p) =n∏
i=1
(1− p)Xip = (1− p)X1+···+Xn pn,
logL(p) = (X1 + · · ·+Xn) log(1− p) + n log p,
∂
∂plogL(p) = −(X1 + · · ·+Xn)
1
1− p+n
p.
Therefore,
p̂MLE =n
n+X1 + · · ·+Xn= p̂MoM.
In our case, p̂MLE = 1/4.
3.) (4 points) Find the Cramer-Rao bound for the variance of unbiased estimators of p.
Solution.
π(X|p) = (1− p)Xp,
log π(X|p) = X log(1− p) + log p,
∂
∂plog π(X|p) = − X
1− p+
1
p= − 1
1− p
(X − 1− p
p
),
E
(∂
∂plog π(X|p)
)2
=1
(1− p)2E
(X − 1− p
p
)2
=1
(1− p)2VarX
=1
p2(1− p).
Therefore, for any unbiased estimator p̂,
Var(p̂) ≥ p2(1− p)n
.
