2010首届丘成桐大学生数学竞赛团体赛解答

n a1 ,··· ,an+1 |θ|≤τ
min
sup E (
i=1
ai Xi + an+1 − θ)2 =
τ 2 n −1 . τ 2 + n −1
Hint: Carefully use the sufficiency principle. 4. The rules for “1 and 1” foul shooting in basketball are as follows. The shooter gets to try to make a basket from the foul line. If he succeeds, he gets another try. More precisely, he make 0 baskets by missing the first time, 1 basket by making the first shot and xsmissing the second one, or 2 baskets by making both shots. Let n be a fixed integer, and suppose a player gets n tries at “ 1 and 1” shooting. Let N0 , N1 , and N2 be the random variables recording the number of times he makes 0, 1, or 2 baskets, respectively. Note that N0 + N1 + N2 = n. Suppose that shots are independent Bernoulli trails with probability p for making a basket. (a) Write down the likelihood for (N0 , N1 , N2 ). (b) Show that the maximum likelihood estimator of p is N1 + 2N2 . p ˆ= N0 + 2N1 + 2N2 (c) Is p ˆ an unbiased estimator for p? Prove or disprove. (Hint: E p ˆ is a polynomial in p, whose order is higher than 1 for p ∈ (0, 1).) (d) Find the asymptotic distribution of p ˆ as n tends to ∞. Proof. (a) The likelihood is l(p) = (1 − p)N0 [p(1 − p)]N1 [p2 ]N2 = (1 − p)N0 +N1 pN1 +2N2 . (Trinomial) (b) Solve the score equation l (ˆ p) = 0. Need to verify that it is the maximizer. (c) No. Suppose that E p ˆ = p for any p ∈ [0, 1]. Note that E p ˆ= n1 +2n2 n0 +n1 n1 +2n2 p is a polynomial in p, (n0 ,n1 ,n2 ):n0 +n1 +n2 =n n−n0 (1 − p) whose order is higher than 1 for p ∈ (0, 1). Then the above inequality does not hold for some p ∈ (0, 1), which contradicts the assumption.
0
|f (t)|dt +
0ห้องสมุดไป่ตู้
|f (t)|dt.
Proof. First use mean value theorem, there exist x0 so that f (x0 ) = 1 x f (t)dt. Then estimate f (x) − f (x0 ) = x0 f (t)dt. 0 3. Consider the equation x ¨ + (1 + f (t))x = 0. We assume that |f (t)|dt < ∞. Study the Lyapunov stability of the solution (x, x ˙ ) = (0, 0). Proof. Use V (x, y ) = 1/2(x2 + y 2 ) as a Lyapunov function, one can get |(log V ) | ≤ f (t), and estimate V . 4. Suppose f : [a, b] → R be a L1 -integrable function. Extend f to be 0 outside the interval [a, b]. Let φ(x) = Show that
1)
|n|=0
ˆ(n)|2 < ∞ implies f ∈ L2 [0, 2π ], |f ˆ(n)| < ∞ implies that f = f0 , a.e., f0 ∈ C 1 [0, 2π ], |nf where C 1 [0, 2π ] is the space of functions f over [0, 1] such that both f and its derivative f are continuous functions.
n
P (Zj < z ) =
k =j n
P {there are exactly k out of n of U1 , · · · , Un ≤ z }
k k Cn z (1 − z )n−k k =j
=
k = where Cn n
n! . k!(n−k)!
Using the equality that n! (j − 1)!(n − j )!
p 0
k k Cn p (1 − p)n−k = k =j
tj −1 (1 − t)n−j dt,
which can be shown by using integration by parts, it follows that n! P (Zj < z ) = (j − 1)!(n − j )!
z 0
n
2)
Proof. 1) Use Parceval identity. (Do we need to prove it?) 2) Use ˆ(n)e−inx and prove f0 ∈ C 1 [0, 2π ] and f = f0 , a.e. f0 = Σf 6. Suppose Ω ⊂ R3 to be a simply connected domain and Ω1 ⊂ Ω with boundary Γ. Let u be a harmonic function in Ω and M0 = (x0 , y0 , z0 ) ∈ Ω1 . Calculate the integral: 1 ∂u ∂ 1 II = − u ( )− dS, ∂n r r ∂n Γ ∂ 1 1 and where = denotes the out 2 2 2 r ∂n (x − x0 ) + (y − x0 ) + (z − x0 ) normal derivative with respect to boundary Γ of the domain Ω1 . (Hint: ∂v ∂v ∂v use the formula ∂n dS = ∂x dy ∧ dz + ∂y dz ∧ dx + ∂v dx ∧ dy .) ∂z
a b ∞
1 2h
x+ h
f
x− h b
|φ| ≤
a
|f |.
1
2
Proof.
b a
φ=
b a
1 −1
f (x + sh)dsdx, exchange the intergal and estimate.
ˆ(n) = 5. Suppose f ∈ L1 [0, 2π ], f
∞
1 2π
2π 0
f (x)e−inx dx, prove that
1
2
2. Let X1 , · · · , Xn be i.i.d. random variable with a continuous density f at point 0. Let 3 Yn,i = (1 − Xi2 /b2 n )I (|Xi | ≤ bn ). 4bn Show that n i=1 (Yn,i − EYn,i ) L −→ N (0, 3/5), 1 (bn n i=1 Yn,i ) /2 provided bn → 0 and nbn → ∞. 3. Let X1 , · · · , Xn be independently and indentically distributed random variables with Xi ∼ N (θ, 1). Suppose that it is known that |θ| ≤ τ , where τ is given. Show
S.-T. Yau College Student Mathematics Contests 2010
Analysis and Differential Equations
Team Solution
1. a) Let f (z ) be holomorphic in D: |z | < 1 and |f (z )| ≤ 1 (z ∈ D). Prove that |f (0)| − |z | |f (0)| + |z | ≤ |f (z )| ≤ . (z ∈ D) 1 + |f (0)||z | 1 − |f (0)||z | b) For any finite complex value a, prove that