University of California, Los Angeles Department of Statistics
Instructor: Nicolas Christou
Practice 2 – solutions
36
√
b. According to the central limit theorem T ∼ N (36(3), 0.28 36) or T ∼ N (108, 1.68). The histogram should be centered
around 108 with spread (103, 113).
EXERCISE 2
It is given X ∼ N (2700, 400). The total supply for n = 12 weeks is 4000 + 12(2500) = 34000. We want the supply to be below 2000 pounds or the total sugar use in these 12 weeks to be more than 32000 pounds:
Statistics 100B
EXERCISE 1
This population has mean μ = 3, and standard deviation σ = 0.28.
̄ ̄ 0.28 ̄
) or X ∼ N(3,0.047). We expect that the range 3 ± 3(0.047) or 3 ± 0.141 or (2.86, 3.141) will cover almost all the area. We conclude that the histogram should
a. According to the central limit theorem X is distributed as X ∼ N(3, √ have been much narrower.
P(T >32000)=P EXERCISE 3
32000 − 12(2700)
Z > √ =P(Z >−0.29)=0.6141.
400 12
We know the the moment generating function of N(μ,σ) is MX(t) = e 2 .
a. Moment generating function of X + Y :
MX+Y (t) = MX(t)MY (t) = e2μt+t2σ2
Moment generating function of X − Y : MX−Y (s) = MX(s)M−Y (s) = es2σ2
b. Joint moment generating function of X + Y, X − Y :
MX+Y,X−Y (t, s) =
Ee(X+Y )t+(X−Y )s
= EeX(t+s)+Y (t−s)
= MX(t+s)MY (t−s)
μ(t+s)+ 1 (t+s)2 σ2 μ(t−s)+ 1 (t−s)2 σ2 =e2e2
= e2μt+t2σ2 et2σ2 = MX+Y (t)MX−Y (s).
c. Since the joint moment generating function of X + Y and X − Y can be expressed as the product of the moment
generating functions of X + Y and X − Y we conclude that X + Y and X − Y are independent. EXERCISE 4
a. We can write X1 − 2X2 + X3 as a′X where a′ = (1, −2, 1). Therefore
5231 var(aX)=a′Σa= 1 −2 1 2 3 0 −2 =18.
303 1
b. LetA= 1 1 0 . ThenY= Y1 =AX. Therefore, 111 Y2
′ 1 1 05 2 31 1 12 15 var(Y)=AΣA= 1 1 1 2 3 0 1 1 = 15 21 .
303 01
μt+ 1 t2σ2
EXERCISE 5
a. Letkbetheminimumnumberofplaysthecasinomustwin. Then1×k−(10000−k)×35>400. Solveforktoget k = 9734. Let Y be the number of games the casino must win, so Y ∼ b(10000, 37 ). The casino must win at least 9734
38
plays and this probability is P (Y ≥ 9734) = 10000 10000 37 y 1 10000−y . Using normal approximation to binomial:
y=9734 y 38 38 P(Y ≥9734)=P(Z > 1000037 1 38 )=P(Z >−0.21)=0.5832.
9733.5−10000 37 38 38
b. If we view the 10000 outcomes as a random sample from the following distribution then we can use the central limit theorem: Let T = X1 + . . . X10000 be the sum of the 10000 outcomes.
X P(X)
37
38 1
38
This distribution has μ = 0.05263 and σ = 5.76.
400−10000(0.05263)
P(T >400)=P(Z > √ )=P(Z >−0.22)=0.5871.
5.76 10000
EXERCISE 6
X t′μ+1t′Σt
1 -35
WehaveZ= Y ∼ N2(μ,Σ) and therefore the joint moment generating function of (Xi,Yi) is e 2 moment generating function of (X ̄,Y ̄) is:
. The joint
̄ ̄ X1 +…+Xn Eet1X+t2Y = Eet1 n
Y1 +…+Yn n
+t2
= Eet1 n +t2 n × … × Eet1 n +t2 n , because the pairs (Xi,Yi) are independent.
X1
Y1
Xn Yn
1 0 X
Each one of these expectations is the joint moment generating function of AZ with A = n 1 and Z = . Since
0nY μ Σ t′ μ + 1 t′ Σ t
E(AZ) = n and var(AZ) = n2 , it follows that the joint moment generating function of AZ is e n 2 n2 . But we have n
′ μ 1 ′ Σ n independent pairs, therefore te joint moment generating function of (X ̄,Y ̄) is et n +2t n2 t
that the joint distribution of (X ̄,Y ̄) is bivariate normal N2(μ, Σ). n
EXERCISE 7
′1′Σ
= et μ+2t n t. This shows
Y1 WewriteY= Y2 Y3
X1 √3 √3 √3 11
111
=AX,whereX=
var(AX) = AA′ = I3, which means Y1, Y2, Y3 are i.i.d. N(0, 1).
112
X2 ∼N3(0,I)andA= √2 −√2 0 . Therefore,var(Y)=
X3
√ √ −√ 666
EXERCISE 8
Let X = (X1,X2,X3) has joint moment generating function
MX(t1, t2, t3) = (1 − t1 + 2t2)−4(1 − t1 + 3t3)−3(1 − t1)−2. Answer the following questions:
a. Find the moment generating function of (X1 , X3 ).
MX1,X3 (t1, t3) = MX(t1, 0, t3) = (1 − t1)−6(1 − t1 + 3t3)−3.
b. Find the moment generating function of X1. MX1 (t1) = MX(t1, 0, 0) = (1 − t1)−9.
c. Find the moment generating function of X3. MX3 (t3) = MX(0, 0, t3) = (1 + 3t3)−3.
d. Are X1, X3 independent?
No, because MX1,X3 (t1, t3) ̸= MX1 (t1) × MX3 (t3).
e. Find the moment generating function of (X2 , X3 ). MX2,X3 (t2, t3) = MX(0, t2, t3) = (1 + 2t2)−4(1 + 3t3)−3.
f. Are X2, X3 independent?
MX2 (t2) = MX(0, t2, 0) = (1 + 2t2)−4. Yes, X2, X3 are independent because MX2,X3 (t2, t3) = MX2 (t2) × MX3 (t3).