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Queen’s University
Department of Mathematics and Statistics
STAT 353
Final Examination April 21, 2011
Instructor: G. Takahara
• “Proctors are unable to respond to queries about the interpretation of exam questions. Do your best to answer exam questions as written.”
• “The candidate is urged to submit with the answer paper a clear statement of
any assumptions made if doubt exists as to the interpretation of any question that
requires a written answer.”
• Formulas and tables are attached.
• An 8.5 × 11 inch sheet of notes (both sides) is permitted.
• Simple calculators are permitted. HOWEVER, do reasonable simplifications.
• Write the answers in the space provided, continue on the backs of pages if needed.
• SHOW YOUR WORK CLEARLY. Correct answers without clear work showing
how you got there will not receive full marks.
• Marks per part question are shown in brackets at the right margin.
Marks: Please do not write in the space below.
Problem 1 [10]
Problem 4 [10]
Problem 2 [10]
Problem 5 [10]
Problem 3 [10]
Problem 6 [10]
Total: [60]
STAT 353 -- Final Exam, April 24, 2010
1.
Page 2 of 9
Let X and Y be independent Exponential random variables with mean 1. Let > 0
be a given positive number. Compute P (|X − Y | > ) by setting up and evaluating the
appropriate double integration.
[10]
Solution: The appropriate double integral would be
Z ∞Z
P (|X − Y | > ) =
e−(x+y) dxdy
0
Z
∞
{x>0:|x−y|>}
∞
−(x+y)
Z
Z
∞
Z
y−
dxdy +
e−(x+y) dxdy
0
Z0 ∞ y+
Z ∞ =
e−y−(y+) dy +
e−y (1 − e−(y− )dy
=
e
0
1 −
1
=
e + e− − e e−2
2
2
= e− .
STAT 353 -- Final Exam, April 24, 2010
Page 3 of 9
2. (a) Let X and Y be two zero mean random variables, each with the same variance σ 2 .
Note that we are not assuming that X and Y are independent or identically distributed.
Let ρ(X, Y ) denote the correlation coefficient between X and Y . Show that
ρ(X, Y ) ≥ −1 + P (|X + Y | ≥
√
Hint: Use Chebyshev’s inequality to bound P (|X + Y | ≥
2σ).
√
2σ).
Solution: Since the mean of X + Y is 0, by Chebyshev’s inequality, we have
P (|X + Y | ≥
√
Var(X + Y )
2σ 2
Var(X) + Var(Y ) + 2Cov(X + Y )
=
2σ 2
2
2
2σ + 2σ ρ(X, Y )
=
2σ 2
= 1 + ρ(X, Y ).
2σ) ≤
Therefore,
ρ(X, Y ) ≥ −1 + P (|X + Y | ≥
√
2σ).
[5]
STAT 353 -- Final Exam, April 24, 2010
Page 4 of 9
(b) As a generalization of part(a), suppose that X1 , . . . , Xn are zero mean random
variables, each with the same variance σ 2 . Also assume that the correlation coefficient
between any pair of the X’s is the same, and given by ρ. Show that
p
1
+ P |X| ≥ σ 1 − 1/n ,
ρ≥−
n−1
P
[5]
where X = n1 ni=1 Xi .
Solution: Proceeding as in part(a), since X has mean 0 we have again by Chebyshev’s
inequality that
P (|X| ≥ σ
p
1 − 1/n) ≤
=
=
=
=
=
=
Var(X)
− 1/n)
P
P
Cov( n1 ni=1 Xi , n1 ni=1 Xi )
σ 2 (1 − 1/n)
P
n Pn
1
i=1
j=1 Cov(Xi , Xj )
n2
σ 2 (1 − 1/n)
1
(nσ 2 + n(n − 1)σ 2 ρ)
n2
σ 2 (1 − 1/n)
1
(1 + (n − 1)ρ)
n
1 − 1/n
1 + (n − 1)ρ
n−1
1
+ ρ.
n−1
σ 2 (1
Therefore, we obtain
ρ≥−
p
1
+ P (|X| ≥ σ 1 − 1/n).
n−1
STAT 353 -- Final Exam, April 24, 2010
Page 5 of 9
3. Let X be a random variable with 0 mean and let Y be a random variable with mean µ
and variance σ 2 . Assume that X and Y are independent.
(a) If µ = 0 show that X and XY are uncorrelated.
[3]
Solution: We compute
Cov(X, XY ) = E[X 2 Y ] − E[X]E[XY ]
= E[X 2 ]E[Y ] − E[X]E[X]E[Y ]
(by independence)
= 0−0=0
if E[Y ] = µ = 0. Hence, X and XY are uncorrelated.
(b) In general, show that the square of the correlation coefficient between X and XY
is ρ2 (X, XY ) = µ2 /(σ 2 + µ2 ).
[7]
Solution: Computing Cov(X, XY ) and Var(XY ), again using independence, we have
Cov(X, XY ) = E[X 2 Y ] − E[X]E[XY ] = E[X 2 ]µ − E[X]2 µ = µVar(X)
and
Var(XY ) = E[X 2 Y 2 ] − E[XY ]2 = E[X 2 ]E[Y 2 ] − E[X]2 µ2
= E[X 2 ](σ 2 + µ2 ) − E[X]2 µ2
= Var(X)(σ 2 + µ2 ),
since E[X] = 0 (and so Var(X) = E[X 2 ]). Therefore,
Cov2 (X, XY )
µ2
µ2 Var2 (X)
ρ (X, XY ) =
=
=
.
Var(X)Var(XY )
σ 2 + µ2
Var2 (X)(σ 2 + µ2 )
2
STAT 353 -- Final Exam, April 24, 2010
Page 6 of 9
4. We have 2 coins and n flips are performed as follows. For the first flip we pick a coin at
random and flip it. For i = 2, . . . , n we flip coin 1 if the (i − 1)st flip was heads and we
flip coin 2 if the (i − 1)st flip was tails. Suppose the probability of heads for coins 1 and
2 are, respectively, p1 and p2 . Let X denote the number of heads that are flipped in the
n flips. Find E[X].
[10]
Solution: Let Xi = 1 if the ith flip is heads and Xi = 0 if the ith flip is tails, for
i = 1, . . . , n. Then we have X = X1 + . . . + Xn and E[X] = E[X1 ] + . . . + E[Xn ]. Since
we pick the first coin at random we have (using the law of total probability) E[X1 ] =
P (X1 = 1) = 21 (p1 + p2 ). For i = 2, . . . , n we may condition on Xi−1 to obtain
E[Xi ] = P (Xi = 1)
= P (Xi = 1 Xi−1 = 1)P (Xi−1 = 1) + P (Xi = 1 Xi−1 = 0)P (Xi−1 = 0)
= p1 E[Xi−1 ] + p2 (1 − E[Xi−1 ])
= p2 + (p1 − p2 )E[Xi−1 ].
Recursively, then, we obtain
E[Xi ] = p2 + (p1 − p2 )E[Xi−1 ]
= p2 + p2 (p1 − p2 ) + (p1 − p2 )2 E[Xi−2 ]
..
.
= p2 + p2 (p1 − p2 ) + . . . + p2 (p1 − p2 )i−2 + (p1 − p2 )i−1 E[X1 ]
1
= p2 + p2 (p1 − p2 ) + . . . + p2 (p1 − p2 )i−2 + (p1 − p2 )i−1 (p1 + p2 )
2
1
1 − (p1 − p2 )i−1
= p2
+ (p1 − p2 )i−1 (p1 + p2 )
1 − (p1 − p2 )
2
p2
p1 + p2
p2
i−1
− (p1 − p2 )
−
=
.
1 − (p1 − p2 )
1 − (p1 − p2 )
2
The above holds also for i = 1. Summing over i from 1 to n we obtain
p2
np2
1 − (p1 − p2 )n
p1 + p2
E[X] =
−
−
.
1 − (p1 − p2 )
1 − (p1 − p2 ) 1 − (p1 − p2 )
2
The answer is fine in this form.
STAT 353 -- Final Exam, April 24, 2010
Page 7 of 9
5. Let X1 , X2 , . . . be a sequence of random variables satisfying E[Xn ] = µn and Var(Xn ) ≤
M/np for n ≥ 1, where M and p are positive constants. Suppose that µn → µ as n → ∞.
(a) If p > 0 show that Xn → µ in probability as n → ∞. Hint: Use the triangle
inequality |a + b| ≤ |a| + |b| and Chebyshev’s inequality.
[7]
Solution: Let > 0 be given. Then
P (|Xn − µ| > ) = P (|Xn − µn + µn − µ| > ) ≤ P (|Xn − µn | + |µn − µ| > ),
since by the triangle inequality |Xn − µn + µn − µ| > implies |Xn − µn | + |µn − µ| > .
Since µn → µ as n → ∞, choose N so that |µn − µ| < for all n ≥ N . For such n, using
Chebyshev’s inequality, we have
P (|Xn − µ| > ) ≤ P (|Xn − µn | + |µn − µ| > )
= P (|Xn − µn | > − |µn − µ|)
Var(Xn )
≤
( − |µn − µ|)2
M
,
≤
np ( − |µn − µ|)2
(1)
which converges to 0 as n → ∞ if p > 0. Hence Xn → µ in probability.
(b) If p = 2 show that Xn → µ almost surely.
[3]
Solution: If p = 2 then the last term in Eq.(1) above is convergent when summed over n
and since this upper bounds P (|Xn − µ| > ), we have
∞
X
P (|Xn − µ| > ) < ∞.
n=1
This implies that Xn → µ almost surely (from a theorem in class).
STAT 353 -- Final Exam, April 24, 2010
Page 8 of 9
6. Let X1 , X2 , . . . be independent and identically distributed random variables with mean
P
√
µ and variance σ 2 < ∞. Let Zn = (X n − µ)/(σ/ n) for n ≥ 1, where X n = n1 ni=1 Xi .
Let Z be another random variable with a N (0, 1) distribution and suppose that Z is
independent of Zn for all n. Show that Zn does not converge to Z in probability. [10]
Solution: Let > 0 be given. We need to show that P (|Zn − Z| > ) does not converge
to 0 as n → ∞. One way to see this is to first condition on Z, giving
Z ∞
1
2
P (|Zn − Z| > Z = z) √ e−z /2 dz
P (|Zn − Z| > ) =
2π
Z−∞
∞
1 −z2 /2
dz
P (|Zn − z| > ) √ e
=
2π
−∞
Z ∞h
i 1
2
P (Zn < z − ) + P (Zn > z + ) √ e−z /2 dz
=
2π
Z−∞
i 1
∞h
2
P (Zn ≤ z − ) − P (Zn = z − ) + P (Zn > z + ) √ e−z /2 dz,
=
2π
−∞
where the second inequality loses the conditioning on Z because Zn is assumed to be
independent of Z. By the central limit theorem, as n → ∞ we have P (Zn ≤ z − ) →
Φ(z − ) and P (Zn > z + ) → 1 − Φ(z + ) (and P (Zn = z − ) → 0), where Φ is the
standard normal cdf. Hence we obtain
Z ∞h
i 1
2
Φ(z − ) + 1 − Φ(z + ) √ e−z /2 dz
P (|Zn − Z| > ) →
2π
Z−∞
∞
1
2
=
P (|Y − z| > ) √ e−z /2 dz,
2π
−∞
where Y has a N (0, 1) distribution. The above is clearly nonzero.
STAT 353 -- Final Exam, April 24, 2010
Page 9 of 9
Formula Sheet
Special Distributions
Exponential with parameter λ:

λe−λx if x > 0
f (x) =
0
otherwise.
E[X] =
1
,
λ
Var(X) =
1
.
λ2
Normal (Gaussian) with mean µ and variance σ 2 :
Z x
(x−µ)2
(t−µ)2
1
1
−
2
2σ
f (x) = √ e
e− 2σ2 dt.
and F (x) = √
σ 2π
σ 2π −∞