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The solutions to exam 2 of math 213, including problems on probability, recurrence relations, and combinatorics. Students can use this document as a study resource to review concepts covered in the exam.
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Math 213 Exam 2 9Solutions) Prof. I.Kapovich April 4, 2008 Problem 1[20 points] For each of the following statements indicate whether it is true or false. You DO NOT have to explain your answers.
(1) For any random variables X and Y we have V (X + Y ) = V X + V Y. (2) For S = {− 5 , 3 , 1 , 5 }, ordered as listed, the permutation 3, 1 , 5 , − 5 is a derangement. (3) Let A be the set of all bit-strings of length 10. Let R be a relation on A consisting of all pairs (x, y) ∈ A × A such that the strings x and y agree in their first two bits. Then R is an equivalence relation with exactly two equivalence classes. (4) Consider a recurrence relation an = 3an− 1 + n. Then for every c, α, β ∈ R, an = c 3 n^ + α + βn is a solution of the above recurrence relation.
Answers: (1) False. (The statement would be true if we knew that X and Y are independent random variables).
(2) True. (3) False. (Although R is an equivalence relation, it has four distinct equivalence classes, corresponding to 4 distinct binary strings of length two that can occur as initial segments of length two of strings from A).
(4) False. (One can find a particular solution of this recurrence relation of the form an = α + βn, but the values of α and β that work are not arbitrary. In this case these specific values are α = − 34 , β = − 12 and the
general solution of the original recurrence relation is an = c 3 n^ − 34 − 12 , where c ∈ R is arbitrary.).
Problem 2[20 points] Let n ≥ 1 be an integer. A fair die is rolled n times, independently. Let Z = 3X + 2Y where X is the number of tosses that came 1 up and where Y is the number of tosses that came 6 up. Find EZ and V Z. Give all the details of your work.
Solution. For i = 1,... , n let
Zi =
3 , if the i-th toss is 1, 2 , if the i-th toss is 6, 0 , otherwise. 1
2
Then Z = Z 1 +· · ·+Zn and the random variables Z 1 ,... , Zn are independent. For each i = 1,... , n we have
EZi = 3
and
E(Z i^2 ) = 3^2
Hence V Zi = E(Z^2 i ) − (EZi)^2 = 136 − 2536 = 5336. Since Z = Z 1 + · · · + Zn, it follows that
EZ = EZ 1 + · · · + EZn =
5 n 6 and, since Z 1 ,... , Zn are independent,
V Z = V Z 1 + · · · + V Zn =
53 n 36
Problem 3[20 points] Find the general solution of the recurrence relation
(†) an = 2an− 1 − an− 2 + 4.
Give all the details of your work.
Solution. The associated homogeneous relation is an = 2an− 1 − an− 2. Its charac- teristic equation is r^2 = 2r − 1 which factors as (r − 1)^2 = 0. Hence the homogeneous problem an = 2an− 1 − an− 2 has the general solution
a( nh )= α 1 n^ + βn 1 n^ = α + βn
where α, β ∈ R are arbitrary constants. Since 4 = 4 · 1 n, we should look for a particular solution of (†) of the form
a( np )= cn^2 ,
where c ∈ R. To find the specific value of c, we substitute a( np )= cn^2 into (†):
cn^2 = 2c(n − 1)^2 − c(n − 2)^2 + 4 cn^2 = 2cn^2 − 4 cn + 2c − cn^2 + 4cn − 4 c + 4 2 c = 4, c = 2.
Thus a( np )= 2n^2 is a particular solution of (†). Therefore the general solution of (†) is
an = a( nh )+ a( np )= α + βn + 2n^2 ,
where α, β ∈ R are arbitrary constants.
Problem 4[20 points]