5.6. EXERCISES
177
5.14. A branching process can be turned into a random walk if we only allow
one individual to die and be replaced by its ospring on each step. If the
ospring distributions is pk and
15 MOMENT GENERATING FUNCTIONS*
Uniform(a, b) rv: X () = (eb ea )/(b a)
Exponential() rv: X () = ( )1
Gamma(, ) rv: X () =
Normal(, 2 ) rv: X () = exp( +
2 2
2 )
The moment generating function (mg
21
A.2. SOLUTIONS FOR CHAPTER 2
The critical part of these calculations is the calculation of the volume, and we can do this
inductively by guessing from the previous equation that the volume, given s
20
A.2
APPENDIX A. SOLUTIONS TO EXERCISES
Solutions for Chapter 2
Exercise 2.1: a) Find the Erlang density fSn (t) by convolving fX (x) = exp(
x), x
0 with itself n
times.
Solution: For n = 2, we conv
13
A.1. SOLUTIONS FOR CHAPTER 1
Since this is nonnegative from (a), we see that
00
X (r)
00
= gX (r)gX (r)
0
[gX (r)]2
0.
d) Assume that X is non-atomic, i.e., that there is no value of c such that Pr
11
A.1. SOLUTIONS FOR CHAPTER 1
b) Is pairwise statistical independence enough to ensure that
hY n
i Yn
E
Xi =
E [Xi ] .
i=1
i=1
for a set of rvs X1 , . . . , Xn ?
Solution: No, (a) gives an example,
6
APPENDIX A. SOLUTIONS TO EXERCISES
To spell this out in greater detail, let Y = Y + + Y where Y + = maxcfw_0, Y and Y =
mincfw_Y, 0. Then Y = Y + +RY and |Y | = Y + Y = Y + + |Y |. Since E [Y + ] =
10
APPENDIX A. SOLUTIONS TO EXERCISES
is continuous to the right. The Riemann sum used to dene the Stieltjes integral is not
sensitive enough to see this step discontinuity at the step itself. Thus, t
7
A.1. SOLUTIONS FOR CHAPTER 1
b) The minimum of n IID rvs, each with CDF FX (x).
Solution: Let M be the minimum of X1 , . . . , Xn . Then, in the same way as in (a),
M > y if and only if Xj > y for 1
33
A.2. SOLUTIONS FOR CHAPTER 2
Note that the question here is not precise (there are obviously many sample paths, and
which are reasonable is a matter of interpretation). The reason for drawing such
34
A.3
APPENDIX A. SOLUTIONS TO EXERCISES
Solutions for Chapter 3
Exercise 3.1: a) Let X, Y be IID rvs, each with density fX (x) = exp( x2 /2). In (b), we show
p
that must be 1/ 2 in order for fX (x)
35
A.3. SOLUTIONS FOR CHAPTER 3
2
X1
=
2
X2
= 1.
Z 1
fZ (z) = fX1 (z) fX2 (z) =
fX1 (x)fX2 (z
1
Z 1
1
1
2
2
p e x /2 p e (z x) /2 dx
=
2
2
1
Z 1
1
2 xz+ z 2 )
2 dx
=
e (x
2 1
Z 1
z2
z2
1
2
=
e (x xz+
38
APPENDIX A. SOLUTIONS TO EXERCISES
2
by multiplying out the left two terms in [K][K 1 ] = [I]. We get X B + X Y C = 1
2 C = 0. From the second equation, C =
and X Y B + X
X B/ Y . Substituting thi
39
A.3. SOLUTIONS FOR CHAPTER 3
Solution: The quantity on the left above is [C] as derived in (a). By using the symmetry
T
between X and Y , we see that [DKX Y KX 1 ] is C T , and taking the transpose
32
APPENDIX A. SOLUTIONS TO EXERCISES
b) Use (a) to nd the joint density of S1 , . . . , Sn conditional on N (t) = n. Verify that your answer agrees
with (2.38).
Solution: For each i, the conditional
40
APPENDIX A. SOLUTIONS TO EXERCISES
p
Since the functions cfw_ 2B sinc(2Bt n); n 2 Z are orthonormal, the energy equation,
R1 2
P
(3.64), says that 1 h (t)dt = n 2Bh2 n/2B) . For the special case of
37
A.3. SOLUTIONS FOR CHAPTER 3
of the reciprocals of the eigenvalues. Since the eigenvalues are positive, their reciprocals are
also, so [K 1 ] is also positive denite. From (a), then, the matrices [
42
APPENDIX A. SOLUTIONS TO EXERCISES
Finally, suppose a cycle contains M states. If there is any transition Pim > 0 for which
(i, m) is not a transition on that cycle, then that transition can be add
8
APPENDIX A. SOLUTIONS TO EXERCISES
a) Show that the set of ! for which Z(!) is innite or undened has probability 0.
Solution: Note that Z can be innite (either 1) or undened only when either X or
Y
4
APPENDIX A. SOLUTIONS TO EXERCISES
Sample points
1
2
3
4
5
6
7
8
A1
1
1
1
1
0
0
0
0
A2
1
1
0
0
1
1
0
0
A3
1
0
1
0
0
0
1
1
b) Show that, for your example, A2 and A3 are not independent. Note that the
15 MOMENT GENERATING FUNCTIONS*
Proposition 1.15.1
Let the Xi s be independent rvs, and put Sn = X1 + + Xn . Then,
n
Sn () =
Xi ().
i=1
Proof:
Note that
Sn () = E(exp(X1 + + Xn )
n
exp(Xi )
= E(
i=1
n
16 APPLICATION OF THE LLN AND CLT TO FINANCIAL MATHEMATICS
Theorem 1.15.1
If, for each ,
Let (Xn : 1 n ) be a sequence of rvs with mgfs (Xn () : 1 n ).
Xn () X ()
as n , then
Xn X
as n .
Proof of the
13 THE LAW OF LARGE NUMBERS
Example 1.3.3(continued) Suppose that Sn is the number of defectives found in the rst n
items. The rv Sn is binomially distributed with parameters n and p. If n = 100 and p
Real World Applications of
Stochastic Models
June 24, 2005
Nathan Hardiman
Rebecca Scotchie
Seems like every actuarial
publication has articles on
requirements involving stochastic
models
This project
Life companies that formerly used a socalled factor approach to establish
reserveswill now have to become
experts in stochastic modeling.
Discussion Topics
Why use stochastic models?
Stochastic modeli
Why use Stochastic Models?
Regulatory requirements driving the use
of Stochastic Models:
Move toward principles-based regulation
Proposal for C-3 Phase II RBC
Proposed AG VACARVM
Regulation suggests u
Value Added Applications
Develop risk/reward profile of new or
existing products
Assess effect of assumptions in
extreme scenarios
Only as good as assumed relationship
E.g., assumed formula to model
the utilization of the GLBs is much
tougher to predict since it is driven by
policyholder behavior. Experience on the
utilization of these riders is still limited and
should be analyzed under a varie
16 APPLICATION OF THE LLN AND CLT TO FINANCIAL MATHEMATICS
n
log Vn = log V0 +
log Ri .
(1.16.2)
i=1
Because the Ri s are iid, it follows that the log Ri s are also iid. So, the LLN guarantees that
n