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 the generating function is then the random
walk increm
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 (mgf) of a rv X gets its name from the fact that the momen
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 sn , of the n 1
dimensional space where 0 < s1 < < sn 1
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 convolve fX (x) with itself.
fS2 (t) =
Z
t
0
fX1 (x)fX2 (t
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 Prcfw_X = c = 1. Show that the
inequality sign
may be r
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, i.e., E [XY Z] = 0 and E [X] E [Y ] E [Z] = 1/8.
Exerci
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 + ] =
R
c
y 0 FY (y) dy and E [Y ] =
y<0 FY (y) dy, the abov
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, the stipulation that
Z be continuous on the right must b
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 j n and for all choice of y. We could make the
same st
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 sketches
is to acquire understanding to guide the solut
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) to integrate to 1, but in this part, we leave undetermi
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+ 4 ) 4 dx
2 1
Z 1
z 2
1
1
z 2 /4
p e
p e (x 2 ) dx
=
2
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 this
2 (1
2 )]. Substituting this into the equation for C,
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 completes
the argument.
c) Show that [KV 1 G] = [H T K
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 probability in (a) is clearly independent of Si 1 , Si
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 t = 0, this is the
same as what is to be shown (summin
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 [B] and [D] are also
positive denite.
Exercise 3.13: a)
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 added to the cycle and
all the transitions between i and m
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 are innite or undened. Since these are events of zero p
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 denition of statistical independence would be very str
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
=
E(exp(Xi ) (due to independence)
i=1
n
Xi ().
=
i=1
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 CLT:
Let
n () = E[exp(Sn E(X1 )/ n)]
and note that
n ()
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 = 0.01, the
approximation (1.12.3) establishes that th
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 takes another step down an
evolutionary path that incl
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 modeling mini-tutorial
Value-added applications of
stochastic
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 use of stochastic
modeling
GAAP SOP 03-1
Coming Soon
If
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 policyholder
behavior
Two formulas thought to be equal
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 variety of market
scenarios.
2
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
1
n
i=1
p
log Ri E(log R1 )
as n , leading to the appro