Assignment 3 ANSWER KEY Assigned February 11, Due February 18 PubH 8462 Advanced Survival Analysis Instructor: David M. Vock

1. Verify that the optional variation process is an unbiased estimators of the variance. Let m(t) be a martingale with respect to the filtration Ft . The optional variation process is: X [m](t) = = [4m(s)]2 s≤t

=

X {m(s) − m(s− )}2

=

X

s≤t

m2 (s) − 2m(s)m(s− ) + m2 (s− )

s≤t

note E{[m](t)}

X = E{ m2 (s) − 2m(s)m(s− ) + m2 (s− )} s≤t

=

X

=

X

=

X

=

X

E[E{m2 (s) − 2m(s)m(s− ) + m2 (s− )|Fs− }]

s≤t

E{m2 (s)} − E[2E{m(s)|Fs− }m(s− )] + E{m2 (s− )}

s≤t

E{m2 (s)} − 2E[m2 (s− )] + E{m2 (s− )}

s≤t

E{m2 (s)} − E{m2 (s− )}

s≤t

Telescoping sum if we think of the above as limit

Pn

i=1

where n → ∞ and m(s0 ) = 0

2

Thus, E{m (t)} = var(m(t)). Alternatively if we think of m(t) as a counting process martingale in continuous time then X [m](t) = [4m(s)]2 s≤t

X = [4N (s)]2 s≤t

= N (t)

1

Note that E{N (t)}

= = = =

Z t E{ dN (u)} 0 Z t E{dN (u)} 0 Z t E[E{dN (u)|Fu− }] 0 Z t E{yi (u)λi (u)du} 0

= E[< m > (t)] = var{m(t)} 2. Using any result given in lecture, prove that the Doob-Meyer decomposition must be unique (as). Let m(t) ¯ be a submargingle. Assume that there exists martingales m1 (t) and m2 (t) wrt Ft− and predictable process C1 (t) and C2 (t) wrt Ft− so that C1 (0) = 0 and C2 (0) = 0 and m(t) ¯

=

m1 (t) + C1 (t) = m2 (t) + C2 (t)

So m1 (t) − m2 (t)

=

C2 (t) − C1 (t)

Now the difference of two martingales wrt the same filtration is a martingale. Similarly the difference of 2 predictable processes is a predictable process. Thus m1 (t)−m2 (t) is a martingale and C2 (t)−C1 (t) is a predictable process. Thus X(t) = m1 (t) − m2 (t) = C2 (t) − C1 (t) is both a predictable process and a martingale. We showed in class that if X(t) is both a martingale and predictable process it must be a constant a.s. Thus m2 (t) = m1 (t) + k C2 (t) = C1 (t) − k for some constant k. If C2 (t) = 0 and C1 (t) = 0 then k=0. Thus the decomposition is unique (a.s.). 3. K&P 5.2: Note that for part (b) we are looking for the unconditional mean. Also for part (c) plot five of the realizations of N· (t) and the corresponding martingale before plotting the average of the 30 realizations. Also, part(c) does not specify a value for β so please use β = 1. (a)Intensity function for counting process N. (t)

= E{dN. (t)|Ft− } n X = E{ dN. (t)|Ft− } i=1

= =

n X i=1 n X i=1

2

E{dN. (t)|Ft− } yi (u) exp{xi (t)β}α

(b) Note that over the interval where xi (t) is constant, the counting process is a homogeneous poisson process. That is if xt (t) is constant between t1 and t2 then E{Ni (t2 ) − Ni (t1 )} = exp(β)α(t2 − t1 ). For any t ≤ τi E{Ni (t)|Ai } = note E{I(t > Ai )}

=

E{I(t > Ai )|Ai }

=

E[E{Ni (t)|Ai }]

I(t > Ai ){exp(β)αAi + α(t − Ai )} + I(t ≤ Ai ) exp(β)αt τ −t t , E{I(t ≤ Ai )} = τ τ Z t 1 Ai dAi τ 0

=

Ai 2 t t2 |0 = 2τ 2τ

=

{exp(β) − α}

= =

t2 α τ − t t2 + + exp(β)αt 2τ 2τ τ t2 exp(β)α t2 α t2 α τ exp(β)αt t2 exp(β)α − + + − 2τ 2τ τ τ τ 2 2 t α t exp(β)α − + t exp(β)α 2τ 2τ

for t > τi , E{Ni (t)} = E{Ni (τi )} thus E{N. (t)} =

n X t∗2 α i=1





t∗2 i exp(β)α + t∗i exp(β)α 2τ

where t∗i = min(t, τi ) 4. K&P 5.3

Z < u > (t)

t

var{du(t)|Ft− }

= 0

Z

t

E{du(t)du(t)T |Ft− }

= 0

Z

t

=

n n X X E[{ Gi (u)dmi (u)}{ Gi (u)dmi (u)}T |Ft− ]

0

= =

i=1

Z tX n 0 i=1 Z tX n

i=1

E[{Gi (u)dmi (u)}{Gi (u)dmi (u)}T |Ft− ], m1 , ...mn are orthogonal Gi (u)Gi (u)T E[{dmi (u)}2 |Ft− ], Gi (u) is predictable

0 i=1

= =

Z tX n 0 i=1 n XZ t i=1

Gi (u)Gi (u)T d < mi > (u) Gi (u)Gi (u)T d < mi > (u)

0

5. Let M (t), t ∈ [0, τ ] be a martingale with respect to the filtration {Ft , t ∈ [0, τ ]}. Suppose M (0) = 0 (wp1). For 0 < s < t ≤ u < v < τ show that cov{M (t) − M (s), M (v) − M (u)} = 0. Because M (t) has zero mean for all t, cov(M (t)−M (s), M (v)−M (u))=E[(M (t)−M (s))(M (v)−M (u))]. Now condition on Fu . 3

E{E[(M (t) − M (s))(M (v) − M (u))|Fu ]} = E{[M (t) − M (s)]E[M (v) − M (u)|Fu ]} But E[(m(v) − M (u))|Fu ] = 0 .s. by the martingale property. By taking expectations, we see the covariance is 0. 6. Let Y and X1 , X2 , . . . be random variables defined on (Ω, F, P ) with EY 2 < ∞. Let Fn = σ(X1 , X2 , . . . , Xn ). Let Yn = E(Y |Fn ). Prove that {Yn } is a margingale sequence with respect to the filtration {Fn , n ≥ 1}. Since this is a discrete time martingale we do not need to worry about the cadlag property. For each n, Yn is adapted to Fn by its definition. The key martingale property is easy to establish. E(Yn |Fn−1 )

=

E(Y |Fn |Fn−1 ), definition of Yn

= E(Y |Fn−1 ), Tower property of conditional expectation = Yn−1 , definition of Yn−1

4

Answer Key

E[E1m2(s) - 2m(s)m(s−) + m2(s−)|Fs− l]. = ∑ ... n i=1 where n → с and m(s0)=0. Thus, E1m2(t)l = var(m(t)). Alternatively if we think of m(t) as a counting process ...

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