HW4

Our tasks

Based on below derivation, you need to finish:

1. Simulate A(t):

• Generate A(t) numerically with nsim≥50,000.

• Summarize the simulated mean and variance.

2. Compute Theoretical E[A(t)]:

• Calculate E[A(t)] in R using the same parameters as the simulation.

3. Compute Theoretical E[A(t)2]:

• Compute E[A(t)2] with consistent parameter values.

4. Compare Results:

• Calculate theoretical variance Var(A(t)) as:

Var(A(t))=E[A(t)2]−(E[A(t)])2.

• Compare the theoretical mean and variance with simulation results.

• Try multiple combination of parameters for comparing the theoretical mean and variance with

simulation results.

To fully show your efforts, please submit html or pdf output with r markdown file.

Problem description

For a homogeneous Poisson process with rate λ, we aim to derive the variance of the random variable A(t),

which is defined as:

N(t)

X

A(t)= A′(t,t ), t∈[0,T]

i

i=1

where A′(t,t )= C exp(−γ(t−t )).

i 1+αi i

The derivation follows from the most basic definitions, starting step by step to achieve the final result. Key

assumptions are:

• N(t) denotes the number of events occurring in [0,t], such that N(t)∼Poisson(λt).

• Event times T ,T ,...,T who has values of t ,t ,...,t are uniformly distributed in [0,T],

1 2 N(T) 1 2 N(T)

without conditional on N(t)=n. If we conditional on N(t)=n, T has marginal distribution of

k

Beta(k,n+1−k)

• Parameters C >0,α>0,γ >0 are given.

1

Derivation for unconditional expectation and variance

Conditional Event times distribution

ByAppendixA,theorderedstatisticsU hasmarginaldistributionofBeta(k,n−k+1). Therandomvariable

(k)

for event times T in the homogeneous Poisson process defined as T =T ·U . Using the transformation

k k (k)

T =T ·U , we find the probability density function (PDF) of T by change of variables:

k (k) k

T du 1

T =T ·u =⇒ u= k, = .

k T dT T

k

Substitute into the Beta PDF:

(cid:18) t (cid:19) 1

f (t)=f · .

Tk U(k) T T

This becomes:

(cid:0)x(cid:1)i−1(cid:0)1− x(cid:1)n−i 1

f (x)= T T · , x∈[0,T],

T(i) B(i,n+1−i) T

xi−1(T −x)n−i

= , x∈[0,T],

TnB(i,n+1−i)

n! xi−1(T −x)n−i

= , 0

(i−1)!(n−i)! Tn

where B(k,n+1−k)=R1 uk−1(1−u)n−kdu= (k−1)!(n−k)!.

0 n!

Similarly, the joint density function of T and T , k ≤l is:

(k) (l)

n!

f (x,y)= xk−1(y−x)l−k−1(T−y)n−l, 0

T(i),T(j)|N(t)=n Tn(k−1)!(l−k−1)!(n−l)!

Conditional Expectation E[A(t)|N(t) = n]

First we derive the conditional expectation E(A(t)|N(t)=n):

n !

X C

E(A(t)|N(t)=n)=E e−γ(t−T(i)) |N(t)=n

1+αi

i=1

n

X C h i

= E e−γ(t−T(i)) |N(t)=n

1+αi

i=1

Ce−γt (cid:20) (cid:18) α (cid:19) (cid:18) α (cid:19)(cid:21) Xn

= ψ +n+1 −ψ +1 · E(eγT(i) |N(t)=n)

α γ γ

i=1

where the digamma function ψ(x)= d ln(Γ(x))= Γ′(x), and Γ(x)=R∞ zx−1e−zdz

dx Γ(x) 0

For E(cid:0) eγT(i) |N(t)=n(cid:1), by f T(i)|N(t)=n(x)= (i−1)n !(! n−i)!xi−1(t t− nx)n−i , 0

2

Z t

E(cid:0) eγT(i) |N(t)=n(cid:1)= eγxf T(i)|N(t)=n(x)dx

0

Z t n! xi−1(t−x)n−i

= eγx dx.

(i−1)!(n−i)! tn

0

Setting x = t·u, dx = tdu, 0 ≤ u ≤ 1, then substituting xi−1 = (tu)i−1 = ti−1ui−1, and (t−x)n−i =

(t−tu)n−i =tn−i(1−u)n−i into the integral:

n! Z 1 ti−1ui−1tn−i(1−u)n−i

E(eγT(i) |N(t)=n)= eγtu tdu

(i−1)!(n−i)! tn

0

n! Z 1

= eγtuui−1(1−u)n−idu

(i−1)!(n−i)!

0

= F (i;n+1;γt),

1 1

where F is the confluent hypergeometric function of the first kind, defined as: F (a;c;z) =

1 1 1 1

Γ(a)Γ Γ( (c c) −a)R 01 ezuua−1(1−u)c−a−1du. And similarly we can have E[emγT(i)|N(t)=n]=

1

F 1(i;n+1;mγt)

Thus we obtained E[A(t)|N(t)=n]=Pn Ce−λt F (i;n+1;γt).

i=1 1+αi 1 1

Unconditional Expectation E[A(t)]

Using the law of total expectation:

∞

X

E[A(t)]= E[A(t)|N(t)=n]P(N(t)=n).

n=1

Substituting P(N(t)=n)= (λt)nexp(−λt):

n!

X∞ e−λt(λt)n

E[A(t)]= E(A(t)|N(t)=n)

n!

n=1

=

X∞ e−λt(λt)n " Ce−γtXn 1F 1(i;n+1;γt)#

.

n! 1+αi

n=1 i=1

Unconditional second moment E[A(t)2]

First, let’s derive the conditional moment:

Define Q

(i)

=A′(t,T (i))= 1+C αie−γteγT(i):

n !2

X

A2(t)= Q

(i)

i=1

n

X X

= Q2 +2 Q Q .

(i) (i) (j)

i=1 1≤i

3

thus,

n

X X

E(A2(t)|N(t)=n)= E(Q2 |N(t)=n)+2 E(Q Q |N(t)=n)

(i) (i) (j)

i=1 1≤i

=(cid:18) C (cid:19)2 e−2γtE(cid:2) e2γT(i) |N(t)=n(cid:3)+2 C2e−2γt Eh eγ(T(i)+T(j)) |N(t)=ni

1+αi (1+αi)(1+αj)

(cid:18) C (cid:19)2 C2e−2γt h i

=

1+αi

e−2γt 1F 1(i;n+1;2γt)+2 (1+αi)(1+αj)E eγ(T(i)+T(j)) |N(t)=n

 

=C2e−2γt

Xn 1F 1( (i 1;n ++ αi1 ); 22γt)

+2

X E[e (γ 1(T +i+ αTj i) )(| 1N +(t α) j=

)

n]

.

i=1 1≤i

where

Z tZ t

E[eγ(T(i)+T(j)) |N(t)=n]= eγ(x+y)f T(i),T(j)|N(t)=n(x,y)dydx

0 x

n! Z tZ t

= eγ(x+y)xi−1(y−x)j−i−1(t−y)n−jdydx.

(i−1)!(j−i−1)!(n−j)!tn

0 x

by the law of total expectation and substituting the conditional expectation derived previously:

  

E[A(t)2]=

X∞

C2e−2γt

Xn 1F 1( (i 1;n ++ αi1 ); 22γt)

+2

X E[e (γ 1(T +i+ αTj i) )(| 1N +(t α) j=

)

n] e−λt(λ nt !)n

.

n=1 i=1 1≤i

Combining terms:

 

E[A(t)2]=C2e−(λ+2γ)tX∞ (λ nt !)n Xn 1F 1( (i 1;n ++ αi1 ); 22γt)

+2

X E[e (γ 1(T +i+ αTj i) )(| 1N +(t α) j=

)

n]

.

n=1 i=1 1≤i

where 1F 1(i;n+1;2γt) and E[eγ(Ti+Tj) |N(t)=n] are defined before.

Variance Var(A(t))

By definition of variance we have:

Var(A(t))=E[A(t)2]−(E[A(t)])2.

where E[A(t)] and E[A(t)2]are defined previously.

In summary,

E[A(t)|N(t)=n]=Ce−γtXn 1F 1(i;n+1;γt)

.

1+αi

i=1

4

E[A(t)]=Ce−(λ+γ)tX∞ (λt)n Xn 1F 1(i;n+1;γt)

.

n! 1+αi

n=1 i=1

h i n! Z tZ t

E eγ(T(i)+T(j)) |N(t)=n = eγ(x+y)xi−1(y−x)j−i−1(t−y)n−jdydx.

(i−1)!(j−i−1)!(n−j)!tn

0 x

 

E[A(t)2 |N(t)=n]=C2e−2γt

Xn 1F 1( (i 1;n ++ αi1 ); 22γt)

+2

X E(cid:2) eγ ((T 1(i +)+ αT( ij )) () 1| +N α(t j) )=n(cid:3)

.

i=1 1≤i

 

E[A(t)2]=C2e−(λ+2γ)tX∞ (λ nt !)n Xn 1F 1( (i 1;n ++ αi1 ); 22γt)

+2

X E(cid:2) eγ ((T 1(i +)+ αT( ij )) () 1| +N α(t j) )=n(cid:3)

.

n=1 i=1 1≤i

Var(A(t))=E[A(t)2]−(E[A(t)])2.

Appendix A: Order Statistics: Marginal and Joint Distributions

Let X ,X ,...,X ∼ i.i.d.F, where F is the cumulative distribution function (CDF) of the sample with

1 2 n

probability density function p(x).

Denote X ≤X ≤···≤X as the ordered statistics of the sample.

(1) (2) (n)

We derive the marginal and joint probability density functions of the ordered statistics X and X (k ≤l)

(k) (l)

under a uniform distribution U(0,1).

Marginal Distribution of X

(k)

ToderivethedensityofX inasmallinterval[x,x+∆x], thefollowingmustoccur: k−1observationsmust

(k)

lie strictly less than x, n−k observations must lie strictly greater than x+∆x. And exactly one observation

must fall in [x,x+∆x].

The probability of this configuration is proportional to the multinomial distribution:

n!

.

(k−1)!(n−k)!

The probabilities of the specific events are: [F(x)]k−1 for probability that k−1 observations fall below x,

[1−F(x+∆x)]n−kfor probability that n−k observations exceed x+∆x, and F(x+∆x)−F(x)≈p(x)∆x

for probability that exactly one observation lies in [x,x+∆x].

The total probability is:

n!

P(X ∈[x,x+∆x])≈ [F(x)]k−1[1−F(x+∆x)]n−kp(x)∆x.

(k) (k−1)!(n−k)!

Taking the limit as ∆x→0, the marginal density is:

n!

p (x)= [F(x)]k−1[1−F(x)]n−kp(x).

k (k−1)!(n−k)!

5

Joint Distribution of X and X , k ≤ l

(k) (l)

To derive the joint density of X and X , consider:

(k) (l)

P(X ∈[x,x+∆x],X ∈[y,y+∆y]), x

(k) (l)

The following conditions must hold: k−1 observations lie strictly less than x, l−k−1 observations lie

strictly in (x,y), n−l observations lie strictly greater than y+∆y. And exactly one observation lies in

[x,x+∆x] and one in [y,y+∆y].

And total Ways to Arrange n Observations:

n!

.

(k−1)!(l−k−1)!(n−l)!

The probabilities of the events are: [F(x)]k−1: k−1 for observations less than x, [F(y)−F(x)]l−k−1 for

l−k−1 observations in (x,y), [1−F(y+∆y)]n−l: n−l for observations greater than y+∆y. p(x)∆x and

p(y)∆y for contributions from the densities at x and y.

Combining these components, the joint probability is:

n!

P(X ∈[x,x+∆x],X ∈[y,y+∆y])≈ [F(x)]k−1[F(y)−F(x)]l−k−1[1−F(y)]n−lp(x)p(y)∆x∆y.

(k) (l) (k−1)!(l−k−1)!(n−l)!

Taking the limit as ∆x,∆y →0, the joint density is:

n!

p (x,y)= [F(x)]k−1[F(y)−F(x)]l−k−1[1−F(y)]n−lp(x)p(y).

k,l (k−1)!(l−k−1)!(n−l)!

Special Case: Uniform Distribution U(0,1)

For U(0,1), the CDF and PDF are:

F(x)=x, p(x)=1, 0≤x≤1.

Substituting F(u)=u and p(u)=1 into the Marginal Density of X :

(k)

n!

p (x)= xk−1(1−x)n−k, 0

u(k) (k−1)!(n−k)!

This is the Beta distribution with parameters k and n−k+1:

U ∼Beta(k,n−k+1).

(k)

Substituting F(u)=u and p(u)=1 into the joint density formula we have Joint Density of U and U :

(k) (l)

n!

p (x,y)= xk−1(y−x)l−k−1(1−y)n−l, 0

U(k),U(l) (k−1)!(l−k−1)!(n−l)!

In summary,

• The marginal density of X is:

(k)

6

n!

p (x)= [F(x)]k−1[1−F(x)]n−kp(x).

k (k−1)!(n−k)!

For U(0,1):

xk−1(1−x)n−k

p (x)= , U ∼Beta(k,n−k+1).

u(k) β(k,n−k+1) (k)

• The joint density of X and X (k ≤l) is:

(k) (l)

n!

p (x,y)= [F(x)]k−1[F(y)−F(x)]l−k−1[1−F(y)]n−lp(x)p(y).

k,l (k−1)!(l−k−1)!(n−l)!

For U(0,1):

n!

p (x,y)= xk−1(y−x)l−k−1(1−y)n−l, 0

U(k),U(l) (k−1)!(l−k−1)!(n−l)!

7