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MTH6141 Random Processes, Spring 2012

Solutions to Exercise Sheet 7

1. (a) Under assumption (i) the fact that I arrive at a random time means thatthe time I wait (in minutes) is a random variable uniformly distributedon [0, 10]. Such an r.v. has expectation 5. So I expect to wait 5minutes. Under assumption (ii) we have that, by the lack of memory ofthe Poisson process, the buses arrive as a Poisson process (even giventhat I arrive at a random time). The time I wait is T1 (or S1) for theprocess starting from when I arrive. So (in minutes) the waiting timeis an Exp(6/60) = Exp(1/10) random variable. The expectation of thisr.v. is 10. So I can expect to wait 10 minutes.

(b) This result appears paradoxical since in both cases the average interval

between buses is 10 minutes. Intuitively the average time I wait shouldbe half this. This argument is correct in case (i). However in the Poissonprocess case I am more likely to arrive during a long interval betweenbuses (because the long intervals take up more time) and so the averagetime I wait is longer.

2. Applying the same calculation for general n 1:

FTn(t) = Pr(X(t) n)

= 1 Pr(X(t) = 0) Pr(X(t) = 1) Pr(X(t) = n 1)

= 1

1 + t +

(t)2

2! + +

(t)n2

(n 2)! +

(t)n1

(n 1)!

e

t

.

Differentiating,

fTn(t) =

+ 2t + +n2tn3

(n 3)!+

n1tn2

(n 2)!

et

+

1 + t +(t)2

2!+ +

(t)n2

(n 2)!+

(t)n1

(n 1)!

et

=ntn1

(n 1)!et

(Notice how all the terms bar one cancel out in pairs.)

3. As usual with continuous random variables (and as the hint suggests), thecdf is easier to work out than the pdf.

FT1|X(t)=n(u) = P(T1 u | X(t) = n)

= 1 P(T1 > u | X(t) = n)

= 1 P(X(u) = 0 | X(t) = n)

= 1

1 u

t

n

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for 0 < u t, since the conditional distribution of X(u), given X(t) = n, isBin(n, u

t))

Differentiating this with respect to u gives the pdf:

fT1|X(t)=n(u) =n

t

1

u

t

n1

for 0 < u t.

The expectation can be found in the usual way:

E(T1 | X(t) = n) =

t

0

xn

t

1

x

t

n1dx

= x1 xtnx=t

x=0

+ t01 x

tn dx (integrating by parts)

= 0

t

n + 1

1

x

t

n+1x=tx=0

=t

n + 1.

4. In each part of this question we are told that X(t) = n for some t and n. Weconsider random variables Ui for 1 i n where the Ui are independent andeach is distributed uniformly on [0, t]. We know from lectures (Theorem 2.5)that if s is a symmetric function of the Ti (as all the functions given are)

thenE(s(T1, . . . , T n) | X(t) = n) = E(s(U1, . . . , U n)).

(a) By the above, and linearity of expectation,

E(T1 + T2 + T3 | X(4) = 3) = E(U1 + U2 + U3) = E(U1) +E(U2) +E(U3),

where Ui U[0, 4]. Then E(Ui) = 2, and

E(T1 + T2 + T3 | X(4) = 3) = 2 + 2 + 2 = 6 .

(b) By linearity of expectation,

E(T1+T2+T3+T4 | X(4) = 3) = E(T1+T2+T3 | X(4) = 3)+E(T4 | X(4) = 3).

We saw in part (a) that the first term is 6. Now T4 is the time of thefirst arrival after time 4, and so T4 4 is distributed Exp(). So E(T4 |X(4) = 3) = 4 + 1 and E(T1 + T2 + T3 + T4 | X(4) = 3) = 10 +

1.

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(c) As with part (a),

E(T21 + T22 + T

23 + T

24 + T

25 | X(2) = 5) = E(U

21 + U

22 + U

23 + U

24 + U

25 )

= E(U21 ) + E(U22 ) + E(U23 ) + E(U24 ) + E(U25 )

where Ui U[0, 2]. Then E(U2i

) =20

12

x2 dx = [16

x3]20 =43

and

E(T21 + T22 + T

23 + T

24 + T

25 | X(2) = 5) =

203

.

5. Processing occurs every T minutes and costs k each time. This gives a costof k

Tper minute.

If there are n requests waiting at time T then the total waiting cost theyhave incurred is a random variable with] expectation

E n

i=1

c(T Ti) X(T) = n

We can use the same trick as Question 2 (replacing Ti with Ui U[0, T]) toget that

E

ni=1

c(T Ti)

X(T) = n

= E

n

i=1

c(T Ui)

=

ncT

2.

Now, conditioning on the number of requests waiting at time T, the expectedtotal waiting cost is

n0

E

n

i=1

c(T Ti)

X(T) = nP(X(T) = n) =

cT

2

n0

neT(T)n

n!=

cT2

2

(where the last identity comes from the fact that the expectation of a Po(T)random variable is T).

So the expectation of the total waiting cost per minute is cT2

. The totalexpected cost per minute (processing and waiting) is

k

T+

cT

2

as required.

Sketching the graph of this function against T we see that it has a singleminimum. Differentiating we get that the minimum is the solution to

c

2

k

T2= 0

So we should take T =

2kc

.

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