Transcript ST3236: Stochastic Process Tutorial

ST3236: Stochastic Process
Tutorial 2
TA: Mar Choong Hock
Email: [email protected]
Exercises: 3
Question 1
Four nickels and six dimes are tossed, and
the total number N of heads is observed.
If N = 4, what is the conditional probability
that exactly two of the nickels were heads?
Question 1
A is the event that N = 4.
B is the event that exactly two of the nickels
were heads.
A

B
Question 1
Let X1 be the number of heads of the nickels and X2 the
number of heads of the dimes. Then, N = X1 + X2
P(X1 = 2 | N = 4) = P(X1 = 2,X1 + X2 = 4) / P(X1 + X2 = 4)
= P(X1 = 2, X2 = 2) / P(X1 + X2 = 4)
= P(X1 = 2)P(X2 = 2) / P(X1 + X2 = 4)

 
 C42 0.52 0.52  C62 0.52 0.54
 C
4
10
0.54 0.56
(assuming the probabilities of heads are 0.5)

Question 2
A dice is rolled and the number N on the
uppermost face is recorded. From a jar
containing 10 tags numbered 1, 2, ..., 10 we
then select N tags at random without
replacement.
Let X be the smallest number on the drawn
tags.
Determine P(X = 2).
Question 2
P X  2   PN  n PX  2 N  n 
6
n 1
Note that,
# n tags from1 to10 with smallest number 2
P X  2 N  n  
# n tags from1 to10
# 1 tagis 2 and n - 1 tags from3 to10 

# n tags from1 to10
C11C8n1 n10  n
P X  2 N  n  

n
C10
90
Question 2
 P X  2   PN  n PX  2 N  n 
6
n 1
  PN  n PX  2 N  n 
6
n 1
1 6 n10  n 
 
 0.2204
6 n 1 90
Question 3
Let X be a Poisson random variable with
parameter, .
Find the conditional mean of X, given that X
is odd.
Question 3
(define that 0 is even) Let pk = P(X = k).

E X X  odd   kPX  k X  odd
k 0
Note thatby p.g.f.,


P X  odd  1  P X  even  1  G X  1 2  1  e  2 2
Similarly by p.g.f.,
G X s   p1  2 p 2 s  3 p3 s 2  4 p 4 s 3  
 G X 1  p1  2 p 2  3 p3  4 p 4  
 G X  1  p1  2 p 2  3 p3  4 p 4  
Question 3
Note thatG X s   e   1 s . We have,


p1  3 p3    GX 1  GX  1 2   1  e  2  2
if k is even,thatis k  2i, i  0,1,2, ,
P  X  k X  odd  0
if k is odd, thatis k  2i  1, i  0,1,2, , ,
P X  k 
k e  k!
P  X  k X  odd 

2
P X  odd 1  e
2


Question 3
T hus,

E  X X  odd   2i  1P  X  2i  1 X  odd
i 0

  2i  1
2i 1e  2i  1!
i 0
1  e  2
2

1
2i  1P X  2i  1


2
1 e
2 i 0


1
 p1  3 p3  

2
1 e
2



 1  e  2   2
1  e  2
2
  coth 
Note: coth (.) denotes hyperbolic cotangent.
Question 3 - Optional
An alternative solution without using p.g.f.:

E  X X  odd   2i  1P  X  2i  1 X  odd
i 0

  2i  1
i 0
2i 1e  2i  1!
1  e  2
2
3
5
7
 







 e .  3e .  5e .  7e .  
3!
5!
7!



2e  .  32 54 76
1 



 
2 
1 e
3!
5!
7!


2

1  e 2




2e  .

1  e 2


 2 4 6

1     
2! 4! 6!


Question 3 - Optional
2
But, cosh   1 
2!

4
4!

6

6!
Or,
e  1   
2
2!
e   1   



2
2!
3

3!
3
3!

4

4!
4
4!


 e   e  2  cosh   1 
2
2!

4
4!

6
6!


1 e 
 E X  x X  odd  
  coth
1  e 
2
2
Note: cosh (.) denotes hyperbolic cosine.
Question 4
Suppose that N has density function:
P(N =n) = (1-p) n-1p for n = 1, 2,…
where p in (0, 1) is a parameter.
This defines the geometric distribution with
parameter p.
Question 4a
Show that: G(s) = sp/[1 - s(1 - p)] for s < 1/(1 - p).
 

G s   E s N   s r PN  r 
r 0

  s p1  p 
r
r 1
r 1

 sp  s1  p 
sp
Proved

1  s1  p 
r 1
r 1
Question 4b
Show that: E(N) = 1/p

d 
sp
E N   G 1  
ds 1  s1  p   s 1



p1  s1  p   1  p sp
1  s1  p 2
s 1
p  sp 1  p   1  p sp
1  s1  p 2
s 1
p
1  s1  p 
2
s 1
1
 P roved
p
Question 4c
Show that: var(N) = (1 - p)/p2

d 
p
E N  N  1  G 1 


ds  1  s 1  p 2 
s 1

 2p
1  s1  p 
2 p1  p 

1  s1  p 3
3
 1  p 
s 1
s 1
21  p 

p2
 var N   E N  N  1  E  N   E  N 
2
21  p  1
1
21  p   p  1 1  p

  2 
 2 P roved
2
2
p p
p
p
p
Question 4d
Show that: P(N is even) = (1 - p)/(2 - p).
1
1
p 
P N  even  1  G  1  1 
2
2  1  1  p  
1  2  p  p  1  21  p  
 
 

2  2  p  2  2  p 
1 p
P roved

2 p