Download ECSEB Probability and Random Signals Midterm I Solution h .h

ECSEB Probability and Random Signals
Midterm I Solution
h .h. Wednesday, th February, Prof. Peter E. Caines
Closed book examination. No notes or calculators permitted.
Question points
a State the three axioms which a probability function P ES R must satisfy, where S
is a sample space and ES is a set of events.
b Let A S, B S, where A and B are not necessarily disjoint. Let A
i
A, i lt
be pairwise disjoint, and let B
i
B, i lt be pairwise disjoint.
i Let X A and assume PA
i
gt , i lt .
Set PX
i
A
i
, and set PX
i
PA
i
and
i
PX A
i
.
State, but do not prove, which of or , or both, is equal to for all possible X A.
ii Give a formula for P
i
A
i
B
i
in terms of PA
i
B
i
, i lt do this by
use of one of the probability axioms and justify its application.
iii In case A B , show what the value of P
i
A
i
B
i
must be.
Solutions
a A probability function P ES on a sample space S and event set ES must
satisfy the following axioms
PA , A ES,
PS ,
P
i
A
i
i
PA
i
, where A
i
A
j
,ij
b
i In general, PX
i
A
i
P
i
XA
i
i
PX A
i
, and in
general .
ii Since A
i
A
k
, i k, B
j
B
l
, j l, A
i
B
i
A
j
B
j
, i j.
Hence Axiom III implies P
i
A
i
B
i
i
PA
i
B
i
.
iii In case A B , A
i
B
i
A B , i lt .
Hence PA
i
B
i
and P
i
A
i
B
i
i
PA
i
B
i
.
Question points
G V, E is a clique graph, i.e. G is such that every distinct pair v, w of vertices in V is
connected by exactly one undirected edge e
vw
e
wv
in E. Assume V n , i.e. V has
n elements, and E m.
i What is the value of m E as a function of n
ii For each r, r n, a distinct unordered set V
r
V of r vertices is chosen from
V . A clique subgraph, G
r
, is constructed from the nodes V
r
using all the edges EV
r
E
connecting all pairs of vertices in V
r
.
How many distinct subgraphs G
r
can be constructed for each r
iii Assume there is only one empty clique graph G
V
, EV
. Including G
and G,
give an expression for the total number N
c
of possible clique subgraphs of G for r n.
Solutions
Let G E, V be a clique graph.
i Since there is exactly one edge e
vw
for each unordered pair of edges v, w,
mEC
n
nn
ii There is a distinct clique subgraph V
r
, EV
r
for each distinct set of r vertices in V .
Hence there are C
n
r
distinct such clique subgraphs.
iii Total possible number of clique subgraphs N
c
n
r
C
n
r
Question points
An Ehrenfest molecular Markov chain model E has three states x
,x
,x
, corresponding
to two urns L, R containing x
,,x
, and x
, particles, respectively.
The probability of transition from x
j
to x
i
, i, j , is given by
Px
i
x
j
P
ij
for i , j and i , j
Px
i
x
j
P
ij
for i , j and i , j
Px
i
x
j
otherwise
i Find the transition matrix T for the Markov chain E.
ii At the initial instant k , the occupancy probability vector P
x
P
x
,
P
x
. Find the occupancy probability vector at the instant k , namely
P
P
x
P
x
P
x
iii Which of P
,P
is a possible steady state occupancy probability
for the Markov chain E
Solutions
i Ehrenfest Markov chain E
x
,
x
,
x
,
has the transition matrix
T
p
p
p
p
p
p
p
p
p
ii
p
Tp
T
p
Substituing with the matrix obtained in i we get
p
iii
T
T
So only
T
is a possible steady state probability.
Question points
Let A and B be disjoint subsets of a sample space S. The probability P on S, ES is such
that PA
, PB
, and hence A B A B
c
S with PA B
c
.
i Dene the property of independence for the events A and B on S, ES, P.
ii State, but do not prove, whether A and B can be independent.
iii Events C and D are such that
PCA PDA
PCB PDB
Given that C occurs what is PAC
Solutions
On the probability space S, ES, P, the events A, B are disjoint, i.e. AB . Further,
PA
,PB
.
i A is independent of B if and only if
PA B PAPB
ii A and B cannot be independent.
iii For events C,D S
PCA PDA
PCB PDB
Given C occurs, Bayes rule gives
PAC
PCAPA
PC
PCAPA
PCAPA PCBPB PCA B
C
PA B
C
is constructed from the nodes Vr using all the edges EVr E connecting all pairs of vertices in
Vr . and E m. Ai Bi A B . P X general .e. Including G and G. and in Ai Bi i P Ai Bi . A clique
subgraph. E is a clique graph. How many distinct subgraphs Gr can be constructed for each r
iii Assume there is only one empty clique graph G V . Solutions Let G E. V be a clique graph.
j l.e. give an expression for the total number Nc of possible clique subgraphs of G for r n. iii
Total possible number of clique subgraphs Nc n r n Cr . V has n elements. iii In case A B . i k.
EVr for each distinct set of r vertices in V . ii There is a distinct clique subgraph Vr . Hence
Axiom III implies P i i Ai P i X Ai i P X Ai . Ai Bi Aj Bj . a distinct unordered set Vr V of r
vertices is chosen from V . Hence P Ai Bi and P Question points G V. ii Since Ai Ak . i j. w of
vertices in V is connected by exactly one undirected edge evw ewv in E. n m E C nn i Ai Bi i
P Ai Bi . Bj Bl . i What is the value of m E as a function of n ii For each r. w. Assume V n . n
Hence there are Cr distinct such clique subgraphs.b i In general. Gr . i. i. r n. G is such that
every distinct pair v. i lt . EV . i Since there is exactly one edge evw for each unordered pair
of edges v.
Find the occupancy probability vector at the instant k . j P xi xj Pij for i . x . x .Question points
An Ehrenfest molecular Markov chain model E has three states x . ii At the initial instant k . x
. x . and x . R containing x . j and i . The probability of transition from xj to xi . P x . . x .
corresponding to two urns L. P is a possible steady state occupancy probability for the
Markov chain E i Ehrenfest Markov chain E x . is given by P xi xj Pij for i . j . the occupancy
probability vector P x P x . j P xi xj otherwise i Find the transition matrix T for the Markov
chain E. has the transition matrix p p p T p p p p p p . particles. j and i . i. namely P x P P x P
x iii Which of P Solutions . respectively.
whether A and B can be independent. P B . i Dene the property of independence for the
events A and B on S.ii p T p T p Substituing with the matrix obtained in i we get p iii T T So
only T is a possible steady state probability. The probability P on S. but do not prove. and
hence A B A Bc S with P A Bc . ES. ii State. ES is such that P A . . P . Question points Let A
and B be disjoint subsets of a sample space S.
A B . iii For events C. P A .e.iii Events C and D are such that P CA P DA. ES. P . i. Given
that C occurs what is P AC Solutions On the probability space S. P CB P DB i A is
independent of B if and only if P A B P AP B ii A and B cannot be independent. B are
disjoint.D S P CA P DA P CB P DB Given C occurs.P B . Further. Bayes rule gives P AC P
CAP A P C P CAP A P CAP A P CBP B P CA BC P A BC . the events A.