I INTERNATIONAL SCIENTIFIC CONFERENCE OF YOUNG RESEARCHERS
Baku Engineering University
11
27-28 April 2018, Baku, Azerbaijan
If we continue to calculate the probability of being certain position after n steps, we will find
binomial characteristics of simple random walk. Let us denote position with m, number of steps to the
right with a and to the left with b, and number of total steps with n. Then,
paths
step
n
of
number
Total
goes to m
that
paths
step
n
Number of
n
m
P
)
,
(
Which is equal to
)!
(
!
!
2
1
)
,
(
a
n
a
n
n
m
P
n
For instance, the probability of being
2
x
after 2 steps is equal to
4
1
)!
2
2
(
!
2
!
2
2
1
)
2
,
2
(
2
P
, as
we get above on tree algorithm. In order to find expected value formula, let us denote it as
n
S
:
n
n
n
n
P
P
P
P
x
P
x
P
x
S
....
....
2
1
2
2
1
1
Where
n
x
x
x
,....
,
2
1
are positions and
n
P
P
P
,....
,
2
1
are probabilities corresponded to the given
positions. Because
1
,....
2
1
n
P
P
P
, we can simplify the expectation value formula,
n
n
n
P
x
P
x
P
x
S
,....
2
2
1
1
The finite additivity property of expectation gives:
n
i
i
n
x
S
1
We can get expectation value of the squared value of random variables as follow:
n
i
i
n
x
S
1
2
2
There are some interesting problems pertain to random walk. One of them is Gambler’s ruin. This
problem states that two players ,
A and
B, play a game with independent rounds where, in each round,
one of the players wins one 1 dollar from his opponent ; A with probability p and B with probability
p
q
1
. A starts the game with a dollar and B with b dollar. The game ends when one of the
players is ruined. The main question for this problem is “What are the player’s ruin probabilities?”
Sometimes it is called as absorption probabilities. Answer of indicated question corresponds to a
random walk where a particle starts at
0
x
and is absorbed in the states
b
x
and
a
x
or,
equivalently, starts at
a
x
and is absorbed at
0
x
and
b
a
x
. Let
dollars)
has
he
when
wins
(
k
A
p
A
k
Then,
0
0
A
and
1
b
a
A
.Our aim is to find
a
A
.
1
1
k
k
k
A
q
A
p
A
(2)
This homogenous equation can be solved by determining the zeros
of characteristic polynomial,