frog

1. Introduction
In the mythological age, it was told that the frog, after having perfected plans to employ
the services of its offsprings, went to challenge the hare for a 12km race. He had boasted
that it would win the race at the hare’s expense, after having ambushed each offspring at
every kilometre end of course. The hare, ignorant of the intrigues, counted the challenge
as an insult but accepted all the same to partake in the race. It was at the finish-point
that he realised that he had been fooled. I read this ”frog-hare” mythology as a young
man and had ever since thought that any life race could be so won. The fun of the frog’s
victory, in part, motivated me in writing this paper, dedicated to its random movement
in a quadrangle.
The life personae of frog and hare can be encapsulated in the following. The frog is
an amphibian small animal with long back legs for jumping. It is tailess, no wonder then,
”it is God that provides protection for its fly menace”, so goes the dictum. The hare is
an animal like a large rabbit with very strong back legs, known for runnning fast. From
the foregoing, it could be concluded that the frog cannot in any form be a match for the
hare in any race whatsoever, except possibly, in the water.
The other leg of motivation for this work is that the random jumps constitute an aspect
of random motion/fractal movement. Theory of random walk, fractal motion or brownian
movement have close relationship with mathematical theory of Graphs which has wide applications
in electrical and civil engineering, communication networks, industrial management,
operations research, computer science, economics, management science, marketing,
sociology and others.
Infact, concept of graphs can be encountered under variety of names such as network
1
in electrical/electronics engineering, structures in civil engineering, molecular structures
in chemistry, organisational structures in economics or management science, sociograms,
road maps in transportation, etc. Some of the specific problems being examined include
minimum cost or time in transportation, best assignment of workers to jobs, most efficient
use of telephone networks, shortest path to a destination. Essentially, most of these lead
to optimisation processes. Hence, there is the need to obtain essential statistical parameters
to authenticate correctness or efficiency of models. This work examines the random
jumps of a frog in a quadrangle and introduces a method for obtaining the associated statistical
parameters such as expectation, mean deviation, moment, skewness and kurtosis.
The obtained parameters are compared with those obtained by other standard methods.
2. Statement of problem
A frog displaces randomly by following the lignes of a a quadrangle with nine points as
indicated in the figure below. At each of these nine points, it randomly chooses a direction
of movement. Hence, from point O, the four points OAi has each a probability of 1/4 of
being followed by the frog. From point Ai, the three directions has each a probability of
1/3 to be followed, and from point Bi, each of the two directions has a probability of 1/2.
The passage from one point to another in the quadrangle is called ”a path” of the frog.
We assume that the frog starts from 0. We are interested in the probability that the frog
returns to 0 for the first time after K movements, where K is a positive integer.
3. Methodology:
Path graph of the problem:
2
Now, we set out to propose (expose) the mathematical theory to obtain the vital statistical
parameters (Moment generating function, skewness and kurtosis) for the jumps
We consider the probability of each path of the frog as follows, using arrows to indicate
directions of movement in the quadrangle.
Simplification of path graph:
We construct a simpler (representative) path for the frog. For this, we consider the frog’s
movement to and fro each of the following:
3
(a) central point O;
(b) a lateral point A;
(c) a corner B
The probability graph of these points is represented below as:
where
(i) probability of out-movement from O is 1;
4
,
(ii) probability of out-movement from A to B is 2/3;
(iii) probability of out-movement from B is 1;
(iv) probability of out-movement from A to O is 1/3.
Probability of jump:
Now, define: P(k), the probability that the frog returns to point O after k steps since
it departed from it.
Clearly, P[OA] = P[AB] = 1.
P[AO] = 1/3
P[AB] = 2/3
Theorem 1:
Arrival of the frog to point O (after its initial departure from it) takes place after even
number of movements i.e K = 2n , n 2 N ,and P(k) = 1/3(2/3)n−1.
5
Proof:
For the frog to jump from O to A and return to O, for example, it needs to jump OA and
back-jump AO: K = 2.
For the frog to want to return from B to O, it must have made the trajectories: OA, AB,
BA and AO: K = 4.
Generally, to return to O (from B), it must undergo the trajectory OA, next (n − 1)
go-come of AB, BA and lastly, AO: K = 2n.
P[K] = P[2n] = P(OA) × P(AB) × P(BA) × P(AO)
= 1 × (2/3)n−1 × 1n−1 × 1/3
= 1/3 × (2/3)n−1
Verification:
1X
n=1
P(k) =
1X
n=1
1
3
(
2
3
)n−1
=
1
3
+ (
2
3
) + . . .
=
1
3
1 − 2
3
= 1
Introduction of differential calculus technique. Before continuing this study, it is
here appropriate to expose and explain a mathematical approach used to obtain the parameters
of the frog’s random jumps.
Now, consider the probaility P(k) as a sequence of probability function f(x) where x is a
random k variable. Then we invoke the theorem of calculus on sum of function on series
of {fk(x)}:
Theorem M: The derivatives of a sum of functions is equal to the sum of derivatives of
6
functions: i.e.
d
dx

X
k
fk(x)
!
=
X
k
d
dx
f(x)
Proof: From l.h.s.
X
k
fk(x) = f1(x) + f2(x) + · · · + fk(x) + · · · (1)
Let the operator d
dx (.) be used on (1), we have:
d
dx

X
k
fk(x)
!
=
d
dx
f1(x) +
d
dx
f2(x) + · · · +
d
dx
fk(x) + · · · (2)
=
X
k
d
dx
f(x) · · · (3)
(3) constitutes the r.h.s. of the theorem
Remark: Theorem M can be extended to cover the case dh
dxh (
P
k fk(x)), where dh
dxh (•) is the
hth derivative of the function fk(x). In this case, we have
dh
dxh

X
k
fk(x)
!
=
X
k
dh
dxh (fk(x))
Statistical parameters of jumps.
Moment Generating Function of X:
Theorem 2: Define X: number of jumps required of the frog to have been made to return
to point O, after having departed from it.
Then E(X) = 6, X 2 {2, 4, . . . , 2n, . . .}
Proof:
E(X) =
1X
n=1
2n
1
3
(
2
3
)n−1
=
2
3
1X
n=1
n(
2
3
)n−1
7
let x = 2/3
E(X) =
2
3
1X
n=1
d
dx
(xn)
=
2
3
d
dx
1X
n=1
xn
!
=
2
3
d
dx

x
1 − x

=
2
3
"
(1 − x) + x
(1 − x)2
#
=
2
3
"
1
(1 − x)2
#
=
2
3
×
1
(1/3)2
=
2
3
×
9
1
= 6
Definition: Since X is a discrete random variate, moment generating function of X is
defined for any real number h:
M(h) = E

ehx

(a)
=
1X
n=1
e2nh 1
3
(
2
3
)n−1
=
1X
n=1

1 + 2nh +
(2nh)2
2!
+ . . .
!
1
3
(2/3)n−1
=
1X
n=1
1
3
(2/3)n−1 +
1X
n=1
(2n)r hr
r!
1
3
(2/3)n−1
= 1 +
1X
n=1
(2n)r hr
r!
1
3
(2/3)n−1
= 1 +
1X
n=1
(X)r hr
r!
P (X = 2n) (b)
Definition: Define μ0
r = E (Xr) =
P
(2n)r 1
3 (2/3)n−1, the rth moment of X about the
origin.
8
μ0
r: first moment about the origin, i.e μ0
r = E (X) = μ, the expectation of X obtained in
theorem 2.
4. New Approach for Evaluating Statistical Parameters of Jump:
Hitherto, evaluation of the moment generating functions of random variates is carried
out using the infinite geometric approach. But here, we propose an alternative approach
which adopts the linearity of summation 0P0 property and the differentiation technique.
We subsequently elucidate on this. In fact, using the ”new” approach, we define
μ
0
r = E

X2

=
1X
n=1
4n2 1
3
(2/3)n−1
=
4
3
X
n (n − 1 + 1) (2/3)n−1
=
4
3
X
n (n − 1) (2/3)n−1 +
4
3
X
n (2/3)n−1
=
4
3
d2
dx2 (xn) +
4
3
× 9
=
4
3

2
(1 − x)3
!
+ 12
=
4
3
×
2
1
×
27
1
+ 12
= 72 + 12 = 84.
Hence, the population variance of the frog’s number of steps required of it to return to
O,for the first time after departure is:
2 = μ2 − μ2 = 84 − 36 = 48
Now, the : population standard deviation of the frog’s number of return jumps to the
origin
 =
p
48 = 4
p
3.
9
Skewness of Jump
It is desired to study whether the distribution of these jumps is symmetrical or skewed
about its mean. That is, it is of interest to veriff whetehr or not the natural assumption
that the frog makes random (unconstrained) movement at each point could lead to an
absolutely normal (i.e. bell-shaped) distribution, without skewness of any sort.
Now, define
μr = E[(x − μ)r] =
X
(2n − 6)r 1
3
(
2
3
)n−1 (c)
This is called the rth moment of the variate X about the mean.
Now, for a symmentrical distribution, μ3 = 0. Also,
skewness =
μ3
3 ,
where
μ3
3 > 0
means a skew to the right in which case, the distribution tapers off to the right more than
to the left, and vice-versa, otherwise. Now,
μ3 = μ
0
3 − 3μ
0
2μ + 2μ3,
obtained by expanding equation (c). Now, using (b),
μ
0
3 =
8
3
X
n3(
2
3
)n−1.
Using the identity
n3 = (n + 1)n(n − 1) + n,
10
μ
0
3 =
8
3
X
(n + 1)n(n − 1)(
2
3
)n−1 +
8
3
X
n(
2
3
)n−1
μ
0
3 =
16
9
d3
dx3 (xn+1) +
8
3
d
dx
(xn) = 888
for x = 2
3 . Hence,
μ3 = 888 − 3(84)(6) + 2(6)3 = 192.
Hence, the distribution of X is skewed to the right, as dividing μ3 by 3 does not change
its positivity.The division merely divests the distribution of its units of measurement.
Hence, the skewness = μ3
3 is a pure number. Thus
192
(4
p
3)3
=
192
332.55
= 0.577
is the skewness factor of the distribution.
Kurtosis It is of much interest also to calculate the kurtosis of the distribution of X,
the number of steps needed by the frog to have been realized before finding itself back at
11
the origin, after initial departure. Again the natural assumption of normal distribution,
encoded in randomness assumption of the jumps at every point, presuposes normalized
kurtosis. Thus, it is of valuable information to ascertain if this is so or not.
The fourth moment, μ4, divided by μ22
is a measure of kurtosis, the peakedness or ”pinpointedness”
of the central peak of the curve of the distribution. It is generally observed
that the bigger the μ4 is, the higher the central peak of the distribution becomes.
Now,
μ
0
4 =
X
(2n)4 1
3
(
2
3
)n−1 =
16
4
X
n4(
2
3
)n−1
From the identity
n4 = (n + 2)(n + 1)n(n − 1) − 2(n + 1)n(n − 1) + n2
we have
μ
0
4 =
16
3
X
(n+2)(n+1)n(n−1)(
2
3
)n−1 −
32
3
X
(n+1)n(n−1)(
2
3
)n−1 +
16
3
X
n2(
2
3
)n−1
=
32
9
d4
dx4 (xn+2) −
64
9
d3
dx4 (xn+1) + 4μ
0
2 = 2064
μ4 = μ
0
4 − 4μ
0
3μ + 6μ
0
2μ2 − 3μ4
= 2064 − 4  888  6 + 6  84  36 − 3  362
= 2064 − 21312 + 18144 − 3888
= −4992
Hence,
μ4
μ22
=
−4992
482 = −2.167
12
a negative central peak.
Results:
The following results have been obtained from this study:
(1) It is revealed that the distribution of the frog’s first return to the ’origin’ is not
symmetrical about its mean;
(2) It is also found out that the kurtosis of the movement is negative;
(3) This study paves way for expressing nk (k, a positive integer) as sum of factorial
products of n as exemplified by
1. n1 = n
2. n2 = n(n − 1) + n
3. n3 = (n + 1)n(n − 1) + n
4. n4 = (n + 2)(n + 1)n(n − 1) − 2(n + 1)n(n − 1) + n2
= (n + 2)(n + 1)n(n − 1) − 2(n + 1)n(n − 1) + n(n − 1) + n
5. n5 = (n+3)(n+2)(n+1)n(n−1)−5(n+2)(n+1)n(n−1)+5(n+1)n(n−1)+n
...
...
...
...
... ...
...
nk = ?
Conclusion
That the distribution of the frog’s first return to the ”origin” after its initial departure
is not symmetric, is strange. A zero kurtosis would probably have meant that the distribution
is ”peakedly normal”. But here, we have a negative kurtosis. Effort would be
13
made in future studies to sort for an explanation on how and why all these ”strange”
observations come about.
Literature
(1) T.J.Adesakin, A.A.Osuntunyi and M.A.Olagunju: The distributional properties of the
family of logistic Distributions: Ife Journal of science. Special Edition 2008, Vol.10 no 1
pgs 245−
(2) Introduction to statistical inference E.S.Keeping.(university of Alberta)
(3) Outils Mathematiques probabilities: Bernard Vauquois(Hermann collection Paris methods)
(4) Aide memoire de mathematiques superieures M.Vygodski(Edition de Moscov).
14