Introduction
Some variables take non-negative values and its probability distribution is positively skewed. In this situation, a normal distribution is not an appropriate distribution to use. Life time of an electronic component, time until decay of a radioactive atom, interarrival time between customers in a queue are the examples of such random variables. These random variables follow exponential distribution. Thus, exponential distribution plays an important role in theory of queues, reliability theory, survival analysis etc.
Exponential Distribution
Definition: A continuous r.v. X taking non-negative values is said to follow exponential distribution with parameter & if its probability density function (p.d.f.) is given by,
f(x) = α e-xxx = 0
x ≥ 0, α 0 otherwise
We shall denote it as X→ Exp (α).
Note: (1) We shall verify that f (x) is a p.d.f.
(i) Obviously f (x) 0x20, a > 0 and e-xx > 0
(ii) () f(x) dx =
0
αe-ax
0
= 1
[
= [-e-x]
From (i) and (ii), it is clear that f (x) is a p.d.f.
00
(iii) Exponential distribution has a nice application in the theory of queueing model. If number of customers arriving at a system has Poisson distribution then interarrival times are exponentially distributed with arrival rate α per unit of time. In this case average time required to arrive a customer is 1/α.
(2) The density curve of an exponential distribution is as follows.
It is a J shaped curve. From the Fig. 5.1, it is clear that mode of the distribution is 0. In this case the value of mode cannot be taken as an average.
Graph of Exponential distribution with mean 1 by R-Software. Exponential probability curve
d
2.0
ヨレ
1.3
15
0.5
alpha=2
alpha=1
0
1
2
3
1
5
X
Fig. 5.1
(3) If a 1, then the distribution is called standard exponential distribution. Hence, its probability density function is given by,
f(x) = ex ;
= 0
;
x 20 otherwise.
(4) Another form of p.d.f.: Alternatively the p.d.f. of exponential distribution with mean a can be taken as,
f(x) =
a e-xa.
!
x > 0, α > 0
= 0
;
otherwise.
(5) We need to use following results in further discussion.
e- mx xn-1 dx =
I (n) ;m > 0, n > 0. mn
0
(ii)
e- ax2 xb-1 dx =
2 ab/2i
; a > 0, b > 0.
0
Mean and Variance of the Distribution
x f (x) dx = α xe-¤x dx
Mean :
E (X) =
0
r (2)
= (α)
α
1
=
α
- m = α, n =
Thus E(X) mean = 1/α
Variance: Now in order to evaluate variance, we find
E (x2) = ( x2 f (x) dx = α) x2 e-xx dx
0
Г (3)
α
α3
2!α 2
=
a3
0
m = α, n =
Var (X) = E (X2) - [E (X)]2 = 2-22-2
S.D. (X) = α
Remarks: (1) Note that if X follows exponential distribution with
p.d.f.
f(x) = α e-αx = 0
; ;
x 20, α > 0
otherwise
then we have mean = S.D.
(2) If we take another form of p.d.f. of exponential distribution
given by,
f(x) =
e-x/α
;
= 0
x ≥ 0, α > 0 otherwise
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I
then mean and variance a2. Hence, in this case we write it as
X → Exp
The expressions for various functions in this case can be 1
obtained replacing a by
Moments of Exp (α)
We derive the expression for rth raw moment of the distribution as follows:
H = E(X) = 0
x' f (x) dx = axe-xx dx
[Here, m = α, n = r + 1]
r(r + 1)_r!
= α
= ar + 1 απ
24
=
..
H=r!/α
putting r = 1, 2, 3, 4 the first four raw moments are given by,
1 Ha'
H2
6
2. ਟ = 2 - 3 ਕ ਕ = 4!
α
Из
Hence, the first 4 central moments are,
..
απ
12 = 12-14)2 = 2/2-1/2 ==1/2
Из = 43-342 4 + 2
На = 2-4 міні+64 (-3 (4)
=
M.g.f. of Exp (α)
00
My (t)= E [et] etx f (x) dx = a etxe-xx dx
0
= a e-(a-t) x dx
0
The integral exists if at > 0, i.e. a> t
0
r (1)
Mx (t) = a(at)1
α
a-t
=
1 1-t/α
t<α
Thus,
My (t) =
1-a
[ Here n = 1, m = α-t]
In order to find raw moments, we expand the m.g.f. in the powers of
t as follows:
・()+(1)+()+
Mx (t) = 1 +
2 t2 3 t3 4! t4
= 1 +
+
a2 2! a3 3! a44! + ...
Hence, the first four raw moments are given by,
t 1
H1 = coefficient of
1! α
t2 2! 2
H2 = coefficient of 21 = a2
t3 3! 6
Ма
= coefficient of 3! = α3 = α3
14 24
На
= coefficient of
4! 04
In general μ1 =
r! ar
The first four central moments can be obtained using the relations between central and raw moments as usual.
Note: If X→ Exp
then it can be shows that My (t) = (1 - at)-1
C.G.f. of Exp (α)
Kx (t) = loge Mx (t)
=loge (1-2) = -loge (1-4) () () ()
(計)
t
2
3
4
r
(r-1) 1
+...
+
Hence, the cumulants are given as follows:
k1 = coefficient of t =
k2 H2 coefficient of
t2 1 2!
k3= 3 coefficient of
= 2!
3!
14
3!
6
K4H4-3 μ coefficient of
04-04
(r− 1)!
In general
k, =
α
..
6 3 9
+
Coefficients of skewness and kurtosis:
4/06
P131/06=4>0
μ2
√1 = √B1 = 2
(μ3 is +ve as a > 0)
Hence, exponential distribution is positively skew distribution.
μ4 9/04
We have,
B2 =
1/04
9
H2
12 B2-39-3=6
Thus, exponential distribution leptokurtic.
Note: The values of B1, Y1, B2 and 2 are same for the another form
of p.d.f. i.e. for Exp(1/)
Distribution function of Exp (α)
Let X → Exp (a) Then the distribution function is given by,
Fx (x) = P(X ≤ x]
f (t) dt =
=0
a e-αt dt = a
e-at dt
0
0
0
= [-ē
Fx (x) = 1-e-xx> 0, α > 0.
Graph of Distribution function of Exponential distribution with
mean 1
Distribution function of exponential distribution
0.4
ZO
0
alpha=1
Note:
(1) When X is life time of a component P [X > x] is taken as reliability function or survival function.
Here, P [X > x] = 1 - P (X ≤ x] = 1 - Fx (x) = e-xx ; x > 0, α > 0. Hence, it is reliability function if exponential distribution is used as life time distribution.
(2) In reliability theory
f(x) 1-F(x)
If X Exp (a) then hazard rate =
is called as hazard rate. f(x) 1- F(x)
a constant.
(3) If X → Exp then Fx (x) = 1 − e−x/α ; x > 0, α > 0.
Quartiles
la
Suppose Q1, Q2 Q3 denote lower quartile, 2nd quartile (median) and P [X ≤ Q2] =
upper quartile respectively. Then we have P [X ≤ Q1] =
and P (X ≤ Q3]
F (Q1) =
F(Q2) = 2
, F (Q3) = 23
F (Q1) =
1-e
⇒e
4
3
Q1 =
loge
F(Q) = gives
1-e-α1
1
= 5⇒e
2
1
-αQ2 1
αQ2
e
= 2
2
Q2=log2e
F (Q3) =
e+ αQ3 = 4
1-e
·αQ3
-αQ3
=
e
=
1
4
..
Q3 = loge
Thus Qi =
i = 1, 2, 3
Note: (1) Quartile Deviation =
Q3-Q1 2
1
loge 109.4-109.109. (3)
2
loge
loge 4/3
=
loge 3
=loge3=1
0.5493
α
Q.D. =
Mean Deviation About Mean
Let X → Exp (a). Then Mean = E (X) =
put,
M.D. about mean = X-
=
e-xxx dx
E(x-1)= |∞x-1 e-x dx
α
0
Y
dy
.. dx = α
xxx = y.. x = α
00
- b-11e7-3
M.D. about mean =
=20ly-11edy + +40
0
dy
ly-1 e dy = I + I2
... (1)
we have, I1 =
(1-y) e y dy
y<1
After integrating by parts
1 - 2 {1-1-ent-0 1-(-294y}
I1 =
(1-y)
e-y dy =
1
1 1
-----
e-1 + 1] =
12 = 0 (y-1) e-dy
α α
e-1
-----
-(y-1) e]-
- {10 +01+1)-(0)-
=
.. From equation (1), (2) and (3), we get,
e-1... (3)
M.D. about mean =
1
e
9-1
α
a
2
=
œe
= 0.7358/α > Q.D. = 0.5493/α
Note:
(1) M.D. about mean > Q.D.
œe
(2) If X Exp (1/α) then M.D. about mean =2
Additive Property
Additive property does not hold good for exponential variates. This means if X1→ Exp (α1), X2 → Exp (α2) and X1 and X2 are independent then (X1 + X2) does not follow exponential distribution.
Proof: X1 Exp (a) the m.g.f. is given by Mx(t) = = (1-4)
similarly,
Mx,(t) =
= (1-4).
(t) Mx, (t) [X, and X2 are independent]
= (1-4)'(1-4)
It is not the m.g.f. of an exponential distribution. Hence, by uniqueness property, X1 + X2 does not have exponential distribution.
Note: (1) Generalization of the above property is also true. It means
n
if X1, X2 ... Xi ... Xn are i.i.d. exponential with mean a then X; does not
i=1
follow exponential distribution. It follows gamma distribution which will be discussed in next chapter.
(2) It can be shown that if X1 and X2 are i.i.d. variates, X1 - X2 does not follow exponential distribution.
Lack of Memory Property (Forgetfulness Property)
We have already seen that in case of discrete type distributions, geometric distribution possesses this property. The exponential distribution is the continuous probability distribution for which the lack of memory property holds good.
Statement: If X→ Exp (a) then
PXzs+t|X 2 s] = P(X 2 t] for s > 0, t> 0
Proof: We have already seen that
if X→ Exp (a) then P [X > x] = P [X > x] = 1 − F(x) = e-ax
L.H.S.
P [Xs + t|X ≥ s]
P[(Xs+t), (X ≥ s)]
P [X ≥ s]
P(X ≥s+t)
e-a(s+t)
P [X ≥ s]
e-as
= p-at
. (1)
R.H.S.
p [X t] = e-αt
From equation (1) and (2) it is clear that
... (2)
P[Xs+t|Xs] = P [X t] for s > 0, t > 0
... (3)
Note: (1) The equality (3) can be written in form
P[X2s+t] = P [X ≥ s] P [X≥ t]; s > 0, t> 0
(2) Converse of the above statement is also true i.e. If X is a continuous random variable taking non-negative values such that (3) holds good then X→ Exp (a). Hence, this is the characteristic property of exponential distribution.
(3) Interpretation: Equation (3) is equivalent to
P[Xs+tX>s] = P [XtX>0]
Suppose, X is life time of an electronic component. Then above equation means that the probability that the component will survive t time units more given that it has already survived s time units is same as the probability that a newly installed component will survive up to t time units.
Thus, whatever may be current age of the component, the distribution of remaining life time is same as the original life time distribution. In short so long as such a component is working it is as good as a new component. It neither improves not deteriorates due to ageing. It is observed in electronic equipments.
Practical situations (1) Some kinds of electrical components like fuses, safety valves, glass wares, transistors etc. do not wearout. They do not experience aging process. Hence, their life time distributions can be reasonably assumed be exponential.
(2) When atoms of radioactive isotopes like carbon, uranium, stronium split or turn into some other kind of atom and emit a pulse of radiation, the process is called 'radio active decay'. It can be detected by an instrument called 'Geiger Counter.' Suppose X denotes time till decay of an atom continues from the starting time (life time of decay). Then distribution of continuous r.v. X possesses lack of memory property. Hence, it is appropriate to take distribution of X as exponential.
Result (1): If X is continuous r.v. with p.d.f.
then, Y =
-
with mean a
0
f(x) = 1; 0 < x < 1
= 0
otherwise
log (1X) where, a > 0 follows an exponential distribution.