Calculus I
Limits
Derivative
Exercices
Applications
Marginal analysis
© The scientific sentence. 2010
 Probability: Phenomena decay
1. Equation of decay
1.1. Definitions
A decay of phenomenon stands for a process in which
a set of objects loses one or more objects during
a certain time. This event of losing is known as decay
process.
Let's consider a group containing N identical objects
at the time t. Within the time dt, the group loses dN
objects, and becomes with the remaining N  dN objects.
If we perform several measurements of the number
of object lost, one after another. Each
measurement takes the same time dt.
First measurement: decay from t to t + dt
(during the period t + dt  t = dt):
dN(t) = N(t)  N(t + dt)
dN(t)/N(t)dt = constant
Second measurement: decay from t + dt to t + 2dt:
(during the period t + 2dt  t  dt = dt):
dN(t + dt) = N(t + 2dt)  N(t + dt)
dN(t + dt)/N(t + dt)dt = constant
...
Nth measurement: decay from t + (n  1)dt to t + ndt:
(during the period t + ndt  t  (n  1)dt = dt):
dN(t + ndt) = N(t + ndt)  N(t + (n  1)dt)
dN(t + ndt)/N(t + ndt)dt =
During the identical period of time dt, the phenomenon
loses its object with the same mode of decay, that is
at the same rate dN(t)/N(t)dt.
It is clear the dN(t) is negative, because there is
a lass of objects.
Therefore , with dN(t) > 0 :
dN(t)/N(t)dt = constant = λ
dN(t)/N(t) =  λ dt
λ is called decay constant.
In unit time, we have:  dN(t)/N(t) = λ.
That is, λ is also the probability that an object
in the group will be lost.
Integrating the previous equation gives:
ln (N(t)) =  λt + const.
Therefore:
N(t) = N_{o} exp{ λt}
Where N_{o} is the initial number of objects of the group
( at t = 0).
N(t) = N_{o} exp{ λt}
1.2. Example
We make a deposit of $1.00 in an account. The related bank
charges 20% per year on the sold. Will the sold be zero at
a time?
Here λ = 20% = 0.2
The sold at the time t is: N(t) = $1.00 exp{ 0.2t} . The sold
will never be zero, but tends to zero. Just at the 10th year,
the sold will be: $ 1/exp{2} = ¢13.53.
N.B.
The formula N(t) = N_{o}(1  20%)^{t}
will gives $1.00 x (0.8)^{t}
For 10 years, we have $(0.8)^{10} = ¢10.74.
The little difference is due to the fact that the rate
20% is not very small to 1. If the rate is 0.002, that
will correspond.
We recall at λ<<1, we have N(t) = N_{o} exp{ λt} = N_{o}(1  λ)^{t}
λ <<1, exp{ λt} = (1  λ)^{t}
1.3. Statistical basis of change
We can find the previous formula N(t) = N_{o} exp{ λt},
another way.
q = dN(t)/N(t) is the probability that the phenomenon
loses dN(t) objects from the remaining objects within
the group, during the time dt. Therefore the chance that
this phenomenon does not lose any object (survive) during
this time dt is p = 1  q = 1  dN(t)/N(t).
At first, we have N0 = N0
After dt: N1 = N0 = No  pNo = q N0
After 2dt: N2 = N1  pN1 = q^{2}No
... = ..
At the time n dt: N(t) = q^{n}No =
[1  dN(t)/N(t)]^{n}
After a long time t = n dt: the chance to survive is:
lim [1  dN(t)/N(t)]^{n}
n → ∞
=
lim [1  λdt]^{n}
n → ∞
=
lim [1  λt/n]^{n} = exp { λt}
n → ∞
Therefore
N(t) = N_{o} exp{ λt}
2. The activity of a phenomenon
If we derive the previous equation with respect
to the time t, we obtain:
dN(t)/dt =  λ N_{o} exp{ λt}
The rate  dN(t)/dt, at which the phenomenon decays, is called
its activity A(t), so
A(t) = λ N_{o} exp{ λt} = A_{o} exp{ λt}
A(t) = A_{o} exp{ λt}
A_{o} = λ N_{o}
A_{o} is the initial (at t = 0) activity of the group.
If the phenomenon changes probabilistically in different
modes λ_{i}, then the activity of the whole group is:
A(t) = A_{o} exp{ λt}
With λ = Σλ_{i}, the sum of the constant probabilities
λ_{i} related to a partial activity of mode i.
3. Half life a phenomenon
We define the half life t_{1/2} of a process
as the time for which the half of the number of objects in
the group survive.
If No is the number of objects at t = 0, then at the time
t, N(t) is just the half of No. That is:
N(t_{1/2}) = No/2
N(_{1/2}) = N_{o} exp{ λt_{1/2}} = No/2
exp{ λt_{1/2}} = 1/2
ln exp{ λt_{1/2}} = ln (1/2)
 λt_{1/2} =  ln 2, so t_{1/2} = ln 2/λ
t_{1/2} = ln 2/λ
4. Mean life time of a process
On average, from t = 0 to t, the group of objects loses:
No λ t_{mean}, that is equal to No. Therefore
No λ t_{mean} = No, thus t_{mean} = 1/λ
t_{mean} = 1/λ
We can find this formula another way. We calculate
the mean value of the time lived by the group before
to disappear. We have:
t_{mean} = ∫ t (P)
P = dN(t)/N0 =  λ dt exp { λ t} is the probability
for that dN(t) objects are lost during the time dt. Therefore:
t_{mean} = ∫ t ( λ dt exp { λ t})
t: 0 → +∞
=
 ∫ t exp { λ t} d(λ t) =
( 1/λ) ∫ (λt) exp { λ t} d(λ t)
∫ x exp { x} dx = (x  1)exp { x} + const.
lim (x e^{ x})= lim (x/e^{x}) = 0 (Hospital's rule)
x → +∞
Therefore
t_{mean} = ( 1/λ) [(x  1)exp { x}]
x: 0 → +∞
= ( 1/λ) [(x + 1)exp { x}] = ( 1/λ) [0  1] = 1/λ
x: 0 → +∞
t_{mean} = 1/λ

