How to calculate the sum and the subtraction of many random variables that follow exponential distributions and have different parameters ?
(The value of Lambda is different for all or some variables).
example :
L(t) = f(t) + g(t) - h(t)
with
f(t) = a.expo(-a.t)
g(t) = b.expo(-b.t)
h(t) = c.expo(-c.t)
st:
a = Lambda_1
b = Lambda_2
c = Lambda_3.
(continued)
In case of more terms (all with different means m_j>0, j=1,2,...,n) the formulas are as follows (ti replaced by -s)
ch.f.(X_1+X_2+...+X_n)(t) = 1/ [ (1+m_1 s) (1 + m_2 s) ... (1 + m_n s)]
= \sum_{j=1}^n A_j / (1+m_j s),
where A_j = \prod_{k\ne j} [ 1 - m_k / m_j]^{-1}
Therefore, in such cases the density of the sum is equal to
\sum_{j=1}^n A_j / m_j \exp( - x/m_j ), for x>0.
If X_j in the sum is preceded by sign -, then the first two formulas remain valid after replacing m_j by - m_j. The last requires replacing the exponential density for positive variable by the opposite one.
Applying characteristic functions, under independence of the summands the result is a rational function. Decomposing it into simple F ructions, we arrive at a linear combination of characteristic functions of the exponential pd-s.
EXAMPLE. If X and Y obey exponential pds with expectations a and b resp. then
ch.f.(X+Y)(t)= 1/(1-ati)(1-bti)=
C(1-ati)+D(1-bti)
with some constants C and D
(continued)
In case of more terms (all with different means m_j>0, j=1,2,...,n) the formulas are as follows (ti replaced by -s)
ch.f.(X_1+X_2+...+X_n)(t) = 1/ [ (1+m_1 s) (1 + m_2 s) ... (1 + m_n s)]
= \sum_{j=1}^n A_j / (1+m_j s),
where A_j = \prod_{k\ne j} [ 1 - m_k / m_j]^{-1}
Therefore, in such cases the density of the sum is equal to
\sum_{j=1}^n A_j / m_j \exp( - x/m_j ), for x>0.
If X_j in the sum is preceded by sign -, then the first two formulas remain valid after replacing m_j by - m_j. The last requires replacing the exponential density for positive variable by the opposite one.
Assuming the exponentially distributed random variables X1, x2, ..., Xn are all independent, the mean of the sum or difference is
E(X1+/-X2+/- ... +/- Xn) = E(X1)+/-E(X2) +/- ... +/- E(Xn) and the variance is V(X1+/-X2+/-... +/-Xn) = V(X1)+V(X2)+...+V(Xn). The sum or difference is also exponentially distributed. That means if their means are b1, b2, ..., bn, respectively, the probability density function of the sum or difference is
f(x) = 1/(b1+/-b2+/-...+/_ bn)*exp(1/b1+1/b2+...+1/bn)x, x > 0, zero elsewhere.
Elementary (but important) note: The sums/differences of independent exponential distributed variables are not exponentially distributed!!!
Joachims answer (as I interpret it) is correct -- although his last sentence has room for improvement with respect to precision and clarity 😉😌
Dear Ette Etuk
Thank you for your answer
Are you sure of that !
What is true for sums and differences is also true for sums only ...
I can give you this book where the author proofs that the sum doesn't obey exponential distribution :
"Introduction to Probability Models | 11th Edition"
https://www.elsevier.com/books/introduction-to-probability-models/ross/978-0-12-407948-9
you find that in chapter 5.
What is known: if all variables have the same parameter value than the sum obeys Erlang distribution.
In general, we can use the characteristic function, which exists for all distributions. As we know the density of exponential distribution, therefore we can find the characteristic function for the linear combinations of exponential random variables.
If they are not independent, they you need to use for variances:
Var[X±Y] = Var[X] + Var[Y] ± 2Cov[X,Y]
provided that you know how to calculate the latter quantity.
(the addition/subtraction of means present no problem).
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