Let
and
be two independent random variables. In this
lecture we discuss how to derive the distribution of their sum, i.e. the
distribution of the random variable
defined
as:
We
first characterize the distribution function
of
;
then, we characterize the probability mass
function and the probability density
function for the two special cases in which
is respectively discrete or
absolutely continuous.
The following proposition characterizes the distribution function of the sum in terms of the distribution functions of the two summands:
Proposition (Distribution function of a
sum)_
Let
and
be two independent random variables and denote by
and
their respective distribution functions.
Let:
and
denote the distribution function of
by
.
The following
holds:
or:
The first formula is derived as
follows:
The
second formula is symmetric to the first.
Example_
Let
be a uniform random variable with
support
and probability density
function:
and
another uniform random variable, independent of
,
with support
and probability density
function:
The
distribution function of
is:
The
distribution function of
is:
There
are four cases to consider:
If
,
then:
If
,
then:
If
,
then:
If
,
then:
Therefore, combining these four possible cases, we
obtain:
When the two summands are discrete random variables, the probability mass function of their sum can be derived as follows:
Proposition (Probability mass function of a
sum)_
Let
and
be two independent discrete random variables and denote by
and
their respective probability mass functions and by
and
their supports.
Let:
and
denote the probability mass function of
by
.
The following
holds:
or:
The first formula is derived as
follows:
The
second formula is symmetric to the first.
The two summations above are called convolutions (of two probability mass functions).
Example_
Let
be a discrete random variable with support
and probability mass
function:
and
another discrete random variable, independent of
,
with support
and probability mass
function:
Define
Its
support is:
The
probability mass function of
,
evaluated at
is:
Evaluated
at
,
it
is:
Evaluated
at
,
it
is:
Therefore,
the probability mass function of
is:
When the two summands are absolutely continuous random variables, the probability density function of their sum can be derived as follows:
Proposition (Probability density of a
sum)_
Let
and
be two independent absolutely continuous random variables and denote by
and
their respective probability density functions.
Let:
and
denote the probability density function of
by
.
The following
holds:
or:
The distribution function
of a sum of independent variables
is:
Differentiating
both sides (and using the fact that the density
function is the derivative of the distribution function), we
obtain:
The
second formula is symmetric to the first.
The two integrals above are called convolutions (of two probability density functions).
Example_
Let
be an exponential random variable with support
and probability density
function:
and
another exponential random variable, independent of
,
with support
and probability density
function:
Define:
The
support of
is:
When
,
the probability density function of
is:
Therefore,
the probability density function of
is:
We have discussed above how to derive the distribution of the sum of two
independent random variables. How do we derive the distribution of the sum of
more than two mutually independent random
variables? Suppose
,
,
...,
are
mutually independent random variables and let
be their
sum:
The
distribution of
can be derived recursively, using the results for sums of two random variables
given above:
first,
define:
and
compute the distribution of
;
then,
define:
and
compute the distribution of
;
and so on, until the distribution of
can be computed
from: