Transforms of Probability Density Functions
For the remainder of the course we will be interested in sums of random variables.
In
general, given a finite sequence of random variables {X
N
}
N
1
≥
, we define the random
variable Z by Z = X
1
+ X
2
+
...
+ X
N
.
We will want to understand how Z is distributed,
particularly in terms of the random variables that comprise its sum.
We will be
particularly interested in computations involving sums of independent random variables,
or sums of random variables that are both independent and identically distributed.
We consider now the transform of a probability density function as a means of
simplifying computations.
There are three motivations for doing this.
1. First, when attempting to compute the probability density function for the sum
of a pair of random variables, we found that when the variables are
independent
,
they distribute according to the convolution of the individual random variables.
Thus, setting Z = X + Y, we have:
dx
x
z
f
x
f
z
f
Y
X
Z
)
(
)
(
)
(

=
∫
∞
∞

.
2. We will find that there is a very simple relationship between the transform
function and the moments of a distribution.
Recall that the moments of a
distribution were the expectations E(X
n
), where n = 1, 2, 3, . . .
3. Transforms will give us a fairly straightforward means of proving the Central
Limit Theorem.
Definition of the Transform
There are several ways to define the transform, all sharing the common trait of
being defined in terms of an expectation.
Your text chose to compute the quantity E(e
sX
),
calling it a Laplacelike transform, but without paying any attention to the details of
integration.
Here lies a serious problem.
Unless the variable s is a complex number with
s
≤
1, the integrals will not converge without invoking a considerable body of theory
that is well beyond the scope of the course.
We can avoid this complication by defining
the transformation in terms of a Fourierlike integral by computing E(e
i
ϖ
X
).
Note that this
expectation is defined since the probability functions all have finite integrals over all
Euclidean space (by definition!!).
To see this we need the following facts:
1. e
i
ϖ
X
 = 1.
(Recall that the magnitude of a complex number is just the square
root of the product of the number times its complex conjugate.)
2. Using this fact, we have
dx
x
f
e
dx
x
f
e
X
X
i
X
x
i
)
(



)
(

∫
∫
ℜ
ℜ
≤
ϖ
ϖ
= 1.
We don’t need
absolute values around
)
(
x
f
X
since it is a nonnegative function, and once we use
the fact that the magnitude of the exponential term is 1, the integral is just that of
the probability density function over
ℜ
.
We could choose to ignore the problems and adopt the text’s approach simply as a
bookkeeping exercise in symbol manipulation, but we lose almost nothing in switching to
1
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the Fourier approach.
As you read the text, you will need to substitute
i
ϖ
for s, and
remember that
i
2
= 1.
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 Fall '08
 Britt
 Remainder, Normal Distribution, Probability, Probability theory, probability density function, lim P

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