Expectation(X), characteristic function

Hi

       Let  f(x) be a prob density function.

Is there any significance to the value  f(E(x)), where 
E(.) is expectation operator ?

Why is the Fourier transform of f(x)  ( known as characteristic function )
significant ?  Note that Fourier transform  indicates  the frequency content 
of f(x).

A link explaining practical application of these concepts in 
engineering / science will be helpful.

               shankar

In article <a382521e.0410102239.4cda50d7@posting.google.com>,
kbc <kbc32@yahoo.com> wrote:


>       Let  f(x) be a prob density function.

>Is there any significance to the value  f(E(x)), where 
>E(.) is expectation operator ?

I assume you're talking about E[X], the expected value of the random variable of which f 
is the density, rather than E[x] where x stands for a number: E[x] = x.  

The answer is no, in general.  In fact, no particular value of f is significant
(a density is only defined up to sets of measure 0 anyway; change its value at a 
single point and it's still just as good as a density for the same random variable).

>Why is the Fourier transform of f(x)  ( known as characteristic function )
>significant ?  Note that Fourier transform  indicates  the frequency content 
>of f(x).

Because a lot of analysis can be done using Fourier transforms.

One example: the characteristic function of a sum of independent random variables
is the product of their characteristic functions.

Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            Vancouver, BC, Canada

kbc32@yahoo.com (kbc) wrote in message news:<a382521e.0410102239.4cda50d7@posting.google.com>...
> Hi
> 
>        Let  f(x) be a prob density function.
> 
> Is there any significance to the value  f(E(x)), where 
> E(.) is expectation operator ?

Does not make much sense to me. Becasue for a random variable x, E(x)
is a number. So, f(E(x)) is another number if you consider f(x) as a
function of x.

> Why is the Fourier transform of f(x)  ( known as characteristic function )
> significant ?  Note that Fourier transform  indicates  the frequency content 
> of f(x).

Because it is called the moment generating function. Looking at the
fourier formula as a mathematical transformation than something that
gives you the frequency content is much more helpful.

Intutively, the higher order moments in someway tells you about the
higher order harmonics of the signal in hand. But remember that
frequencies as we speak of it, does not make much sense for random
signals.

Mathematically, the characteristic function of a random variable X is
just
phi(X) = E(exp(sX)) or E(exp(iwX)), and this function is helpful in
calculating the moments of X, like, E(X) = d(phi(X))/dx, E(X^2) =
d^2(phi(X))/dx^2 (not precise formulas, but can be derived easily).

For literature, see Proakis - Digital Communications, Papoulis: Random
variables and stochastic process, or any descent book on probability
theory. I recommend Wozencraft and Jacobs, "Principles of
communications engineering", but it is out of print.

"kbc" <kbc32@yahoo.com> wrote in message
news:a382521e.0410102239.4cda50d7@posting.google.com...
> Hi
>
>        Let  f(x) be a prob density function.
>
> Is there any significance to the value  f(E(x)), where
> E(.) is expectation operator ?
>
> Why is the Fourier transform of f(x)  ( known as characteristic function )
> significant ?  Note that Fourier transform  indicates  the frequency
content
> of f(x).
>
> A link explaining practical application of these concepts in
> engineering / science will be helpful.
>
>                shankar


> Is there any significance to the value  f(E(x)), where
> E(.) is expectation operator ?

Let the lowercase Greek letter "mu" denote E(X) (which I will write as "u"
in this post) (assuming you meant E(X) where X denotes the random variable
under examination). I don't see any special significance to the density
value at "u" unless X possesses some extra special properties. For example,
if X is a unimodal and (perfectly) symmetric distribution such as a Gaussian
distribution, then (u, f(u)) will be where the density function peaks.


> Why is the Fourier transform of f(x)  ( known as characteristic function )
> significant ?  Note that Fourier transform  indicates  the frequency
content
> of f(x).

As I recall, the density of the sum of two independent random variables is
the convolution of their densities (assuming their densities exist).
Furthermore the characteristic function of a sum of two independent random
variables equals the product of their characteristic functions.

These mathematical properties allow one to rephrase the Central Limit
Theorem (CLT) in the following manner: The product of the characteristic
functions of a finite number of density functions approaches the
characteristic function of the Gaussian density as the number of density
functions approaches infinity (under the usual conditions stated for the
central limit to hold such as independent and identically distributed random
variables from the same population, etc, etc). Furthermore, as the
characteristic function of a random variable is essentially a Fourier
transform of its density, one can cast the CLT in the language of Fourier
transforms.

Shedar

On Mon, 11 Oct 2004 12:06:07 GMT, "shedar" <nobody@nonesuch.com>
wrote:

>[...]
>
>Let the lowercase Greek letter "mu" denote E(X) (which I will write as "u"
>in this post) 

Priceless.




************************

David C. Ullrich

kbc32@yahoo.com (kbc) wrote in message news:<a382521e.0410102239.4cda50d7@posting.google.com>...
> Hi
> 
>        Let  f(x) be a prob density function.
> 
> Is there any significance to the value  f(E(x)), where 
> E(.) is expectation operator ?

I'm not sure, but I *think* you might find an interesting 
property of the probability  

                   E(f(x))
 P(x < E(f(x))) = integral f(y) dy.
                   -inf

I don't think the integral as above can be evaluated 
analytically in general, but you might be able to 
do so for particular f's and E(f(x))'es.

Rune

kbc32@yahoo.com (kbc) wrote in message news:<a382521e.0410102239.4cda50d7@posting.google.com>...
> Hi
> 
>        Let  f(x) be a prob density function.
> 
> Is there any significance to the value  f(E(x)), where 
> E(.) is expectation operator ?

Not that I can think of. The mean of a random variable
need not correspond to anything significant, such as the
peak (if any) of the pdf.

> 
> Why is the Fourier transform of f(x)  ( known as characteristic function )
> significant ?  Note that Fourier transform  indicates  the frequency content 
> of f(x).

It's closely related to the moment-generating function,
which uniquely specifies all the moments, and the
pdf itself. It's often easier to do calculations and
proofs in terms of the MGF or characteristic function
than in terms of the pdf.

> 
> A link explaining practical application of these concepts in 
> engineering / science will be helpful.
> 

I don't know about scientific applications. It's a powerful
mathematical tool for doing difficult probability calculations.
Here are a couple of links on characteristic functions:

http://www2.sjsu.edu/faculty/watkins/charact.htm
http://planetmath.org/encyclopedia/CharacteristicFunction2.html

             - Randy

"kbc" <kbc32@yahoo.com> wrote in message 
news:a382521e.0410102239.4cda50d7@posting.google.com...
> Hi
>
>       Let  f(x) be a prob density function.
>
> Is there any significance to the value  f(E(x)), where
> E(.) is expectation operator ?
>
> Why is the Fourier transform of f(x)  ( known as characteristic function )
> significant ?  Note that Fourier transform  indicates  the frequency 
> content
> of f(x).
>
> A link explaining practical application of these concepts in
> engineering / science will be helpful.

A very simple example from the real world of engineering:

We have a resistor bridge circuit for remote control of a motor-driven or 
solenoid-driven rotary switch.
At the controlled end is the switch and one pole of the switch "rotates" 
over a series string of resistors of "equal" value.
At the controlling end is a manual switch and the pole of the switch also 
"rotates" over a series string of resistors of "equal" value.
The strings of resistors have the "same" voltage applied.
The poles of the controlled and controlling switches are to be compared with 
a difference amplifier.
When the voltages are the "same" the drive to the controlled switch is 
turned off.
When the voltages are "different" the drive to the controlled switch is 
energized.
The controlled switch is on a portable device.
The controller needs to work with any portable device that's connected.
The resistors on both sides were selected from a random selection of 
resistors of nominally the same value and the distribution of actual 
resistor values is known or can be approximated - likely cases and worst 
cases.
What is the Expected value of the proportional voltage on one side of the 
bridge for each setting of the switch?
What is the probability distribution for the proportional voltage on one 
side of the bridge for each setting of the switch?
Is there any situation in the worst case where the voltages will appear to 
be "different" even though the switches are in the same relative position? 
... such that the drive to the controlled switch never goes off?
What is the optimum threshold for the voltage difference for "same position" 
and "different position".

Hints:
The probability distribution of two resistors in series is the convolution 
of the individual probability distributions.
This can be calculated by multiplying the Fourier Transforms of each and 
inverse transforming the result.
Or, one can convolve the two distributions.
So, with a square distribution for one resistor, you would get a triangular 
distribution, twice as wide, for two resistors in series, etc.
It's also a rather neat demonstration of the central limit theorem ... just 
keep increasing the number of resistors in series until you reach something 
very much like a truncated Gaussian distribution - which I'm sure has a 
fancier name...

Fred 

Fred Marshall wrote:

  ...

> It's also a rather neat demonstration of the central limit theorem ... just 
> keep increasing the number of resistors in series until you reach something 
> very much like a truncated Gaussian distribution - which I'm sure has a 
> fancier name...

Binomial distribution? Where the densities are the binomial coefficients
(a row across Pascal's triangle) of order N?

Jerry
-- 
Engineering is the art of making what you want from things you can get.
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"Jerry Avins" <jya@ieee.org> wrote in message 
news:2t05glF1pm1boU1@uni-berlin.de...
> Fred Marshall wrote:
>
>  ...
>
>> It's also a rather neat demonstration of the central limit theorem ... 
>> just
>> keep increasing the number of resistors in series until you reach 
>> something
>> very much like a truncated Gaussian distribution - which I'm sure has a
>> fancier name...
>
> Binomial distribution? Where the densities are the binomial coefficients
> (a row across Pascal's triangle) of order N?
>

In discrete terms - yep.  What about continuous?