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In order to study the characteristics of a random process , let us look at some of the basic properties and operations of a random process. Below wewill focus on the operations of the random signals that compose our random processes. We will denote our random process with X and a random variable from a random process or signal by x .

Mean value

Finding the average value of a set of random signals or random variables is probably the most fundamental concepts we use inevaluating random processes through any sort of statistical method. The mean of a random process is the average of all realizations of that process. In order to find this average, we must look at a random signal over arange of time (possible values) and determine our average from this set of values. The mean , or average, of a random process, x t , is given by the following equation:

m x t x t X X x x f x
This equation may seem quite cluttered at first glance, but we want to introduce you to the various notations used torepresent the mean of a random signal or process. Throughout texts and other readings, remember that these will all equalthe same thing. The symbol, x t , and the X with a bar over it are often used as a short-hand to represent anaverage, so you might see it in certain textbooks. The other important notation used is, X , which represents the "expected value of X " or the mathematical expectation. This notation is very common and will appearagain.

If the random variables, which make up our random process, are discrete or quantized values, such as in a binary process,then the integrals become summations over all the possible values of the random variable. In this case, our expectedvalue becomes

x n x x n
If we have two random signals or variables, their averages can reveal how the two signals interact. If the product of the two individualaverages of both signals do not equal the average of the product of the two signals, then the twosignals are said to be linearly independent , also referred to as uncorrelated .

In the case where we have a random process in which only one sample can be viewed at a time, then we will often not haveall the information available to calculate the mean using the density function as shown above. In this case we mustestimate the mean through the time-average mean , discussed later. For fields such as signal processing that deal mainly withdiscrete signals and values, then these are the averages most commonly used.

Properties of the mean

  • The expected value of a constant, , is the constant:
  • Adding a constant, , to each term increases the expected value by that constant:
    X X
  • Multiplying the random variable by a constant, , multiplies the expected value by that constant.
    X X
  • The expected value of the sum of two or more random variables, is the sum of each individual expectedvalue.
    X Y X Y

Mean-square value

If we look at the second moment of the term (we now look at x 2 in the integral), then we will have the mean-square value of our random process. As you would expect, this is written as

X 2 X 2 x x 2 f x
This equation is also often referred to as the average power of a process or signal.

Variance

Now that we have an idea about the average value or values that a random process takes, we are often interested in seeingjust how spread out the different random values might be. To do this, we look at the variance which is a measure of this spread. The variance, often denoted by 2 , is written as follows:

2 Var X X X 2 x x X 2 f x
Using the rules for the expected value, we can rewrite this formula as the following form, which is commonly seen:
2 X 2 X 2 X 2 X 2

Standard deviation

Another common statistical tool is the standard deviation. Once you know how to calculate the variance, the standarddeviation is simply the square root of the variance , or .

Properties of variance

  • The variance of a constant, , equals zero:
    Var 0
  • Adding a constant, , to a random variable does not affect the variance because the mean increases by the same value:
    Var X X X
  • Multiplying the random variable by a constant, , increases the variance by the square of the constant:
    Var X X 2 X
  • The variance of the sum of two random variables only equals the sum of the variances if the variable are independent .
    Var X Y X Y X Y
    Otherwise, if the random variable are not independent, then we must also include the covariance of the product of the variablesas follows:
    Var X Y X 2 Cov X Y Y

Time averages

In the case where we can not view the entire ensemble of the random process, we must use time averages to estimate thevalues of the mean and variance for the process. Generally, this will only give us acceptable results for independent and ergodic processes, meaning those processes in which each signal or member of the process seems to have thesame statistical behavior as the entire process. The time averages will also only be taken over a finite interval sincewe will only be able to see a finite part of the sample.

Estimating the mean

For the ergodic random process, x t , we will estimate the mean using the time averaging function defined as

X X 1 T t 0 T X t
However, for most real-world situations we will be dealing with discrete values in our computations and signals. Wewill represent this mean as
X X 1 N n 1 N X n

Estimating the variance

Once the mean of our random process has been estimated then we can simply use those values in the following varianceequation (introduced in one of the above sections)

x 2 X 2 X 2

Example

Let us now look at how some of the formulas and concepts above apply to a simple example. We will just look at a single,continuous random variable for this example, but the calculations and methods are the same for a random process.For this example, we will consider a random variable having the probability density function described below and shown in .

f x 1 10 10 x 20 0

Probability density function

A uniform probability density function.

First, we will use to solve for the mean value.

X x 10 20 x 1 10 x 10 20 1 10 x 2 2 1 10 200 50 15
Using we can obtain the mean-square value for the above density function.
X 2 x 10 20 x 2 1 10 x 10 20 1 10 x 3 3 1 10 8000 3 1000 3 233.33
And finally, let us solve for the variance of this function.
2 X 2 X 2 233.33 15 2 8.33

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
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Abhi
Commplementary angles
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or infinite solutions?
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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Source:  OpenStax, Intro to digital signal processing. OpenStax CNX. Jan 22, 2004 Download for free at http://cnx.org/content/col10203/1.4
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