# Correlation confused on integration

• cutesteph
In summary: However, in summary, the conversation discusses the equation x(n)=exp(i(x*n+theta1)) +exp(i(y*n+theta2)), where x and y are constants and theta1 and theta2 are uniform random variables on the interval [0,2*pi]. The conversation then goes on to discuss the expected value of a function R(f), which involves the above equation and its inverse, and concludes that the expected value is equal to zero.
cutesteph
x(n)=exp(i(x*n+theta1)) +exp(i(y*n+theta2)) x and y are constants , theta1, theta2 are uniform random on [0,2*pi]

R(f) =E[{ exp(i(x*n+theta1)) +exp(i(y*n+theta2)) }exp(-i(x*(n-f)+theta1)) +exp(-i(y*(n-f)+theta2)) }]
= 4 products 2 of which will be similar but with the opposite constants

E[exp(i*x*n+i*theta1 -i*x*n+i*n*f0i*theta1] = E[exp(i*x*f)] =integral exp(i*x*f) dtheta1 0 to 2 pi = 2*pi*exp(i(x*f)
Or am I suppose to integral over both theta1 and theta 2 to get 4*pi*exp(i*x*f) ? E[exp(i*n*n+i*theta1 -i*y*n +i*y*f- i*theta2 ] = double integral of the inside but it equals zero since we can separate the exponential and get exp(i*theta1) and the integral of this over 0 2pi = 0.
Right?

You're going to get much more responses if you learn LaTeX. It'll make your posts way more readable. Read this: https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

Anyway, because jbunniii asked me to:

cutesteph said:
Let ##x(n)=e^{i(xn+\theta_1)} +e^{i(yn+\theta_2)}##, where ##x## and ##y## are constants and ##\theta_1## and ##\theta_2## are uniform random on ##[0,2\pi]##.

Let ##R(f) = \mathrm{E}\left[\left(e^{i(xn+\theta_1)} +e^{i(yn+\theta_2)}\right)e^{-i(x(n-f)+\theta_1)} + e^{-i(y(n-f)+\theta_2)}\right]## This will give ##4## products of which ##2## will be similar but with the opposite constants.

We see that

$$\mathrm{E}\left[e^{ixn+i\theta_1 -ixn+inf_0i\theta_1}\right] = \textrm{E}\left[e^{ixf}\right] =\int_0^{2\pi} e^{ixf} d\theta_1 = 2\pi e^{ixf}$$

Or am I supposed to integral over both ##\theta_1## and ##\theta_2## to get ##4\pi e^{ixf}##?

$$\textrm{E}\left[e^{inn+i\theta_1 -iyn +iyf- i\theta_2}\right]$$

is a double integral of the inside but it equals zero since we can separate the exponential and get ##e^{i\theta_1}## and the integral over this with bounds ##0## to ##2\pi## equals ##0##, right?

Last edited:
There seem to be some typos in the formulas, which make it difficult to understand what is being presented.

## What is correlation?

Correlation refers to a statistical measure that determines the strength and direction of the relationship between two variables. It can help us understand how changes in one variable may affect changes in another variable.

## What does it mean if two variables have a high correlation?

If two variables have a high correlation, it means that they are strongly related and tend to change in a similar direction. This indicates a strong relationship between the variables and suggests that changes in one variable may cause changes in the other.

## How is correlation calculated?

Correlation is calculated using a mathematical formula called the correlation coefficient. This coefficient ranges from -1 to 1, with a value of 0 indicating no correlation, a value of 1 indicating a perfect positive correlation, and a value of -1 indicating a perfect negative correlation.

## What is the difference between correlation and causation?

Correlation does not imply causation. This means that just because two variables are correlated, it does not necessarily mean that one variable causes the other to change. There may be other factors at play that influence the relationship between the variables.

## How is correlation used in research?

Correlation is commonly used in research to determine the strength and direction of relationships between variables. It can help researchers identify patterns and make predictions about how changes in one variable may affect changes in another. However, it is important to note that correlation does not necessarily imply causation and should be interpreted carefully in research studies.

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