Correlation confused on integration

[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?

 

[micromass]

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:

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?

 

[mathman]

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