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Sum of two random variables- kind of

  1. Apr 4, 2010 #1
    I'm sitting here with an interesting problem that I can't seem to figure out. I'm given two random variables

    X=a*exp(j*phi)
    Y=b

    where both a and b are known constants.

    phi is uniformly distributed on the interval [0,2pi)

    a third random variable Z=X+Y.

    My goal, is to find the magnitude of the resulting vector.

    At first, I thought that this was an easy problem that could be solved by use of convolution. This doesn't work here since phi makes X a random vector. I tried using MATLAB to help solve it. I wrote an mfile that tried solving it using convolution, and it failed. I tried turning X into a toeplitz matrix and doing matrix multiplication to do the convolution, but that too, failed.

    Can anyone help me out?
     
  2. jcsd
  3. Apr 5, 2010 #2
    Does anyone have an idea of how to do this?
     
  4. Apr 5, 2010 #3

    mathman

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    I believe you should try to clarify what you are trying to get.

    Comments: Y=b, where b is constant, so Y is not particularly random.
    Z is a complex random variable, uniformly distributed over a circle of radius a, centered at b.
    What do want to know any further about Z?
     
  5. Apr 5, 2010 #4
    I'm essentially trying to find the pdf of Z and its' magnitude. I want to run a monte carlo simulation to verify my results and make a histogram of the results from the simulation in MATLAB for various values of A and B. The way phi is declared, (in matlab) is phi=rand(1,1000).*2*pi;
     
  6. Apr 6, 2010 #5
    Does that clarify things? I appreciate the help
     
  7. Apr 6, 2010 #6

    mathman

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    Z is complex so the usual concept of probability distribution [F(x)=P(X≤x)] can't be used, since X (random variable) has to be real. |Z|, being real, will have a probability distribution.
     
  8. Apr 6, 2010 #7
    I understand this, but obtaining the actual solution is where I'm stuck. I'm looking for the probability density function of Z so that I can create a histogram of the values of the magnitude of Z for various a and b values.
     
  9. Apr 7, 2010 #8

    mathman

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    Ordinary probability density functions are derivatives of ordinary distribution functions, which need real valued random variables. For a complex valued random variable you would need a two dimension density function treating the real and imaginary parts as (dependent) random variables.
     
  10. Apr 7, 2010 #9
    I think you're mis understanding my question a bit, I'll try to clarify. In matlab code, phi=rand(1,1000).*2*pi; this makes X a random vector, not a random variable. If it was a random variable, things would be much easier. I'm having trouble of addressing the magnitude of the density function of a complex random vector, X that has a constant, Y being added to it. I appreciate your help though!
     
  11. Apr 8, 2010 #10

    mathman

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    As far as I can tell, the density function for a random vector can only be expressed as a joint density function of its components.
     
  12. Apr 9, 2010 #11
    I've made progress on this one, but I'm confused on part of the theory behind it. Here is my Matlab code.
    >> a = 1;
    >> b = 0.25;
    >> phi = 2*pi*rand(1,10000);
    >> z = a+b*exp(j*phi);
    >> hist(abs(z),100)
    This code produces the histogram I was looking for. It is a U shaped histogram with its' smallest value at .75, largest at 1.25 and it looks to be symmetric at 1. I'm trying to come up with an expression for the |Z| in terms of a and b.

    My biggest question, and what would really help me out the most is if some one could provide like a geometric or linear algebra argument for why this
    problem is relevant to the problem of the eigenvectors of two random
    matrices (the area I'm tip-toeing my way into learning). I'm having trouble understanding this piece.
     
  13. Apr 10, 2010 #12

    mathman

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    z=1 + 0.25(cosφ + isinφ)
    |z|² = (1 + .25cosφ)² + (.25sinφ)² = 17/16 + .5cosφ

    You should be able to do the rest. The min and max for |z| agree with what you observed.
     
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