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The mean and variance

  1. Jul 27, 2007 #1
    Can anybody explain to me how to get the mean and the Variance for a specific function.
    Thanks alot.
     
  2. jcsd
  3. Jul 27, 2007 #2
    If you know exactly the pdf (probability density function) [tex]f(x)[/tex], the formula for the mean is
    [tex]\mu = E[x] = \int x f(x) dx[/tex]
    and for the variance
    [tex]\sigma^{2} = E[(x - E[x])^{2}] = \int (x - E[x])^{2} f(x) dx[/tex]

    If you only have experimental data, you can estimate the mean and variance of the distribution :
    [tex]m = 1/N\times\sum_{i = 1}^{N} x_{i}[/tex]
    [tex]s^{2} = 1/(N - 1)\times\sum_{i = 1}^{N} (x_{i} - m)^{2}[/tex]

    Hope it helps
     
  4. Jul 27, 2007 #3

    EnumaElish

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    For a specific function h of a random variable x with p.d.f. f(x),

    Mean = E[h(x)] = ∫h(x)f(x) dx
    Variance = E[(h(x) - Mean)^2] = ∫(h(x) - Mean)^2 f(x) dx

    both integrated over the domain of f(x).

    m = Σi h(xi)/N
    s^2 = Σi (h(xi) - m)^2/(N-1)
     
    Last edited: Jul 27, 2007
  5. Jul 30, 2007 #4
    for my function
    Hn(t)= (-1)^n cos(2π fc t)* e^[(t^2)/4] *d^n/dt^n *e^[(t^2)/4]

    what it will be h(x) and f(x)?
    Thanks alot!
     
    Last edited: Jul 30, 2007
  6. Jul 30, 2007 #5

    EnumaElish

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    For me to answer this, you should tell me what is random. (You need a random variable for this to work.) Are signal times (t) random? Is the time between two signals random? What is your random variable?
     
    Last edited: Jul 30, 2007
  7. Jul 30, 2007 #6
    n is also random variable.
     
  8. Jul 30, 2007 #7

    chroot

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    T.Engineer,

    Please post the complete problem, exactly as it was given to you. You seem to be leaving out a lot of important information.

    - Warren
     
  9. Jul 31, 2007 #8
    I'd like to find the mean and variance for the following function
    Hn(t)= (-1)^n cos(2π fc t)* e^[(t^2)/4] *d^n/dt^n *e^[(t^2)/4]

    where n=1,2,...,N
    fc=6.5MHz
     
  10. Jul 31, 2007 #9

    EnumaElish

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    You can simulate this for a given n (random t).

    You can simulate it for a given t and random n.

    You can also simulate it with random t and random n.

    You can collect the data and calculate the mean and the variance.
     
  11. Jul 31, 2007 #10
    Thanks alot!
    but I dont know how to start?
    should I use the method which represented by
    http://w3eos.whoi.edu/12.747/notes/lect06/l06s02.html
    and if yes, how to enter my function to this simulation?
    for example in the first equation , what did he mean by
    yi, y
     
  12. Jul 31, 2007 #11
    Also I'd like to find an expression for the mean and variance in general not by just data.
    first of all I want to find a mathematical expression for my function Hn(t).
    Or , should firstly to find the data? and after that to find the mathematical expression for mean and variance of Hn(t)??
     
  13. Jul 31, 2007 #12
    And sure I prefer to simulate with random t and random n.
    But how?

    Thanks alot!!!
     
  14. Jul 31, 2007 #13

    EnumaElish

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    yi is the i'th individual data point (function value). (y1 = first data point, y2 = second, ...)

    "y bar" is the mean yi, calculated as the average of all the yi's:

    y bar = Σi yi / N for i = 1, ..., N.

    To simulate the function, you need to answer:
    1. What variables are random?
    2. Are they independent?
    3. What is the probability distribution function for each random variable?

    Your answers may be:
    1. t and n (see footnote)
    2. Yes
    3. This is the difficult question. What determines the time at which the signal is emitted? Is it a random process like nuclear (radioactive) decay? And what determines n?

    To start simple, you can assume ti is distributed uniformly between ti-1 and Ti, where Ti is an upper bound. Also assume n is uniformly distributed between 0 and M (a large number).

    1. Let i = 0. Assume t0 = 0. Assume n0 = 0.
    2. Let i = i + 1. Generate uniform random value ti between ti-1 and Ti (say, Ti = ti-1 + 1)
    3. Generate uniform random value ni between 0 and M (say, M = 10)
    4. Evaluate H[ni](ti).
    5. Go to step 2.
    _______________________________
    Footnote: Although I don't understand why n is random, I am going with your statement that n is random.
     
    Last edited: Jul 31, 2007
  15. Jul 31, 2007 #14
    The transmitted signal represented by the function Hn(t)
    which it will be transmitted according to time hopping format for kth users and given by :

    S(t)= [tex]\sum^{\infty}_{j=-\infty} A^k Hn(t - jTf-cj Tc - rd^kj[/tex]

    where A: is the signal amplitude
    Hn(t): transmitted signal
    Tf: is the frame time, which is typically a hundre to a thousand times
    the impulse width resulting in a signal with very low duty cycle.
    Each frame is divided into N tim slots with duration Tc
    cj: time-hopping sequence (0<=cj<= N) with period Tc
    This provides an additional shift in order to avoid catastrophic
    collisions due to multipl access interference.
    d: is the sequence of the MN-ary data stream generated by the kth
    source after channel coding.
    r :is the additional time shift utilized by the N-ary pulse positio
    modulation.
    I dont know if the above information is important for what I am going to determine?
    thanks alot!
     
  16. Jul 31, 2007 #15

    EnumaElish

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    It will take me some time to digest this information.

    I thought Hn was only a function of t. See your earlier post https://www.physicsforums.com/showpost.php?p=1390181&postcount=4

    How does the t in your last post relate to the t in your earlier post? Are they the same t? Do you mean to say Hn(#) = (-1)^n cos(2π fc #)* e^[(#^2)/4] *d^n/d#^n *e^[(#^2)/4] for some generic (general) argument # where # = t - jTf - cj Tc - r d^kj ?

    If this is not it, what is it?

    Assuming this is it, I advise you start simple by assuming t is uniformly distributed; you can easily change it later and replace it with a more complicated frequency distribution. I still do not understand why n is random; but if you think it is, then I am not going to argue with you. I will advise that you start simple and also assume n has a uniform frequency distribution.

    Once you attach each of t and n to a frequency distribution, you can easily simulate your function to calculate the AC coefficient. You can also determine it analytically, by applying the formulas under this thread and under this other thread.
     
    Last edited: Aug 1, 2007
  17. Aug 2, 2007 #16
    yes, exactly! that's right!
     
  18. Aug 2, 2007 #17
    You mean firstly I will work for n=1, for example.
    is not that right?
     
  19. Aug 2, 2007 #18
    Now, can you tell me how to start and from where?
    Really I get confused.
    Thanks alot!
     
  20. Aug 2, 2007 #19

    EnumaElish

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