Discrete time white noise

Discussion in 'CT4' started by forza_bologna, Aug 6, 2015.

  1. forza_bologna

    forza_bologna Keen member

    Hello,


    I came across a revision notes exercise which asks if in case of a discrete time white noise process, the increments are stationary.

    And the solution provided is: yes the increments are stationary for white noise process.

    Does somebody know why is that?
    I did not find anything on it in the theory.


    Thank you.
     
  2. Hemant Rupani

    Hemant Rupani Senior Member

    As far as I understand increments of discrete white noise is not stationary, But White noise itself is stationary process.
     
  3. Julie Lewis

    Julie Lewis Member

    I wonder if you're getting confused between independence and stationarity. The increments of a white noise process are not independent but they are stationary since theZn random variables are identically distributed. This means, for example, that the distribution of Z2-Z1 is the same as the distribution of Z3-Z2.
     
  4. Hemant Rupani

    Hemant Rupani Senior Member

    Hi mam, I understand independence and stationarity, but there may be flaw on my other understanding.

    reasoning for I said,"increments of discrete white noise is not stationary" is:-
    suppose for Binomial Stochastic process(a discrete white noise) , \(Z_i \sim Binomial(n,p) \).
    \( Z_{i+1}-Z_i \) can not have stationarity......... (as for \(Z_i=1~and~Z_i=n-2 \), \( Z_{i+1}-Z_i \) have different distribution. )
     
    Last edited: Aug 18, 2015
  5. Julie Lewis

    Julie Lewis Member

    But the Z random variables are identically distributed, so Z2-Z1 has the same distribution as Z3-Z2 and Z4-Z3,etc.

    For example, if Zn is normally distributed with mean mu and variance sigma^2, Z(n+1)-Zn has a normal with mean 0 and variance 2sigma^2. This doesn't depend on the value of n.
     
  6. Hemant Rupani

    Hemant Rupani Senior Member

    ohh Yes! Actually I was mistakenly thinking for conditional distribution as \( f(Z_{n+1}-Z_n|Z_n) \)....
     

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