Discrete time white noise

[forza_bologna]

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.

 

[Hemant Rupani]

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

 

[Julie Lewis]

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.

 

[Hemant Rupani]

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.

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. )

 

[Julie Lewis]

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.

 

[Hemant Rupani]

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.

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