White Noise Process

[Monkey_Mike]

Hi,

Why doesn't a white noise process have independent increments, I thought white noise was just a random process ie et ~ N(0, sigma^2)

How would you prove it is markov?

 

[Scarecrow]

I've got the same question. Any tutors want to get involved?

 

[Monkey_Mike]

I actually worked this out in the end.... Julie helped from memory

You look at the Cov(Zt, Zt-1) etc and you get a sigma... ie not independent

(or something like that)

 

[Julie Lewis]

Let's think about discrete-time white noise...

For the increments to be independent, you need Z(n) - Z(n-1) to be independent of Z(0), Z(1), ... Z(n-1). The problem is Z(n) - Z(n-1) is not independent of Z(n-1).

You can think about the covariance to help you see this:

cov(Z(n) - Z(n-1) , Z(n-1)) = cov(Z(n), Z(n-1)) - cov(Z(n-1), Z(n-1))

= 0 - sigma^2

Since the cov is not equal to 0, Z(n)-Z(n-1) and Z(n-1) are not independent.

As for the Markov property, it holds in a trivial way. Since the Z's are indep, we have:

P(Z(n) = z(n) | Z(n-1) = z(n-1), Z(n-2) = z(n-2), ...., Z(0) = z(0))
= P(Z(n) = z(n))

and also:

P(Z(n) = z(n) | Z(n-1) = z(n-1)) = P(Z(n) = z(n))

So P(Z(n) = z(n) | Z(n-1) = z(n-1), Z(n-2) = z(n-2), ...., Z(0) = z(0))
= P(Z(n) = z(n) | Z(n-1) = z(n-1))

(they are both just equal to the unconditional probability)

Continuous-time white noise can be thought about in the same sort of way, but you have to use the filtration stuff when you think about all the past values since you can't list them all.

 

[Scarecrow]

Thanks, Julie

 

[Dipesh]

Yes we can see mathematically that a discrete time and discrete state space white noise process donot have independent increments but I am not able to understand it logically as in case of simple random walk which is also discrete time & discrete state process , possess independent increments Xt-Xt-1 = Yt then why do the white noise process donot have independent increments and what if the white noise is in continuous time and state space..??

 

[Mazvita Ndawana]

Twelve years later and your contribution has helped yet another prospective Actuary, thank you so much!

Let's think about discrete-time white noise...

For the increments to be independent, you need Z(n) - Z(n-1) to be independent of Z(0), Z(1), ... Z(n-1). The problem is Z(n) - Z(n-1) is not independent of Z(n-1).

You can think about the covariance to help you see this:

cov(Z(n) - Z(n-1) , Z(n-1)) = cov(Z(n), Z(n-1)) - cov(Z(n-1), Z(n-1))

= 0 - sigma^2

Since the cov is not equal to 0, Z(n)-Z(n-1) and Z(n-1) are not independent.

As for the Markov property, it holds in a trivial way. Since the Z's are indep, we have:

P(Z(n) = z(n) | Z(n-1) = z(n-1), Z(n-2) = z(n-2), ...., Z(0) = z(0))
= P(Z(n) = z(n))

and also:

P(Z(n) = z(n) | Z(n-1) = z(n-1)) = P(Z(n) = z(n))

So P(Z(n) = z(n) | Z(n-1) = z(n-1), Z(n-2) = z(n-2), ...., Z(0) = z(0))
= P(Z(n) = z(n) | Z(n-1) = z(n-1))

(they are both just equal to the unconditional probability)

Continuous-time white noise can be thought about in the same sort of way, but you have to use the filtration stuff when you think about all the past values since you can't list them all.