Independent Increments

[Scotty]

Hi,

I was hoping if someone could explain to me how a stochastic process can have independent variables, but not independent increments. Because surely this increment is defined by the difference in value of two consecutive random variables. Therefore if this difference is dependent, then surely so is the random variable itself?....right?

Scotty

 

[didster]

Your question isn't clear, but perhaps this will help.

Consider two decks of cards (randomly shuffled)
One card from each deck is drawn without replacement, and the outcome is some formula of the two cards.
The cards from each deck (read variables) are independent.
However successive cards from each deck (read increments) are not independent - once you draw 4 Aces you must draw a non-ace thereafter etc.
Consequently the outcome's increments are also not independent.

 

[Scotty]

Hmmm doesn't help unfortunately. Let me try again....

Say you have a stochastic process that has independent variables, where we shall say the state space can take the numbers 1 to 5. Let's say I pick the number 3 then 5 then 4, where each time I picked a number, was independent of the number picked before....

Now assume that the increments are dependent.

So having an increment of 2, would then imply that I have affected the probability of then having an increment of -1. But this would then imply that I my choice of choosing 4 is not independent. Contradiction right?

Scotty

 

[Scotty]

Ahh Think I just worked it out...correct me if I'm wrong though

What I have actually concluded in the previous post was wrong.
The random variables I described (a number between 1 and 5) are not dependent on the number chosen before hand...it is dependent on the increment. This increment itself is not dependent on the value of the random variable preceeding it, but dependent on the increment that produced this number.

So the increments are dependent, but the random variables are independent.....I hope.

Scotty

 

[redzer]

I have a quick question about the CT4 Chpt 2 Summary notes.

It says "an increment of a stochastic process (that has a numerical state space) is just the change in the values between two times i.e. Xt2 - Xt1. If this is independent of the past values of the process up to and including time t1 then the process is said to have independent increments."

Why does the summary sheet include the phrase "and including time t1". I can understand the increment being independent of time t0 and further back in the process but the increment is based on the difference between Xt2 and Xt1.

I can't see this phrase mentioned in the notes other than the summary.

T

 

[jenil10]

hi..

Hi,
Suppose the value of the stock price is 30 on day1,32 on day 2 and 35 on day3. So the increment in the daily prices ie. (X3-X2) is 3. It is independent of the value of 32 (the price on day 2). But yes the value of Day 3 is equal to (day2 value + increment between day 3 and 2). So the price of Day 3 is dependent on the price of day 2 but the increment between day 2 and day 3 is independent of all increments upto day 2 (eg increment of 2 between day 1 and 2 at time 2)
I think the confusion was because you considered the value at time Xt1 and not increments upto time Xt1..
Thanks

 

[John Potter]

Consider a white noise process Zn = 1 or -1 with probabilty 0.5
(so toss a coin, 1 for heads, 0 for tails)

The coin tosses are independent of each other. White Noise is a series of IID RVs by definition.

Now consider Xn = Zn - Zn-1 (the increments of Zn)

Q What is E[X2]?
A Zero, it's just one coin toss minus another. Correct

Q What if you knew though that X1 = 2. Now, what's E[X2]?
A Well, the fact that X1 = 2 means we know that Z0= -1 and Z1 = 1, ie that the first coin toss was a tail, the second was a head. So, the E[X2] = E[Z2] - 1 = -1. Correct! The answer changes!

So, the E[X2] has changed when we know the value of X1.

Zn does NOT have independent increments.

This is a good example of a process that has IID RVs but does NOT have independent increments.

Zn is also a good example of a Markov process that does NOT have independent increments. I would know an example like this going into the exam,

Good luck everyone!
John

 

[SURESH SHARMA]

Consider a white noise process Zn = 1 or -1 with probabilty 0.5
(so toss a coin, 1 for heads, 0 for tails)

The coin tosses are independent of each other. White Noise is a series of IID RVs by definition.

Now consider Xn = Zn - Zn-1 (the increments of Zn)

Q What is E[X2]?
A Zero, it's just one coin toss minus another. Correct

Q What if you knew though that X1 = 2. Now, what's E[X2]?
A Well, the fact that X1 = 2 means we know that Z0= -1 and Z1 = 1, ie that the first coin toss was a tail, the second was a head. So, the E[X2] = E[Z2] - 1 = -1. Correct! The answer changes!

So, the E[X2] has changed when we know the value of X1.

Zn does NOT have independent increments.

This is a good example of a process that has IID RVs but does NOT have independent increments.

Zn is also a good example of a Markov process that does NOT have independent increments. I would know an example like this going into the exam,

Good luck everyone!
John

HI,

can you please explain how come E(Z2)=0 SO THAT E[X2]=E[Z2]-1=-1

REGARDS

SURESH SHARMA

 

[KaustavSen]

Hello Suresh,
I believe we can work this out as follows:
We know that \(Z_2\) can take the values \(1\) and \(-1\) with probability \(0.5\). So, we get:
\begin{align*}
\mathbb{E}(Z_2) &= 1\times \mathbb{P}(Z_2=1) + (-1)\times \mathbb{P}(Z_2=-1)\\
&=1\times0.5 -1\times0.5\\
&=0
\end{align*}

Now, we also have the information that \(Z_1 = 1\). So, finally we get:
\begin{align*}
\mathbb{E}(X_2) &= \mathbb{E}(Z_2-Z_1)\\
&=\mathbb{E}(Z_2) - \mathbb{E}(Z_1)\\
&=0-1=-1
\end{align*}
Hope this helps.
Regards
Kaustav

 

[SURESH SHARMA]

Hello Suresh,
I believe we can work this out as follows:
We know that \(Z_2\) can take the values \(1\) and \(-1\) with probability \(0.5\). So, we get:
\begin{align*}
\mathbb{E}(Z_2) &= 1\times \mathbb{P}(Z_2=1) + (-1)\times \mathbb{P}(Z_2=-1)\\
&=1\times0.5 -1\times0.5\\
&=0
\end{align*}

Now, we also have the information that \(Z_1 = 1\). So, finally we get:
\begin{align*}
\mathbb{E}(X_2) &= \mathbb{E}(Z_2-Z_1)\\
&=\mathbb{E}(Z_2) - \mathbb{E}(Z_1)\\
&=0-1=-1
\end{align*}
Hope this helps.
Regards
Kaustav

thanks kaustav

regards
Suresh sharma