# Understanding expectation operator notation

#### sc2

##### New Member
Hi, I am trying to understand how to interpret a problem that uses the expectation operator. Please see the attached pdf for more information.

Could someone explain in words how to read the definition of Yk that uses the summation operator?

Why would the mean of Y equal 0? I think I am misinterpreting the subscripts of X because if subtract 0 to 3 from k, to me that would shift the mean to a negative time index. Why is that 0?

In part b, I'm not even really sure where to start. Both of these questions seem relatively simple. I think my biggest hurdle is just knowing how to properly read this notation and the subscripts. Any resources would be greatly appreciated.

Thank you.

Also, if anyone knows how to insert equations directly into a post, please let me know. I would much rather do that and avoid using an attachment.

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#### JesperHP

##### TS Contributor
You have a discrete stochastic proces from -infinity to infinity in the variable X_k
This means that you are assuming the proces has been going back to the time of the dinosaurs and still further back to the big bang and still further back
Also it means the proces continues infinitely into the future....

Now you choose a value for k fx. k=0 this is simply a specific point in the proces fx. it could be the year of the birth of christ. Then take the expectation of the average of X(0),X(-1),X(-2),X(-3) which is simply the expectation of the average of fx. inhabitants in Jerusalem in the year 0 -1 -2 -3....

The indexation has nothing to do with the mean since the proces in X_k is iid: No matter where in the sequence X(inf),...,X(4),X(3),X(2),X(1),X(0),X(-1),X(-2),X(-3),X(-4),X(-5),....,X(-inf) you pick out 4 neighbouring X's the expectation of the sum of these and hence the average will be zero beacuse they all individually have an expectation of zero and the expectation of the sum is the sum of expectations...
E[sum(X(0),X(-1),X(-2),X(-3))] = E[X(0)]+E[X(-1)]+E[X(-2)]+E[X(-3)]

The definition of Y_k reads: Pick any point in time - that is any value for k - and the take the average of X(k),X(k-1),X(k-2),X(k-3) the average is moving as you alter k but you are consistently looking back only 4 periods (or perhaps three if you only count k-1,k-2,k-3 as "looking back").

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#### sc2

##### New Member
You have a discrete stochastic proces from -infinity to infinity in the variable X_k
This means that you are assuming the proces has been going back to the time of the dinosaurs and still further back to the big bang and still further back
Also it means the proces continues infinitely into the future....

Now you choose a value for k fx. k=0 this is simply a specific point in the proces fx. it could be the year of the birth of christ. Then take the expectation of the average of X(0),X(-1),X(-2),X(-3) which is simply the expectation of the average of fx. inhabitants in Jerusalem in the year 0 -1 -2 -3....

The indexation has nothing to do with the mean since the proces in X_k is iid: No matter where in the sequence X(inf),...,X(4),X(3),X(2),X(1),X(0),X(-1),X(-2),X(-3),X(-4),X(-5),....,X(-inf) you pick out 4 neighbouring X's the expectation of the sum of these and hence the average will be zero beacuse they all individually have an expectation of zero and the expectation of the sum is the sum of expectations...
E[sum(X(0),X(-1),X(-2),X(-3))] = E[X(0)]+E[X(-1)]+E[X(-2)]+E[X(-3)]

The definition of Y_k reads: Pick any point in time - that is any value for k - and the take the average of X(k),X(k-1),X(k-2),X(k-3) the average is moving as you alter k but you are consistently looking back only 4 periods (or perhaps three if you only count k-1,k-2,k-3 as "looking back").
Ok, I think that made some sense. But then in part b, how do I incorporate the values of n -10 to 10? Wouldn't the resulting values all be 0?

#### Dason

Wouldn't the resulting values all be 0?
No. If the variables $$Y_k$$ and $$Y_{k-n}$$ were independent then the value of $$E[Y_k Y_{k-n}]$$ would be 0. But they aren't independent in general.

#### sc2

##### New Member
No. If the variables $$Y_k$$ and $$Y_{k-n}$$ were independent then the value of $$E[Y_k Y_{k-n}]$$ would be 0. But they aren't independent in general.
Ok. If they are not independent then that means I cannot say E(Y_k)*E(Y_k-n) correct? How do I treat them and calculate different values for n = -10:10 ? I'm sorry, this is just very confusing for me.

#### JesperHP

##### TS Contributor
Ok. If they are not independent then that means I cannot say E(Y_k)*E(Y_k-n) correct?
Yes correct....

How do I treat them and calculate different values for n = -10:10 ? I'm sorry, this is just very confusing for me.
Start by noticing that E[Y(k),Y(k-n)] is the covariance because expectations are zero
so what you are being asked is to decide whether and how much the variable covary... When will the moving average covary? When they overlap...

Let k=0 and consider the sum
X(0)+X(-1)+X(-2)+X(-4)
Let k increase by 1 so k=1
X(1)+X(0)+X(-1)+X(-3)
Three variable are the same in these sum ..there is overlap and they will covary if X(0) is highvalued in one sum then it will also be highvalued in the other...

Compare:
Let k=0 and consider the sum
X(0)+X(-1)+X(-2)+X(-4)
Let k increase by 1 so k=2
X(2)+X(1)+X(0)+X(-1)
Only two variable are the same...new covariance to be calculated...

Compare:
Let k=0 and consider the sum
X(0)+X(-1)+X(-2)+X(-4)
Let k increase by 1 so k=3
X(3)+X(2)+X(1)+X(0)
Only one variable are the same...new covariance to be calculated...

Compare:
Let k=0 and consider the sum
X(0)+X(-1)+X(-2)+X(-4)
Let k increase by 1 so k=4
X(4)+X(3)+X(2)+X(1)
No variables are the same...new covariance to be calculated...it is zero, due to independence