# Martingale

#### LED2127

##### New Member
I must find an appropriate value so that the sequence constitutes a martingale wrt {X_n}.

I have attached the problem so it is easier to read then typing it here.

I understand the definition of martingale I am just not understanding how to get what the problem is asking. I feel like they are asking to condition on Y_n and if so I am having trouble understanding how to do that. Any kind of help would be appreciated.

#### Dason

Do you know latex? The forum supports latex in the form of [noparse]$$Put latex here$$[/noparse] type tags.

What is giving you the most trouble? What have you done so far? What would it mean for Yn to be a martingale? We want to see that you've put some effort into this.

#### LED2127

##### New Member
I'm sorry I don't know latex, I attached the problem as a word document, otherwise it would have been a headache for anyone to read.

Regarding the question,

I know the definition of a martingale. Its a stochastic process such that the conditional expected value of an observation at some time t , given all the observations up to some earlier time s, is equal to the observation at that earlier time s.

i.e. it's expected value of the next observation is/should be the same expected value of the current observation independent of the history.

I need to find a value for a that makes Y_n a martingale wrt to {X_n}

Y_n= a*X_n+ X_n-1. n>=1 Y_0=X_0

I am not sure how to start showing that Y_n constitutes a martingale wrt {X_n}.
or how you would find a value for a to make Y_n be a martingale wrt {X_n}

Any help would be appreciated.

#### LED2127

##### New Member
For Y_n to be a martingale it would have to satisfy E[Yn] is finite and E[Y_n+1|X1,...Xn]= Yn. I think I am just not seeing this application to my problem, forgive me I can lack some common sense sometimes.

#### BGM

##### TS Contributor
You say that you know you need to check

$$E[Y_{n+1}|X_n, \ldots, X_1] = Y_n$$

So what can you do? Replace $$Y_{n+1}$$ by the expression of $$X_{i}$$s and then use the given fact of the conditional expectation of those $$X_{i}$$s ...

#### LED2127

##### New Member
So,

$$E[Y_{n+1}|X_n, \ldots,X_1]=Y_n$$

$$E[X_n|X_n,\ldots,X_1]=aX_n+X_{n-1}$$

$$X_n=aX_n+X_{n-1}$$

$$a= [X_n-X_{n-1}]/[X_n]$$

or this isn't what you were getting at.

#### LED2127

##### New Member
if I replace the $$Y_{n+1}$$ by $$X_i 's$$

$$E[Y_{n+1}|X_n, \ldots,X_1]= Y_n$$

$$E[g(X_n, \ldots,X_1)|X_n, \ldots,X_1]=g(X_n, \ldots,X_1)= aX_n + X_{n-1}$$

$$E[Y_n|X_n, \ldots,X_1]= Y_n$$
with the law of total probability

$$E[Y_{n+1}]=E[E[Y_{n+1}|Y_n,\ldots,X_1]]=E[Y_n]$$

now to find a:
$$E[Y_n]= aX_n + X_{n-1}$$

#### BGM

##### TS Contributor
On one hand you know $$Y_n = \alpha X_n + X_{n-1}$$,
on the other hand you should know $$Y_{n+1} = \alpha X_{n+1} + X_n$$ ?

#### LED2127

##### New Member
So $$E[Y_n]=aX_n+X_{n-1}$$
if a=0

then $$E[Y_n]=X_{n-1}$$

is that what you mean

#### BGM

##### TS Contributor
Ok more direct:

$$E[Y_{n+1}|X_n, \ldots, X_1] = Y_n$$

$$\Rightarrow E[\alpha X_{n+1} + X_n|X_n, \ldots, X_1] = \alpha X_n + X_{n-1}$$

#### Dason

I'm never sure if you're continuing something from the previous line or going off on a new line of thought. You're stating things without giving reasons - sometimes I'm not sure if you're just saying what you want to show or you're assuming what you want to show. For example in post #8 are your first two lines connected or were you just writing unconnected thoughts?

#### LED2127

##### New Member
I'm sorry if my posts aren't clear I will write them more clearly from now on. I appreciate anyone who takes the time to help.

#### LED2127

##### New Member
Is it correct if I do the following:

Given $$E[aX_{n+1}+X_n |X_n,...X_1]$$
$$= a E[X_{n+1}|X_n,..X_1] + E[X_n|X_n,..,X_1]$$
$$=a X_n + X_{n-1}$$

and if this is correct and equate it to what is given : $$Y_n=a X_n + X_{n-1}$$

I obtain: $$a X_n + X_{n-1}=a X_n + X_{n-1}$$

and further a=1

#### BGM

##### TS Contributor
The first two lines are fine and is what the question asking you to do.

Have you forgotten the question? Why not use the properties given for the conditional expectation of $$X_{n+1}$$ on the third line??

#### LED2127

##### New Member
Please forgive me I am unsure what you mean use the conditional expectation of $$X_{n+1}$$.

do you mean that after using the linearity property on the second line that

$$aE[X_{n+1}|X_n,...,X_1] + E[X_n|X_n,...,X_1] \neq\ aX_n + X_{n-1}$$

#### BGM

##### TS Contributor
So look at the question statement again. What property is given?

#### LED2127

##### New Member
Do you mean that r.v. $${X_n}$$ has finite expectation and satisfy

$$E[X_{n+1} | X_n,..,X_0]= \alpha X_n +\beta X_{n-1}$$

Last edited:

#### Dason

That would definitely be something to keep in mind.

#### LED2127

##### New Member
Let me try this again,
considering:
$$aE[X_{n+1}|X_n,..,X_1]+E[X_n|X_n,...,X_1] =Y_n$$

and if I use the given property

$$= a[\alpha X_n+\beta X_{n-1}] +X_{n-1} = Y_n$$

and $$Y_n = aX_n+X_{n-1}$$

$$= a[\alpha X_n+\beta X_{n-1}] +X_{n-1} = aX_n+X_{n-1}$$

Last edited: