Martingale

#1
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

Ambassador to the humans
#2
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.
 
#3
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.
 
#4
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
#5
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 ...
 
#6
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.
 
#8
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
#9
On one hand you know \( Y_n = \alpha X_n + X_{n-1} \),
on the other hand you should know \( Y_{n+1} = \alpha X_{n+1} + X_n \) ?
 

BGM

TS Contributor
#11
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

Ambassador to the humans
#12
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?

My point is that it's hard to follow you and thus it's hard to help you.
 
#14
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
#15
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??
 
#16
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

\( aE[X_{n+1}|X_n,...,X_1] + E[X_n|X_n,...,X_1] \neq\ aX_n + X_{n-1} \)
 
#18
Do you mean that r.v. \( {X_n} \) has finite expectation and satisfy

\(E[X_{n+1} | X_n,..,X_0]= \alpha X_n +\beta X_{n-1} \)
 
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#20
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 \( Y_n = aX_n+X_{n-1} \)

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