# Using a Logarithmic Transformation for a Simpler Random Walk Model

• WMDhamnekar
In summary, the conversation discusses various mathematical concepts related to the sequence ##M_n##. These include the computation of ##E(\log(M_n))##, the proof of ##M_{\infty}=0##, and the existence of a finite value ##C## such that ##\mathbb{E}[M^2_n]\leq C## for all ##n##. The conversation also includes a discussion about the correctness of certain answers and the use of the Law of Large Numbers in proving certain results.
WMDhamnekar
MHB
Homework Statement
Let ## X_1, X_2, . . . ## be independent, identically distributed random variables with ##\mathbb{P}\{X_j=2\} =\frac13 , \mathbb{P} \{ X_j = \frac12 \} =\frac23 ##

Let ##M_0=1 ## and for ##n \geq 1, M_n= X_1X_2... X_n ##

1. Show that ##M_n## is a martingale.

2. Explain why ##M_n## satisfies the conditions of the martingale convergence theorem.

3. Let ##M_{\infty}= \lim\limits_{n\to\infty} M_n.## Explain why ##M_{\infty}=0## (Hint: there are at least two ways to show this. One is to consider ##\log M_n ## and use the law of large numbers. Another is to note that with probability one ##M_{n+1}/M_n## does not converge.)

4. Use the optional sampling theorem to determine the probability that ## M_n## ever attains a value as large as 64.

5. Does there exist a ##C < \infty ## such that ##\mathbb{E}[ M^2_n ] \leq C \forall n ##
Relevant Equations
No relevant equations

How would you answer rest of the questions 4 and 5 ?

Why does ##ln(M_n)=ln(M_{n-1})##?

Office_Shredder said:
Why does ##ln(M_n)=ln(M_{n-1})##?
Sorry, if ##\log{M_n} =0, \log{M_{n-1}}=\frac13 \log{2}## or ##\frac23 \log{\frac12}##

Whatever it may be, we can't say ## M_{\infty} =0 ##

WMDhamnekar said:
Sorry, if ##\log{M_n} =0, \log{M_{n-1}}=\frac13 \log{2}## or ##\frac23 \log{\frac12}##

Whatever it may be, we can't say ## M_{\infty} =0 ##

This feels like you're looking backwards - why are you computing what ##\log(M_{n-1})## ?

What is ##E(\log(M_{n+1})-\log(M_n))##

Office_Shredder said:
This feels like you're looking backwards - why are you computing what ##\log(M_{n-1})## ?

What is ##E(\log(M_{n+1})-\log(M_n))##
##(\log\{E[M_{n+1}]=1\} -\log\{E[M_n=1]\}) = 0-0 =0##

The expected value of log of ##M_n## s not 1. Your answer for part three is wrong, the book is right. Try computing that expected value from the definition.

Office_Shredder said:
The expected value of log of ##M_n## s not 1. Your answer for part three is wrong, the book is right. Try computing that expected value from the definition.
##E[X_n] = 2\times \frac13 + \frac12\times \frac23= 1##
##E[M_n]=X_1X_2...X_n= 1= E[M_0]##

WMDhamnekar said:
##E[X_n] = 2\times \frac13 + \frac12\times \frac23= 1##
##E[M_n]=X_1X_2...X_n= 1= E[M_0]##

##E(\log(M_n))\neq \log(E(M_n))##!

WMDhamnekar
Office_Shredder said:
##E(\log(M_n))\neq \log(E(M_n))##!
That means you want to say ##\lim\limits_{n\to\infty}E[\log{M_{\{n+1\}}}=0]=0 ##

##\therefore## by using the Law of large numbers ##M_{\infty}=0##
Author said another way to prove ##M_{\infty}=0## is ## \mathbb{P}[\lim\limits_{n\to\infty}\displaystyle\sum_{n=0}^{n}\frac{M_{n+1}}{M_n}=\infty]## i-e the sequence ##\frac{M_{n+1}}{M_n} ## does not converge.

After using the Optional sampling theorem I determined the ##\mathbb{P}[M_n=64]=\frac{1}{3^6}## Is this answer correct?

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WMDhamnekar said:
That means you want to say ##\lim\limits_{n\to\infty}E[\log{M_{\{n+1\}}}=0]=0 ##

##\therefore## by using the Law of large numbers ##M_{\infty}=0##
Author said another way to prove ##M_{\infty}=0## is ## \mathbb{P}[\lim\limits_{n\to\infty}\displaystyle\sum_{n=0}^{n}\frac{M_{n+1}}{M_n}=\infty]## i-e the sequence ##\frac{M_{n+1}}{M_n} ## does not converge.

This notation doesn't make any sense to me to be honest. Maybe we can start with, what is ##E(\log(M_1))##?

After using the Optional sampling theorem I determined the ##\mathbb{P}[M_n=64]=\frac{1}{3^6}## Is this answer correct?

That's the odds you hit 2 six times in a row at the start, so it has to be too small. Can you show your work?

WMDhamnekar
Office_Shredder said:
This notation doesn't make any sense to me to be honest. Maybe we can start with, what is ##E(\log(M_1))##?
That's the odds you hit 2 six times in a row at the start, so it has to be too small. Can you show your work?
Sorry, Answer to 3 and 4 are wrong. Answer to 4 is ##\frac{1}{2^6}##

Now, Let me move on to answer 3.
##M_n= X_1X_2...X_n \therefore M_1= X_1.## Now ##X_1## may be 2 or ##\frac12 \therefore \log{M_1}=0, \therefore E[M_1]=1=E[M_0] ## So,my answer is still ##M_{\infty}=1## But the author said ##M_{\infty}=0##

How is that?

WMDhamnekar said:
Sorry, Answer to 3 and 4 are wrong. Answer to 4 is ##\frac{1}{2^6}##

Now, Let me move on to answer 3.
##M_n= X_1X_2...X_n \therefore M_1= X_1.## Now ##X_1## may be 2 or ##\frac12 \therefore \log{M_1}=0, \therefore E[M_1]=1=E[M_0] ## So,my answer is still ##M_{\infty}=1## But the author said ##M_{\infty}=0##

How is that?

You haven't done anything with the fact that 2 and 1/2 are not equally likely!

##E(\log(M_1))= \frac{1}{3}\log(2)+\frac{2}{3}\log(\frac{1}{2})## . This is *not* equal to 0

WMDhamnekar
Office_Shredder said:
You haven't done anything with the fact that 2 and 1/2 are not equally likely!

##E(\log(M_1))= \frac{1}{3}\log(2)+\frac{2}{3}\log(\frac{1}{2})## . This is *not* equal to 0

## \because \lim\limits_{n\to\infty} M_n= 2^n\times (\frac13)^n +(\frac12)^n \times (\frac23)^n =0 \therefore M_{\infty}=0##

Yes. There exists a ##(C < \infty ): \mathbb{E} [M^2_n]\leq C \forall n ##

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I think your answer to 4 is correct, but without seeing any work I can't say if you got it the right way.

It still looks like you're just writing random strings of symbols for number 3 (like literally, are you just putting stuff into chatgpt?) The limit you've written doesn't correspond to the limit of any object that depends on n in the problem. We don't have to cover it if you just wanted to focus on the later parts though.

WMDhamnekar
Office_Shredder said:
I think your answer to 4 is correct, but without seeing any work I can't say if you got it the right way.

It still looks like you're just writing random strings of symbols for number 3 (like literally, are you just putting stuff into chatgpt?) The limit you've written doesn't correspond to the limit of any object that depends on n in the problem. We don't have to cover it if you just wanted to focus on the later parts though.

Note: This answer is provided to me by Chat.G.P.T.

I don't understand this answer from second step onwards. If this is the correct answer, would any member explain me this answer?

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I don't agree with chatgpt's answer given by me in #15.
My own computed answer is as follows:

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Wouldn’t it be a lot simpler to write ##Y_i=\log_2(X_i)##, making ##\Sigma_0^n Y_i## a random walk?

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