Using a Logarithmic Transformation for a Simpler Random Walk Model

[WMDhamnekar]

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

Answer to 1.

Answer to 2.

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

 

[Office_Shredder]

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

 

[WMDhamnekar]

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 $M_{\infty} =0$

 

[Office_Shredder]

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 $E(\log(M_{n+1})-\log(M_n))$

 

[WMDhamnekar]

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

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

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

 

[Office_Shredder]

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.

 

[WMDhamnekar]

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.

$E[X_n] = 2\times \frac13 + \frac12\times \frac23= 1$
$E[M_n]=X_1X_2...X_n= 1= E[M_0]$

 

[Office_Shredder]

$E[X_n] = 2\times \frac13 + \frac12\times \frac23= 1$
$E[M_n]=X_1X_2...X_n= 1= E[M_0]$

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

 

[WMDhamnekar]

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

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.

Answer to 4.
After using the Optional sampling theorem I determined the $\mathbb{P}[M_n=64]=\frac{1}{3^6}$ Is this answer correct?

 

[Office_Shredder]

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 $E(\log(M_1))$?

Answer to 4.
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]

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?

 

[Office_Shredder]

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]

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

Answer to 3.

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

Answer to 5.
Yes. There exists a $(C < \infty ): \mathbb{E} [M^2_n]\leq C \forall n$

 

[Office_Shredder]

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]

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.

Answer to 5.

Is the above answer correct?
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?

 

[WMDhamnekar]

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

Is my answer correct?

 

[haruspex]

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