# Finding out the sequence as Martingale

#### Dhamnekar Winod

##### Active member
Consider the sequence $\{X_n\}_{n\geq 1}$ of independent random variables with law $N(0,\sigma^2)$. Define the sequence $Y_n= exp\bigg(a\sum_{i=1}^n X_i-n\sigma^2\bigg),n\geq 1$ for $a$ a real parameter and $Y_0=1.$

Now how to find the values of $a$ such that $\{Y_n\}_{n\geq 1}$ is martingale, submartingale, supermartigale?

Solution:I have no idea to answer this question. I searched on internet, but i didn't get any clew about the way using which one can answer this question till now. If any member knows the correct answer, he may reply with correct answer.

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#### steep

##### Member
Where did this question come from, and what background knowledge do you have? It takes a fair amount of background knowledge to answer questions like this.

For this question: if you know what a martingale is, as well as sub and super martingales, plus how MGFs and cumulants work, basic facts about convexity, and are familiar with the product form of martingales, then this is a straightforward question. If you don't know some of those things, then you really need to study them first as I'd have no clue how to answer this question in a satisfying way without them.
* * * * * *
The key thing is to get this in the form of a product martingale (i.e. mean one which is the identity element for products) and recognize that we are using MGFs of the Gaussian with independent random variables. The MGF for the Gaussian exists everywhere (due to super exponential declines / niceness in the pdf) and for a zero mean normal $X_i$ it is given by

$M(a) = E\big[e^{a X_i}\big] = \exp\big(\frac{\sigma^2 a ^2}{2}\big)$

so consider $Y_1$ and recognize that what you want is

$E\big[Y_1\big] = E\big[e^{a X_1 - \sigma^2}\big]= E\big[e^{a X_1} e^{- \sigma^2}\big] = E\big[e^{a X_1} \big]e^{- \sigma^2} = M(a)\cdot e^{- \sigma^2} :=1$

It's easier to work with the cumulant here so take logs of each side to see

$\sigma^2 = \frac{\sigma^2 a ^2}{2}$

solve for $a$. You should get two candidate values. (I assume $\sigma^2 \neq 0$ as a value of 0 would be a degenerate case and there isn't much to do if these 'normal's are zero almost everywhere.)

* * * * * *
with this in hand you should be able to easily verify that

$E\Big[\big \vert Y_n \big \vert \Big] = E\Big[ Y_n \Big] \lt \infty$ and that
$E\Big[ Y_n\big \vert Y_{n-1}, Y_{n-2}, ..., Y_1, Y_0 \Big] = Y_{n-1}$

and hence this is a martingale

As for submartingales and supermartingales
If you understand basic facts about cumulants, you'll know that the cumulant is zero at $a=0$ (why?) and in this problem with slope of 0 at $a =0$ (again why?) and that the cumulant is always convex over the open interval where it exists -- in this case $(-\infty, \infty)$ This leads to a nice geometric / visual interpretation as to how to get from the equality / martingale case to the sub and super case.

#### Dhamnekar Winod

##### Active member
Where did this question come from, and what background knowledge do you have? It takes a fair amount of background knowledge to answer questions like this.

For this question: if you know what a martingale is, as well as sub and super martingales, plus how MGFs and cumulants work, basic facts about convexity, and are familiar with the product form of martingales, then this is a straightforward question. If you don't know some of those things, then you really need to study them first as I'd have no clue how to answer this question in a satisfying way without them.
* * * * * *
The key thing is to get this in the form of a product martingale (i.e. mean one which is the identity element for products) and recognize that we are using MGFs of the Gaussian with independent random variables. The MGF for the Gaussian exists everywhere (due to super exponential declines / niceness in the pdf) and for a zero mean normal $X_i$ it is given by

$M(a) = E\big[e^{a X_i}\big] = \exp\big(\frac{\sigma^2 a ^2}{2}\big)$

so consider $Y_1$ and recognize that what you want is

$E\big[Y_1\big] = E\big[e^{a X_1 - \sigma^2}\big]= E\big[e^{a X_1} e^{- \sigma^2}\big] = E\big[e^{a X_1} \big]e^{- \sigma^2} = M(a)\cdot e^{- \sigma^2} :=1$

It's easier to work with the cumulant here so take logs of each side to see

$\sigma^2 = \frac{\sigma^2 a ^2}{2}$

solve for $a$. You should get two candidate values. (I assume $\sigma^2 \neq 0$ as a value of 0 would be a degenerate case and there isn't much to do if these 'normal's are zero almost everywhere.)

* * * * * *
with this in hand you should be able to easily verify that

$E\Big[\big \vert Y_n \big \vert \Big] = E\Big[ Y_n \Big] \lt \infty$ and that
$E\Big[ Y_n\big \vert Y_{n-1}, Y_{n-2}, ..., Y_1, Y_0 \Big] = Y_{n-1}$

and hence this is a martingale

As for submartingales and supermartingales
If you understand basic facts about cumulants, you'll know that the cumulant is zero at $a=0$ (why?) and in this problem with slope of 0 at $a =0$ (again why?) and that the cumulant is always convex over the open interval where it exists -- in this case $(-\infty, \infty)$ This leads to a nice geometric / visual interpretation as to how to get from the equality / martingale case to the sub and super case.
Hello,

So, $a=\pm\sqrt{2}$. But how can we say $E\bigg[Y_n|Y_{n-1},Y_{n-2},...,Y_1,Y_0\bigg]=Y_{n-1}$

I know very little about the term 'Cumulant'. If i have any diffculty in understanding last paragaph of your answer, i shall post it under this thread.

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#### steep

##### Member
Hello,

So, $a=\pm\sqrt{2}$. But how can we say $E\bigg[Y_n|Y_{n-1},Y_{n-2},...,Y_1,Y_0\bigg]=Y_{n-1}$
I had a hunch you might ask this --This is very basic martingale stuff and why I asked where this problem came from... In general its easier to start with 'sum form' martingales as opposed to 'product form', and there are easier, more intuitive, problems out there that don't require use of an MGF, and even ones involving random variables taking on finitely many values (easier than uncountably many). My question remains: where did this problem come from?
* * * *

$E\bigg[Y_1|Y_0\bigg]=Y_0 = 1$

then do it for n = 2
$E\bigg[Y_2|Y_{1},Y_0\bigg]=Y_1$
(why is this true?)

you should be able to pick up some intuition from this to figure out how to show it for the general $n$ case. But again, I think your time is better spent on other martingale problems right now.

Hello,
I know very little about the term 'Cumulant'. If i have any diffculty in understanding last paragaph of your answer, i shall post it under this thread.
The fix for this is fairly easy, either (a) ignore the term cumulant and just work explicitly with the exact Gaussian MGF (using logarithms as an analytical tool) -- you can solve this problem with the tradeoff that you'll miss deeper analytical insights this way or (b) read up on cumulants and in particular tease out what their first and second derivatives look like in general.

Also, as I suggested in my original post, the key thing is to figure out that equality case of an actual Martingale. Once you have that then you can figure out qualitative tweaks to get to sub and super martingales -- so to put it a different way: ignore the sub and supermartingales right now and maybe come back to them later.

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