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Consider the outcome of a game played by repeatedly tossing a fair coin, where you win a dollar if heads appears and you lose a dollar if tails appear, the outcome is denoted $X_1$, $X_2$, $X_3$,...,$X_n$. Let $M_n = \Sigma X_i$ be the total earnings after $n$ such tosses.

Consider the process $Y_n = e^{\sigma M_n} \cdot\left( \frac{2}{e^\sigma + e^{-\sigma}} \right)^2 $

How can i show whether this process is martingale or not?

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    $\begingroup$ How is $M_n$ defined? $\endgroup$ – KeSchn Feb 24 at 19:17
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    $\begingroup$ Consider the outcome of a game played by repeatedly tossing a fair coin, where you win a dollar if heads appears and you lose a dollar if tails appear, the outcome is denoted 𝑋1, 𝑋2,𝑋3, … . . 𝑋𝑛. Let 𝑀𝑛 = Σ𝑋𝑖 be the total earnings after n such tosses. Apologies for not mentioning it before $\endgroup$ – quantish Feb 24 at 20:27
  • $\begingroup$ So for all $n$, $X_n$ takes value 1 with probability 0.5 and value -1 with probability 0.5. Is that what you are saying? $\endgroup$ – Daneel Olivaw Feb 24 at 21:29
  • $\begingroup$ And what is $\sigma$ then? Arbitrary constant or something else? $\endgroup$ – noob2 Feb 24 at 21:32
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The process $(Y_n)$ is a Martingale if we assume the coin tosses to be independent.

Indeed, let us show that $E[Y_{n+1}|F_n]=Y_n$ where $(F_n)$ is the filtration generated by the process $(X_n)$.

We have \begin{equation} Y_{n+1}= e^{\sigma M_n}*(\dfrac{2}{e^\sigma + e^{-\sigma}})^{n+1}*e^{\sigma X_{n+1}}=Y_n*(\dfrac{2}{e^\sigma + e^{-\sigma}})*e^{\sigma X_{n+1}} \end{equation}

where $Y_n*(\dfrac{2}{e^\sigma + e^{-\sigma}})$ is $F_n-$measurable.

Therefore

\begin{equation} E[Y_{n+1}|F_n]=Y_n*(\dfrac{2}{e^\sigma + e^{-\sigma}})*E[e^{\sigma X_{n+1}}|F_n] \end{equation}

To conclude, it suffices to see that $E[e^{\sigma X_{n+1}}|F_n]=E[e^{\sigma X_{n+1}}]=\dfrac{e^\sigma + e^{-\sigma}}{2}$. The first equality comes from the independence of the coin tosses.

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  • $\begingroup$ Where does your last equality come from? It seems you are assuming $X_{n+1}$ takes its values in $\{e^\sigma,e^{-\sigma}\}$. $\endgroup$ – Daneel Olivaw Feb 24 at 21:26
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    $\begingroup$ $X_{n+1}$ takes its values in $\{-1, 1\}$. But I forgot some sigmas before $X_{n+1}$. $\endgroup$ – SN76 Feb 24 at 21:30

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