# How to show if this is Martingale or not?

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?

• How is $M_n$ defined? – KeSchn Feb 24 at 19:17
• 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 – quantish Feb 24 at 20:27
• 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? – Daneel Olivaw Feb 24 at 21:29
• And what is $\sigma$ then? Arbitrary constant or something else? – noob2 Feb 24 at 21:32

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.

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