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Let $X_n$ the outcome of the $n^{th}$ coin. Its value is $1$ if it is heads, $0$ otherwise. Let $G_n$ the value of your portfolio after the $n^{th}$ round. If $X_n=1$, I have to check the outcome of $X_{n+1}$. If $X_{n+1}=1$, I make one dollar, if $X_{n+1}=0$, I lose one dollar. If $X_n=0$, I have to check the outcome of $X_{n+1}$. If $X_{n+1}=1$, I make ...


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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_{...


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Recall that any traded asset divided by a numéraire is a martingale under the measure associated to that numéraire. For the 3 interest rates you mention, the natural measure (namely the one that makes those processes martingales) is deduced from the structure of the rate. Always keep in mind that the value at $t_0$ of a cash flow $C$ paid at $T$ is equal to ...


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Via a combination of the Cameron-Martin-Girsanov Theorom and the Martingale Representation Theorom you can find the equivalent martingale measure.


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