American call option is a submartingale

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Consider a complete, arbitrage-free $T$-period market $(S^0_t,...,S^d_t)_{t=0,\dots,T}$ on a filtered probability space with risk-neutral measure $\mathbb P^*$. Furthermore, $S^0_{t+1}\ge S^0_t$ $\mathbb P^*$-almost surely.
We are interested in the american call-option $C_t=(S^1_t-K)^+$ (With strike $K>0$)

Show the following:

1) The discounted option $H=\frac C {S^0}$ is a $\mathbb P^*$-submartingale
2) The Snell envelope $U^{\mathbb P^*}$ of $H$ is a $\mathbb P^*$-martingale.

I need some help with these questions.

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So far I tried the following:

Regarding 1): I know that I need to show that for all $t$ we have $E_{\mathbb P^*} [H_{t+1}|\mathcal F_t] \le H_t$. But with that few information given, I don't know how to start. Of course we could directly use $S^0_{t+1}\ge S^0_t$, but then, we are still left with $E_{\mathbb P^*} [C_{t+1}|\mathcal F_t]\le C_t.$

Regarding 2): I know that the Snell envelope is a supermartingale, so it suffices to show that in this case is also a submartingale. I tried the following induction argument, but i am sceptical if that is true: (I am sceptical about the second line)

$$ E_{\mathbb P^*}[U^{\mathbb P^*}_t|\mathcal F_{t-1}]= E_{\mathbb P^*}[E_{\mathbb P^*}[U^{\mathbb P^*}_{t+1}|\mathcal F_t] \vee H_t|\mathcal F_{t-1}] \\ \ge E_{\mathbb P^*}[E_{\mathbb P^*}[U^{\mathbb P^*}_{t+1}|\mathcal F_t]|\mathcal F_{t-1}] \vee E_{\mathbb P^*}[H_t|\mathcal F_{t-1}] \\ \ge E_{\mathbb P^*}[U^{\mathbb P^*}_{t}|\mathcal F_{t-1}] \vee H_{t-1} = U^{\mathbb P^*}_{t-1} ,$$

where in the last line we used the induction hypothesis that the Snell envelope has the sub-martingale property for $t>t-1$ and that $H$ is a submartingale.
Also the Induction start $t=T$ should simply follow from $H$ being a submartingale..

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  1. $$E_P[H_{t+1}|F_t] = E_P[(\frac{S^1_{t+1}}{S^0_{t+1}}-\frac K{S^0_{t+1}})^+|F_t] \\ \ge \big(E_P[\frac{S^1_{t+1}}{S^0_{t+1}}|F_t]-E_P[\frac K{S^0_{t+1}}|F_t]\big)^+ \\ = \big( \frac {S^1_t}{S^0_t} - E_P[\frac K{S^0_{t+1}}|F_t]\big)^+ \\ \ge \big( \frac {S^1_t}{S^0_t} - E_P[\frac K{S^0_{t}}|F_t]\big)^+ \\ = \big( \frac {S^1_t}{S^0_t} - \frac K {S^0_t}\big)^+ = H_t $$

Where we used Jensen for the convex function $x \mapsto x^+$ for the first inequality, and the martingale property of $S^1$ with respect to $P$ for the second equality.

  1. No reason to be sceptical, this simply follows from $X\le X \vee Y$ and $Y \le X \vee Y$.