Martingales and two-period stock model

Question

I'm currently studying martingales and I encountered an example on two-period stock models.

Let $S_0,S_1,S_1$ represent the value of a stock at $t\leq 0,1,2$. Let $S_0=s_0$, $S_1=s_0Z_1$, and $S_2=s_0Z_1Z_2$ where $Z_i$ are i.i.d. random variables and

$Z_i=\begin{cases} u,\quad \text{with probability p}\\ d, \quad \text{with probability (1-p)} \end{cases}$

where $u\geq 1$ and $d\leq 1$ is assumed so that $up+d(1-p)=1$ is true. Let $\mathcal{F}_0$ be the trivial $\sigma$-algebra.

I'm just having trouble understanding the condition of integrability. How is $\mathbb{E}(S_1)<\infty$ and $\mathbb{E}(S_2)<\infty$? Sure this is true because the example is finite but I want to understand why $S_i$ is bounded by $s_0 u^2$. Could anyone explain to me why this is?

Sorry if this is a dumb question. I'm only starting to get a hang of this.

Answer 1

That $S_i<s_0u^2$ follows simply from the definition of $S_i$.

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Let's do it slowly.

1. We know that $S_0=s_0$ so that $$ S_0=s_0<s_0u^2$$ since $u>1$.

2. Next, consider $S_1$. We have that $S_1 = s_0Z$, where $Z$ can be equal to $u>1$ or $d<1$. So we have that $Z\leq u$. Hence, it follows that $$ S_1=s_0Z\leq s_0u<s_0u^2$$ where the last inequality simply follows from the fact that $u>1$.

3. You can easily use the same argument to show that $S_2\leq s_0u^2$.

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Above I was deliberately detailed. But this result follows from the fact that for each $i\geq 1$ we can represent $S_i$ as $S_i = s_0 \prod_{k=1}^{i} Z_k$. Then $$ S_i = s_0\prod_{k=1}^{i} Z_k \leq s_0 \prod_{k=1}^{i} u = s_0 u^i $$