# Showing the discounted stock is a martingale

Background Information:

This question follows from here

It is tempting to write $$V_0(X) = \beta\left[\left(\frac{\beta^{-1}S_0 - S_1(d)}{S_1(u) - S_1(d)}\right)X(u) + \left(\frac{S_1(u) - \beta^{-1}S_0}{S_1(u) - S_1(d)}\right)X(d)\right]$$ as

$$V_0(X) = E_Q[\beta X]$$ where the expectation is taken with respect to the new purely formal probability measure $Q$ defined by $$Q(u) = \frac{\beta^{-1}S_0 - S_1(d)}{S_1(u) - S_1(d)}$$ and $$Q(d) = \frac{S_1(u) - \beta^{-1}S_0}{S_1(u) - S_1(d)}$$

Note that $Q(u) + Q(d) = 1$; $Q$ will be a probability measure provided these values are non-negative.

Question:

The quantity $\frac{S}{B}$ is called the discounted stock price. It will be useful later to notice that the pricing measure $Q$ has a special property with respect to the discounted stock price: it makes the discounted stock a martingale, i.e.,

$$\frac{S_0}{B_0} = E_{Q}[S_1/B_1]$$

I am curious on how to verify this. I assume we need to show both sides but I am not sure how to proceed.

In your setting, $\beta = B_0 B_1^{-1}$. Then \begin{align*} E_Q(S_1/B_1) &= Q(u)S_1(u)/B_1 + Q(d)S_1(d)/B_1\\ &=\frac{\beta^{-1}S_0 - S_1(d)}{S_1(u) - S_1(d)}S_1(u)/B_1 + \frac{S_1(u) - \beta^{-1}S_0}{S_1(u) - S_1(d)}S_1(d)/B_1\\ &=\frac{\beta^{-1}S_0S_1(u)/B_1 - S_1(d)S_1(u)/B_1 + S_1(u)S_1(d)/B_1-\beta^{-1}S_0S_1(d)/B_1}{S_1(u) - S_1(d)}\\ &=\beta^{-1}S_0/B_1\\ &=\frac{S_0}{B_0}. \end{align*}