Do we need $\tau \leq \nu$ to show $E(X_\tau)=E(X_\nu)$?

Question

My lecture notes claim that if $(X_n)$ is a martingale and $\tau$ is a stopping time bounded by $N$ then $$E(X_\tau)=E(X_{\tau \wedge N})=E(X_{\tau \wedge 0})=E(X_0)$$

and then remarks that if $\tau$ and $\nu$ are both bounded stopping times and $\tau \leq \nu$ a.s. then

$$E(X_\tau)=E(X_\nu)$$

I tried to prove it mentally and couldn't understand why would we need $\tau \leq \nu$ a.s. if we could just write it as above:

$$E(X_\tau)= \ldots =E(X_0)=E(X_{\nu \wedge 0})=E(X_{\nu \wedge M})=E(X_\nu)$$

And then I noticed that $\tau \leq \nu$ a.s. is asymmetric while the result $E(X_\tau)=E(X_\nu)$ is symmetric. So it made even less sense.

Any clarification is hugely appreciated.

Answer 1

If $(X_n)_{n \in \mathbb{N}}$ is a martingale, then

$$\mathbb{E}(X_{\tau}) = \mathbb{E}(X_{\nu}) \tag{1}$$

holds for any two bounded stopping times $\tau,\nu$. In contrast, if $(X_n)_{n \in \mathbb{N}}$ is a submartingale, then

$$\mathbb{E}(X_{\tau}) \leq \mathbb{E}(X_{\nu})$$

for bounded stopping times $\tau \leq \nu$. These results are known as optional stopping theorem.

Remark In fact, one can show that a process $(X_n)_{n \in \mathbb{N}}$ is a submartingale if, and only if,

$$\mathbb{E}(X_{\tau}) \geq \mathbb{E}(X_{\nu})$$

for all bounded stopping times $\tau \leq \nu$. This means that $(1)$ is even an equivalent condition for being a martingale.