# Thread: basic question on martingales definition

1. ## basic question on martingales definition

In many books, a martingale (roughly speaking) is defined as a process X1,X2,... satisfying:

$
\mathbb{E}[X_{t_{n+1}} | X_{t_n},\ldots,X_{t_0}] = X_{t_n}
$

which is equivalent to

$
\mathbb{E}[X_{t_{n+1}} - X_{t_n} | X_{t_n},\ldots,X_{t_0}] = 0
$

While the latter is understandable to me, I interpret the former in this sense: "expectation is equal to a random variable", which is absurd to me.

Now, I am sure this is a stupid question but: how should I interpret the former?
Maybe that X_tn is no more a random variable but a number (because at time tn is known)? If so, why don't textbook put that symbol in lower case as usual... ?

2. Originally Posted by tossifan
In many books, a martingale (roughly speaking) is defined as a process X1,X2,... satisfying:

$
\mathbb{E}[X_{t_{n+1}} | X_{t_n},\ldots,X_{t_0}] = X_{t_n}
$

which is equivalent to

$
\mathbb{E}[X_{t_{n+1}} - X_{t_n} | X_{t_n},\ldots,X_{t_0}] = 0
$

While the latter is understandable to me, I interpret the former in this sense: "expectation is equal to a random variable", which is absurd to me.

Now, I am sure this is a stupid question but: how should I interpret the former?
Maybe that X_tn is no more a random variable but a number (because at time tn is known)? If so, why don't textbook put that symbol in lower case as usual... ?
Hi,
you should read again the chapter on conditional expectations in your textbook: a conditional expectation is indeed a random variable. The conditional expectation $E[X|Y,Z]$ is intuitively "the random variable depending on $Y$ and $Z$ only that best approximates $X$"; formal definition isn't straightforward, cf. textbooks or probably wikipedia.

3. Originally Posted by tossifan
In many books, a martingale (roughly speaking) is defined as a process X1,X2,... satisfying:

$
\mathbb{E}[X_{t_{n+1}} | X_{t_n},\ldots,X_{t_0}] = X_{t_n}
$

which is equivalent to

$
\mathbb{E}[X_{t_{n+1}} - X_{t_n} | X_{t_n},\ldots,X_{t_0}] = 0
$

While the latter is understandable to me, I interpret the former in this sense: "expectation is equal to a random variable", which is absurd to me.

Now, I am sure this is a stupid question but: how should I interpret the former?
Maybe that X_tn is no more a random variable but a number (because at time tn is known)? If so, why don't textbook put that symbol in lower case as usual... ?
They are the same since

$
\mathbb{E}[X_{t_n} | X_{t_n},\ldots,X_{t_0}] = X_{t_n}
$

because you know $X_{t_n}$

$
\mathbb{E}[X_{t_{n+1}} - X_{t_n} | X_{t_n},\ldots,X_{t_0}]$

$=
\mathbb{E}[X_{t_{n+1}} | X_{t_n},\ldots,X_{t_0}] -
\mathbb{E}[ X_{t_n} | X_{t_n},\ldots,X_{t_0}] = X_{t_n}-X_{t_n}=0
$

4. Originally Posted by matheagle
They are the same since

$
\mathbb{E}[X_{t_n} | X_{t_n},\ldots,X_{t_0}] = X_{t_n}
$

because you know $X_{t_n}$

$
\mathbb{E}[X_{t_{n+1}} - X_{t_n} | X_{t_n},\ldots,X_{t_0}]$

$=
\mathbb{E}[X_{t_{n+1}} | X_{t_n},\ldots,X_{t_0}] -
\mathbb{E}[ X_{t_n} | X_{t_n},\ldots,X_{t_0}] = X_{t_n}-X_{t_n}=0
$
So $X_{t_n}$ would be a constant at time t_n ...

But this seems to disagree with what Laurent wrote!

5. Originally Posted by tossifan
So $X_{t_n}$ would be a constant at time t_n ...
Why would it be? In the equality $E[X_{t_n}|X_{t_1},\ldots,X_{t_n}]=X_{t_n}$, both hand sides are random variables. By the way, this equality relates to the intuition I gave: the left-hand side is the best approximation of $X_{t_n}$ in terms of $X_{t_1},\ldots,X_{t_n}$, so in particular in terms of itself; therefore it is $X_{t_n}$ itself (the best approximation is the variable itself...).