0

Let $(A_{n})$ with $n\geq 1$ an non decreasing sequence of random variable such that $A_{n+1}$ dépend of $X_1,...,X_n$ for all $n$. I agree that $$\mathbb E[A_{n+1}\mid X_1,...,X_n]=A_{n+1}$$ But I don't understand why $$\mathbb E[A_n\mid X_1,...,X_{n}]=A_n.$$

Is it a consequence of the fact that $A_n$ doesn't depend of $X_{n}$ and thus

$$\mathbb E[A_n\mid X_1,...,X_{n}]=\mathbb E[A_n\mid X_1,...,X_{n-1}]=A_n\ \ \ ?$$

idm
  • 11,824

1 Answers1

1

$A_n$ depends on $X_1, ..., X_{n-1}$, therefore it is $\sigma(X_1, ..., X_n)$-measurable, because $\sigma(X_1, ..., X_{n-1}) \subset \sigma(X_1, ..., X_n)$. Also, there is no need for $A_n$ to be nondecreasing.

Tlön Uqbar Orbis Tertius
  • 3,194
  • I've never seen proba with measure, so I don't understand. But I modified my last argument, is it correct ? – idm May 27 '15 at 12:15
  • Your argument is correct and is the same as mine. Saying "$A_n$ is $\sigma(X_1, ..., X_{n-1})$-measurable" is exactly the mathematical definition of the intuitive fact $"A_n$ depends of $X_1, ..., X_{n-1}$". Intuitively, if $A_n$ depends on $X_1, ..., X_{n-1}$, then it is perfectly known if you have the values of $X_1, ..., X_{n-1}$, and therefore it is perfectly known if you have the values of $X_1, ..., X_{n}$. – Tlön Uqbar Orbis Tertius May 27 '15 at 12:17
  • Bon travail, btw :) – Tlön Uqbar Orbis Tertius May 27 '15 at 12:21