6
$\begingroup$

Let $X_1$ and $X_2$ be arbitrary random variables defined one the same probability space $(\Omega,\mathcal{F},P)$. I am trying to determine if the event $\{\max\{X_1,X_2\}=X_2\}$ is measurable with respect to $\sigma(\max\{X_1,X_2\})$? Normally I would try to decompose the event into a countable union or intersection of events which are clearly elements of the $\sigma(\max\{X_1,X_2\})$. However, the only way that I can think to rewrite the original set is as $\{\max\{X_1,X_2\}=X_2\}=\{X_1\leq X_2\}$. I'm not sure where to go from here.

$\endgroup$
8
$\begingroup$

It is not true. Let $X_1$ be a Bernoulli(1/2) random variable and set $X_2=1-X_1$. Then $\max(X_1,X_2)$ is the constant random variable 1. The generated $\sigma$-algebra $\sigma(\max\{X_1,X_2\})$ is trivial, but the event $\{\max\{X_1,X_2\}=X_2\}$ is not.

$\endgroup$
  • 1
    $\begingroup$ +1, a nice example! Maybe for OP it would be also useful to know how to check such kind of clues - e.g. $\sigma(\max(X_1,X_2))$ contains only the information about the maximum of two variables but not the whole information about them, in particular it can't tell whether one of them is greater than the other? $\endgroup$ – Ilya Jan 29 '13 at 7:35
  • $\begingroup$ @Ilya, I think your explanation matched what my intuition was telling me, but I had no idea how to make it rigorous. In the example that I was working, $X_1$ and $X_2$ were exponential random variables with rate $\mu$ and $\lambda$, respectively. The problem was to check the independence of $\max(X_1,X_2)$ and the indicator $1_{(X_1\leq X_2)}$. The random variables are independent if and only if $\lambda=\mu$. Someone presented a proof based on the reasoning in my question, and I couldn't verify the validity of this person's proof. $\endgroup$ – Carl Morris Jan 29 '13 at 17:43
  • $\begingroup$ @CarlMorris The random variables are independent if and only if $\lambda =\mu$. Sorry, but that does not make any sense even if you put exponential somewhere there. $\endgroup$ – Ilya Jan 29 '13 at 17:47
  • $\begingroup$ @Ilya, what I mean to say is the following. Consider my example where I use exponential random variables with $\lambda=\mu$. If $F$ is the distribution function of $Y=\max\{X_1,X_2\}$ and $G$ is the distribution function of $Z=1_{\{X_1\leq X_2\}}$, then the joint distribution function $H$ of $Y$ and $Z$ is given by the product of $F$ and $G$. $\endgroup$ – Carl Morris Feb 6 '13 at 16:59
0
$\begingroup$

$X_1$ a Bernoulli(p) $p>0$ r.v. taking the two possible values 0 or 1

$\endgroup$
  • $\begingroup$ How does this answer the question? $\endgroup$ – BCLC Jan 2 '16 at 17:29
-1
$\begingroup$

Convince yourself that $\exists A \in \mathscr F$ s.t. we can write:

$$X: = \max(X_1, X_2) = X_1 1_A + X_2 1_{A^C}$$

Note that:

  1. $\sigma(X) \subseteq \sigma(\sigma(X_1) \cup \sigma(X_2) \cup \sigma(A))$

  2. Almost surely, $1_{A^C} = 1_{\{\max\{X_1, X_2\} = X_2\}}$

  3. $A^C \in \sigma(\sigma(X_1) \cup \sigma(X_2) \cup \sigma(A))$

Q: $A^C \in \sigma(X)$ ? (Alternatively, $A \in \sigma(X)$ ?)

We want to know if we can determine the maximum of $X_1$ and $X_2$ given that we know the value of $X$. We cannot. Just pick $X_1$ and $X_2$ s.t. the value of $X$ is constant.

Let $B \in \mathscr F$ and $a \in \mathbb R$ s.t.

$$X_1 = a \times 1_B$$

$$X_2 = a \times 1_{B^C}$$

Then $X = a \times 1_{A \cap B} + a \times 1_{(A \cap B)^C} = a \times 1_{\Omega} \to$ (almost?) surely, $X \equiv a$.

$\to \sigma(X) = \{\{X=a\}, \{X \ne a\}, \emptyset, \Omega\} = \{\Omega, \emptyset, \emptyset, \Omega\} = \{\emptyset, \Omega\}$

I don't see $A$ or $A^C$ in $\sigma(X)$.

$\endgroup$
  • 1
    $\begingroup$ Why more or less copy @Byron's accepted answer three years later? $\endgroup$ – Did Jan 2 '16 at 20:55
  • $\begingroup$ @Did I wanted to use indicator functions $\endgroup$ – BCLC Jan 3 '16 at 2:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.