Timeline for Process equivalent to conditional probability
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 21, 2010 at 10:04 | vote | accept | Grzenio | ||
| Jun 14, 2010 at 16:35 | comment | added | Jeff Schenker | That should be $d \nu(x)$ in the last integral. (I can't figure out how to edit comments....) Also, your suggestion for constructing $X_j$ wouldn't work. If $X$ has a continuous distribution, then your definition would always give $X_j=X_{j-1}$. | |
| Jun 14, 2010 at 16:33 | comment | added | Jeff Schenker | Let $E=(X_j >tT)\&(j_t=j)$. So $E$ is the intersection of $(X_j>T)$, $(X_j>t)$ and $(X_{j−1}>t)$. Thus $$P(E)=\int_{[0,t]\times (m,\infty)} d \mu(x,y)$$ where $\mu$ is the joint distribution of $X_{j−1}$ and $X_{j}$ and $m=\max(t,T)$. By the definition of $X_j$ we have $$d\mu(x,y)=d\nu(x)d\kappa_x(y) $$ where $\nu$ is the distribution of $X_{j−1}$ and $d\kappa_x$ is the distribution of $X_j$ ``given that $X_{j−1}=x$'' which satisfies $$\kappa_x(S)=P(X>S|X>x)$$ for measurable sets $S$. So $$P(E)=\int_{[0,t]}P(X>m|X>x)d\kappa(x).$$ | |
| Jun 14, 2010 at 14:30 | comment | added | Grzenio |
Another simple question about the construction of the sequence $X_j$: would it also work if we defined $X_j = \inf (Y_i : Y_i > X_{j-1})$, so in a sense to "sort" the set $Y$ and take values in increasing order?
|
|
| Jun 14, 2010 at 14:14 | comment | added | Grzenio | Hi @Jeff, your construction looks really promising! I am now trying to understand the proof that it works. Would you be so kind and explain how did you come up with the line containing integral? | |
| Jun 14, 2010 at 6:36 | history | edited | Jeff Schenker | CC BY-SA 2.5 |
added 4 characters in body
|
| Jun 13, 2010 at 18:22 | history | answered | Jeff Schenker | CC BY-SA 2.5 |