Timeline for Joint distribution of a Normal and Truncated Normal
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 29, 2015 at 0:53 | vote | accept | CommunityBot | ||
| Jan 17, 2014 at 0:19 | comment | added | Alecos Papadopoulos | Formally yes, but for the case $Y=X$ it is equal to unity. | |
| Jan 17, 2014 at 0:16 | comment | added | user30490 | Last question, since Y is a transformation of X do we need a Jacobian term? | |
| Jan 17, 2014 at 0:10 | comment | added | user30490 | Sorry, I read the notation incorrectly $B(y,z;\rho)$ | |
| Jan 17, 2014 at 0:08 | comment | added | Alecos Papadopoulos | I do not state that $(Y,Z)$ are jointly bivariate Normal - I assume that $(X,Z)$ are jointly bivariate normal. The cdf of $(Y,Z)$ is the last expression, which is not a bivariate normal, viewed as a whole. | |
| Jan 17, 2014 at 0:08 | comment | added | whuber♦ | I do not see any such claim in this answer: the bivariate normal distribution is applied to $X$ and $Z$, not to $Y$ and $Z$. Although I have not checked it over in detail, it definitely is the right approach (and therefore received my upvote). | |
| Jan 17, 2014 at 0:04 | comment | added | user30490 | Also, are (Y,Z) jointly bivariate Normal as he states above? | |
| Jan 17, 2014 at 0:03 | comment | added | user30490 | @Whuber, is his solution correct? | |
| Jan 16, 2014 at 23:59 | comment | added | whuber♦ | Do you need to do anything? Alecos has given you an explicit solution in terms of standard mathematical objects; that's a reasonable interpretation of "closed form" and it allows for simple calculation of the probability of any event. | |
| Jan 16, 2014 at 23:46 | comment | added | user30490 | What should I do @Whuber? | |
| Jan 16, 2014 at 23:44 | comment | added | whuber♦ | Strictly speaking, this CDF doesn't have a density because it is singular along a line $Y=0$. | |
| Jan 16, 2014 at 23:31 | history | answered | Alecos Papadopoulos | CC BY-SA 3.0 |