tag:blogger.com,1999:blog-2198942534740642384.post6009005603235998016..comments2023-10-24T03:16:41.009-07:00Comments on Econometrics Beat: Dave Giles' Blog: Spherically Distributed Errors in Regression ModelsDave Gileshttp://www.blogger.com/profile/05389606956062019445noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-2198942534740642384.post-54064114920286428942012-09-18T13:34:16.438-07:002012-09-18T13:34:16.438-07:00Dimitriy - I believe your last sentence gives the ...Dimitriy - I believe your last sentence gives the explanation.Dave Gileshttps://www.blogger.com/profile/05389606956062019445noreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-85646782183628282822012-09-18T10:53:14.247-07:002012-09-18T10:53:14.247-07:00Thanks! I would like to ask one more question if I...Thanks! I would like to ask one more question if I may.<br /><br />I am trying to figure out why I would care about the iso-probability curves being circular. I am thinking of your $x$ and $y$ in a panel-ish context as $\eps_{it}$ and $\eps_{jt}$, and bivariate distribution represents shocks at different times. The expected value of the shocks at each time is zero, but we might have some correlation between units $i$ and $j$. They might be two firms located in the same town, for example. <br /><br />The peak of the distribution (and innermost iso-probability ellipse) is at (0,0). Suppose I get a negative draw of $\eps_{it}=-1$ (or $x=-1$). In the spherical case, the most likely value of corresponding $\eps_{jt}$ is zero. I can see that by looking at where the vertical line from $x=-1$ intersects the highest probability ellipse. In the correlated case, the same logic say that's no longer true. However, in the case where the shocks are uncorrelated, but one has a higher variance, the first geometric intuition still goes through. <br /><br />Can you explain where I am going wrong? Is it by equating the expected value with the highest valued iso-probability ring?Dimitriyhttps://www.blogger.com/profile/02728704178088861714noreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-76257504785186425942012-09-18T08:12:31.129-07:002012-09-18T08:12:31.129-07:00That's right. I drew these using the fMultivar...That's right. I drew these using the fMultivar package in R. See the next post (16 September) for a link to some R code that does this sort of thing.Dave Gileshttps://www.blogger.com/profile/05389606956062019445noreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-58220935308307507022012-09-17T17:05:34.747-07:002012-09-17T17:05:34.747-07:00In the last set, do we have a somewhat positive co...In the last set, do we have a somewhat positive correlation, with Var(y)>Var(x)?<br /><br />What did you draw these with?Dimitriyhttps://www.blogger.com/profile/02728704178088861714noreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-3464216729159208012012-09-17T10:03:09.226-07:002012-09-17T10:03:09.226-07:00Thanks Dan - glad it's helpful.Thanks Dan - glad it's helpful.Dave Gileshttps://www.blogger.com/profile/05389606956062019445noreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-23370521731650453082012-09-17T08:40:22.842-07:002012-09-17T08:40:22.842-07:00Great images! Thanks, this is a great resource. I&...Great images! Thanks, this is a great resource. I've always just said "It looks like a sphere" then tried to draw a picture, but this is far better. Dan Duttonnoreply@blogger.com