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Apr 10, 2013 at 10:58 comment added Druss2k I decided to do the following: Since $E$ depends upon the difference of $MR-P$ and will be -Inf if $MR=P$ and 0 if $MR=$-Inf I do a censored regression onto $y=\begin{cases} min(E)-\epsilon, & \text{if }E < min(E)\\ E, & \text{if }min(E)<=E<0\\ 0, & \text{if } E=0 \end{cases}$ The $\epsilon$ is choosen because $E=$-Inf corresponds to the case where $MR$ is equal to $P$. Hence min$(E)$ corresponds to the case where $MR$ is almost equal to $P$. The difference between those cases is quite marginal hence the $\epsilon$.
Apr 10, 2013 at 9:43 history edited Druss2k CC BY-SA 3.0
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Apr 10, 2013 at 8:49 comment added Druss2k The problem of the $D=0$ case relates to other variables which should be checked for their explanatory power. These "other" variables do indeed change between those $D=0$ cases. Since I really want to explain why the $D=0$ case occurs and when, i.e.,why the $D\ne 0$ case does not occur, the splitting of the problem seems not to work for me? The only thing I could acutally do is to do a regression model for the $D\ne 0$ case and then check if theres is a systematic difference in the independend variables between $D=0$ and $D\ne 0$.
Apr 10, 2013 at 8:45 comment added Glen_b "But the dependend variable for the D=0 case would be a constant because MR is a constant (equals to P all the time)." -- then it doesn't take a very big model to explain those values. I'm not sure of their value to your model otherwise. If you can make your subsequent suggestion work, then it sounds like it might be better.
Apr 10, 2013 at 8:43 comment added Druss2k What about a censored regression model. The variable $E$ can have values between -Inf and 0. In the context of a censored regression model I could use a dependend variable as you described. I'm not quite sure how this dependend variable would look because we assume that a latent variable determines some other variable. Since $E$ is determined by the difference of $MR$ and $P$ I would like to split like: $E$ if $MR\ne P$, and if $MR=P$ I would set something else. I cannot set 0 because $E$ itself ca be 0 since $MR$ itself can be -Inf/Inf.
Apr 10, 2013 at 8:28 comment added Druss2k But the dependend variable for the $D=0$ case would be a constant because MR is a constant (equals to $P$ all the time). How would I know if changes in $P$ do indeed have an impact on $E$? Another thing I'm not sure about is, that in models which concerns a ordinal dependend variable we need some grouping in order to do inference. If I observe a continous variable which is different for every individual, i.e., no grouping can be done, how can there be a fit? Maybe this leads to your suggestion to split the problem into two parts. But then I could not use $P$ as a explanatory variable.
Apr 10, 2013 at 8:03 comment added Glen_b Well, yes and no. I mean that when D=1, you'd have a suitable model for the original dependent variable as a continuous thing. When D=0, you'd do whatever was suitable there. Imagine writing a model for each of those rows.
Apr 10, 2013 at 7:50 comment added Druss2k Thank you for helping me! I thought you meant something like: $D=\begin{cases} 0, & \text{if }E = \text{-Inf/Inf} \Leftrightarrow MR = P\\ 1, & \text{if }E \neq \text{-Inf/Inf} \Leftrightarrow MR < P \end{cases}$
Apr 10, 2013 at 7:40 comment added Glen_b Thanks for the clarification. Why would that put all those cases together?
Apr 10, 2013 at 7:38 comment added Druss2k In general the values of $Q$ are very small but I actually dont observe $Q$ I only got data on $MR$ and $P$. I thought about a possible categorization as well but then I would put all the cases where $\frac{\partial P}{\partial Q}<0$ into one group. Hence I would loose quite a big amount of information how $E$ behaves if $MR<P$.
Apr 10, 2013 at 7:20 comment added Glen_b One possibility would be to split into the cases where $\frac{\partial P}{\partial Q} = 0$ (I'm assuming Q isn't 0, let me know if that's wrong), and the cases where it isn't. The model may be somewhat different there, and then you don't need to categorize the whole thing.
Apr 10, 2013 at 7:17 comment added Glen_b I assume there's an $E$ goes in here: $MR = \frac{P}{MR-P}$ somewhere. In any case, that looks like one is observing P and Q, and then calculating E.
Apr 10, 2013 at 7:07 comment added Druss2k If one calculates the elasicity of demand. Since elasticity of demand ($E$) with price ($P$) and quantity ($Q$) equals $E = \frac{\partial Q}{\partial P}\frac{P}{Q}$ and the marginal revenue (MR) equals $MR = P + \frac{\partial P}{\partial Q}Q$ we can express $E$ in terms of $MR$ and $P$ as $MR = \frac{P}{MR-P}$. For every time $MR=P$ we observe $E$ = -Inf (or Inf) , this depends on the convention one makes about $E$ being negative or positive. The reasoning does not depend on the sign before $E$.
Apr 9, 2013 at 23:45 comment added Glen_b How is it possible to observe "-Inf"?
Apr 9, 2013 at 13:33 history edited Druss2k CC BY-SA 3.0
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Apr 9, 2013 at 13:15 history edited Druss2k CC BY-SA 3.0
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Apr 9, 2013 at 11:24 history asked Druss2k CC BY-SA 3.0