Timeline for Is there a probability density function satisfying the following conditions?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Aug 15, 2013 at 15:40 | comment | added | fedja | @Nilotpal Sure. I posted one. :) | |
| Aug 15, 2013 at 15:40 | history | edited | fedja | CC BY-SA 3.0 |
added 458 characters in body
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| Aug 15, 2013 at 15:00 | comment | added | Nilotpal Kanti Sinha | @Fedja: No I still don't understand. Can you exemplify with one working example of a solution that works for all $y > 0$. | |
| Aug 15, 2013 at 14:36 | comment | added | fedja | Looks like you still don't understand what I'm talking about: $p(x,y)=1$ for $x<1$ and $0$ for $x>1$ satisfies the "limit" requirements and I can create a family of continuous curves satisfying the conditions you posed formally and giving the distributions indistinguishable from this one for any practical purpose, which means that the problem statement is either flawed or incomplete because such solutions would obliterate any meaning of the $y$-dependence. What exactly are you after? | |
| Aug 15, 2013 at 13:05 | comment | added | Nilotpal Kanti Sinha | @Nik, I had the same formulation that you have given and I can draw several distributions through $(0,1)$ and $(1,e^{-y})$. However satisfying condition 5 is the problem. For example $p(x,y) = e^{-yx^a}$ satisfies the first four conditions. Now if conditions 5 is also going to be true the we find that $y$ must be exactly equal to $\Gamma(1+1/a)^a$. But the minimum value of $\Gamma(1+1/a)^a$ is $e^{-\gamma}$. So this solutions is good for $y > e^{-\gamma}$ but it does not hold for $0 < y < e^{-\gamma}$. Likewise I have not been able to find a function $p(x,y)$ which satisfies 5 for all $y > 0$. | |
| Aug 14, 2013 at 23:31 | comment | added | Nik Weaver | That's a slicker way to look at it. I guess it's worth mentioning that if $p$ is decreasing from $1$ at $(0,y)$ to $e^{-y}$ at $(1,y)$, then for small $y$ you've almost used up the entire unit of mass already, and have to decrease fast for $x > 1$. | |
| Aug 14, 2013 at 21:10 | history | answered | fedja | CC BY-SA 3.0 |