Timeline for The min of the mean of iid exponential variables
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 10, 2020 at 18:58 | history | edited | user44143 | CC BY-SA 4.0 |
clarified that these are probability densities
|
| Dec 10, 2020 at 12:32 | comment | added | Viktor B | I meant $k$ instead of $n$. If $f$ is a joint density in $(m,r)$, then it is not necessarily a problem if it grows to infinity for $r = 1$ so the second part of my first comment does not apply. | |
| Dec 9, 2020 at 18:29 | comment | added | user44143 | @Sinusx, yes, a probability density. But what is $n$ in your comment? Whatever it is, I suspect that the asymptotic analysis in the comment is neglecting the factor of $1_{[m<r]}$ in the formula for $f(k,m,r)$. | |
| Dec 9, 2020 at 11:47 | comment | added | Viktor B | Since exponential distribution is continuous, $f(k,m,r) = g(k, m) =0$ for any $m$. Maybe you mean the density function? Still, if we take $r = 1$ and $m <1$ in the expression for $f(k,m,r)$ in your claim, then $f(k,m,r) \sim \sqrt{\frac{n}{2 \pi}}$ grows to infinity, thus it cannot be a density. | |
| Nov 19, 2020 at 22:20 | history | answered | user44143 | CC BY-SA 4.0 |