Timeline for Expectation of reciprocal of a variable
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Jan 7 at 20:40 | history | suggested | Arto | CC BY-SA 4.0 |
fixed some parenthesis and grammar mistake
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| Jan 7 at 14:46 | review | Suggested edits | |||
| S Jan 7 at 20:40 | |||||
| Nov 22, 2020 at 2:03 | comment | added | Math1000 | This is not correct. For example, if $X$ has exponential distribution with mean $\frac1\mu$ then its moment-generating function is $$ \mathbb E[e^{-\lambda X}] = \int_0^\infty e^{\lambda x}\mu e^{-\mu x}\ \mathsf dx = \frac\mu{\mu-\lambda}, $$ but the integral only converges for $\lambda<\mu$. So the integral of the moment-generating function over $(0,\infty)$ cannot possible converge - and indeed, $\mathbb E\left[\frac 1X\right]$ does not exist. | |
| Aug 8, 2017 at 20:19 | comment | added | kjetil b halvorsen♦ | You are right, the problems was less than I thought. Still this answer would be better withm some more details. I will upvote this tomorrow ( when I have new votes) | |
| Aug 8, 2017 at 19:48 | comment | added | whuber♦ | @Kjetil I don't see what the problem is: apart from the inconsequential differences of using $t X$ instead of $-t X$ in the definition of the MGF and naming the variable $t$ instead of $\lambda$, the answer you just posted is identical to this one. | |
| Aug 7, 2017 at 19:56 | comment | added | kjetil b halvorsen♦ | The idea here is right, but the details wrong. Pleasecheck | |
| Aug 7, 2017 at 19:53 | review | Late answers | |||
| Aug 7, 2017 at 21:01 | |||||
| Aug 7, 2017 at 19:38 | review | First posts | |||
| Aug 7, 2017 at 19:45 | |||||
| Aug 7, 2017 at 19:35 | history | answered | user172761 | CC BY-SA 3.0 |