Timeline for Can the differential entropy of a continuous distribution with lebesgue integrable density be negative infinity?
Current License: CC BY-SA 4.0
12 events
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Nov 26, 2022 at 22:55 | comment | added | Iosif Pinelis | Do you have a response to the answers below? | |
Jul 8, 2018 at 16:32 | comment | added | Deane Yang | Quick commentary on how to find this example (using natural log instead of log base 2): Note that $\|f\|_p$ is an increasing function of $p$ and its limit as $p\rightarrow 0$ is $\exp -H(f)$. So $f$ must be in $L^1$ but not $L^p$ for any $p > 1$. $f = x^{-\alpha}$, $\alpha > 0$, does not work, so the next thing to try is $x^{-1}(1-\log x)^{-\beta}$, where we add 1 to $-\log x$ to prevent the function from blowing up at $x = 1$. This blows up at $0$ slower than $x^{-1}$ but faster than $x^{-\alpha}$ for any $\alpha < 1$. Testing this, it works if $1 < \beta \le 2$. | |
Jul 8, 2018 at 15:41 | comment | added | Tejas Bhojraj | @Deane. Yes, your function does work, it has a negative infinity entropy! | |
Jul 8, 2018 at 15:25 | answer | added | R W | timeline score: 1 | |
Jul 8, 2018 at 14:06 | review | Close votes | |||
Jul 9, 2018 at 21:06 | |||||
Jul 8, 2018 at 12:34 | answer | added | Iosif Pinelis | timeline score: 7 | |
Jul 8, 2018 at 5:30 | comment | added | Tejas Bhojraj | Yes, you are right; with $1-log(x)$, atleast $f \in L_{1}([0,1])$. $\int_{0}^{1} f = (1-log(x))^{-1}|_{0}^{1}= 1$. I still need to work out $H(f)$. And my previous comment was wrong, sorry! | |
Jul 8, 2018 at 5:14 | comment | added | Deane Yang | Sorry, it has to be $1-\log x$ and not $1 + \log x$, if it works at all. | |
Jul 8, 2018 at 5:11 | comment | added | Tejas Bhojraj | Thanks. However, $f \notin L_{1}([0,1])$. $\int_{0}^{1} f = -(1+log(x))^{-1}|_{0}^{1}= \infty$, right? | |
Jul 8, 2018 at 4:29 | comment | added | Deane Yang | Try $f(x) = (x(1+\log x)^2)^{-1}$ | |
Jul 8, 2018 at 3:18 | review | First posts | |||
Jul 8, 2018 at 3:38 | |||||
Jul 8, 2018 at 3:14 | history | asked | Tejas Bhojraj | CC BY-SA 4.0 |