Skip to main content
MathOverflow

Return to Answer

added 4 characters in body
Source Link
Tony Huynh
Tony Huynh
  • 32.1k
  • 11
  • 112
  • 187

I suspect that it has to do with the fact that the most likely outcome is $\lfloor 1/(x y) \rfloor=1$ (which happens to be odd), followed with an application of the Strong Law of Small Numbers. Here are some more details. Let $Z$ be the random variable $xy$. Note that $Z$ takes values in $[0,1]$ and $\lfloor 1/Z \rfloor$ is odd if and only if

$$ Z \in (1/2, 1] \cup (1/4, 1/3] \cup (1/6, 1/5] \cup \dots $$

Note that this set has measure more than 1/2$1/2$. However, the distribution of $Z$ is of course not uniform on [0,1]$[0,1]$ (it is actually skewed towards 0$0$ instead of 1$1$). I suspect what ends up happening is that the distribution of $\lfloor 1/Z \rfloor$ is almost perfectlyperfectly split between even and odd in the regimefor say $Z < 1/5$, and the discrepancy is thus a result of what happens for $Z > 1/5$$Z \geq 1/5$ (where one easily sees that odd wins out).

I suspect that it has to do with the fact that the most likely outcome is $\lfloor 1/(x y) \rfloor=1$ (which happens to be odd), followed with an application of the Strong Law of Small Numbers. Here are some more details. Let $Z$ be the random variable $xy$. Note that $Z$ takes values in $[0,1]$ and $\lfloor 1/Z \rfloor$ is odd if and only if

$$ Z \in (1/2, 1] \cup (1/4, 1/3] \cup (1/6, 1/5] \cup \dots $$

Note that this set has measure more than 1/2. However, the distribution of $Z$ is of course not uniform on [0,1] (it is actually skewed towards 0 instead of 1). I suspect what ends up happening is that the distribution of $\lfloor 1/Z \rfloor$ is almost perfectly split between even and odd in the regime $Z < 1/5$, and the discrepancy is thus a result of what happens for $Z > 1/5$ (where one easily sees that odd wins out).

I suspect that it has to do with the fact that the most likely outcome is $\lfloor 1/(x y) \rfloor=1$ (which happens to be odd), followed with an application of the Strong Law of Small Numbers. Here are some more details. Let $Z$ be the random variable $xy$. Note that $Z$ takes values in $[0,1]$ and $\lfloor 1/Z \rfloor$ is odd if and only if

$$ Z \in (1/2, 1] \cup (1/4, 1/3] \cup (1/6, 1/5] \cup \dots $$

Note that this set has measure more than $1/2$. However, the distribution of $Z$ is of course not uniform on $[0,1]$ (it is actually skewed towards $0$ instead of $1$). I suspect what ends up happening is that the distribution of $\lfloor 1/Z \rfloor$ is almost perfectly split between even and odd for say $Z < 1/5$, and the discrepancy is thus a result of what happens for $Z \geq 1/5$ (where one easily sees that odd wins out).

Source Link
Tony Huynh
Tony Huynh
  • 32.1k
  • 11
  • 112
  • 187

I suspect that it has to do with the fact that the most likely outcome is $\lfloor 1/(x y) \rfloor=1$ (which happens to be odd), followed with an application of the Strong Law of Small Numbers. Here are some more details. Let $Z$ be the random variable $xy$. Note that $Z$ takes values in $[0,1]$ and $\lfloor 1/Z \rfloor$ is odd if and only if

$$ Z \in (1/2, 1] \cup (1/4, 1/3] \cup (1/6, 1/5] \cup \dots $$

Note that this set has measure more than 1/2. However, the distribution of $Z$ is of course not uniform on [0,1] (it is actually skewed towards 0 instead of 1). I suspect what ends up happening is that the distribution of $\lfloor 1/Z \rfloor$ is almost perfectly split between even and odd in the regime $Z < 1/5$, and the discrepancy is thus a result of what happens for $Z > 1/5$ (where one easily sees that odd wins out).