Define the Jaccard distance between two continuous vectors $a, b\in [0,1]^p$ as
\begin{equation}
J(a,b) = 1 - \frac{\|a\odot b\|\_1}{\|a\odot b\|\_1+\|a-b\|\_1}
\end{equation}
where $\odot$ is the Hadamard product (element-wise product).


Is it a metric? Note that $a,b \in [0,1]^p$ rather than $\{0,1\}^p$.


I've tried with the naive approach. After some messy algebra, I need to prove the following
\begin{equation}
\|a-b\|\_1\|a\odot c\|\_1\|b\odot c\|\_1 + \|a-b\|\_1\|b-c\|\_1\|a\odot c\|\_1 + \|a-b\|\_1 \|b-c\|\_1 \|a-c\|\_1 + \|b-c\|\_1\|a\odot b\|\_1 \|a\odot c\|\_1 + \|a-b\|\_1\|b-c\|\_1\|a\odot c\|\_1 \geq \|a-c\|\_1\|a\odot b\|\_1 \|b\odot c\|\_1.
\end{equation}



Is the Jaccard distance between continuous vectors a metric?
All norms are supposed to be $1$-norms. Rewrite $J(a,b)$ as
$$
 J(a,b)=\frac{\|a-b\|}{\|a-b\|+\|a\odot b\|}.
$$


Notice that
$$
 \|b\odot c\|=\sum\_i b\_ic\_i
 =\sum\_i\bigl((b\_i-a\_i)c\_i+a\_ic\_i\bigr)
 \leq\sum\_i|b\_i-a\_i|+\sum\_ia\_ic\_i=\|a-b\|+\|a\odot c\|.
$$
Similarly,
$$
 \|a\odot b\|\leq \|b-c\|+\|a\odot c\|.
$$
Hence we have
$$
 J(a,b)+J(b,c)
 =\frac{\|a-b\|}{\|a-b\|+\|a\odot b\|}+\frac{\|b-c\|}{\|b-c\|+\|b\odot c\|}\\
 \geq \frac{\|a-b\|}{\|a-b\|+\|b-c\|+\|a\odot c\|}
 +\frac{\|b-c\|}{\|b-c\|+\|a-b\|+\|a\odot c\|}\\
 =\frac{\|a-b\|+\|b-c\|}{\|a-b\|+\|b-c\|+\|a\odot c\|}
 \geq \frac{\|a-c\|}{\|a-c\|+\|a\odot c\|}=J(a,c),
$$
where the last inequality follows from $\|a-b\|+\|b-c\|\geq \|a-c\|$.


