Just came to know about a fact that the Wasserstein distance between two cumulative distribution functions, $F_1$ and $F_2$, can be expressed as: $$\int_{-\infty}^{\infty} |F_1 (x)-F_2(x)| dx$$.
Is it possible to compute this distance in case $F_1$ and $F_2$ are two normal distributions, with constant variance $1$ and means $\mu_1 \neq \mu_2$ by hand?
I have no clue how to work with the absolute value along with the integral which does not seem to have a closed form!
Best Answer
Suppose that $\mu_2 > \mu_1$.
The graph of density of $N(\mu_1, 1)$ lies to the left of the graph of density of $N(\mu_2, 1)$. As it was already mentioned by Jean Marie, it follows that $F_1 > F_2$.
It's known that if $\xi$ has distribution function $F(x)$ then $\mathbf{E}\xi = \int_{-\infty}^{0} (- F(x))dx + \int_{0}^{\infty} (1 - F(x))dx$.
Suppose that $\xi_i \sim N(\mu_i, 1)$. We have $$\int_{-\infty}^{\infty} |F_1(x) - F_2(x)|dx = \int_{-\infty}^{\infty} (F_1(x) - F_2(x))dx =$$ $$= \int_{-\infty}^{0} (- F_2(x))dx - \int_{-\infty}^{0} (- F_1(x))dx + \int_{0}^{\infty} (F_1(x) - F_2(x))dx =$$ $$= \int_{-\infty}^{0} (- F_2(x))dx - \int_{-\infty}^{0} (- F_1(x))dx + \int_{0}^{\infty} ((1 - F_2(x)) - (1-F_1(x)))dx =$$ $$=\int_{-\infty}^{0} (- F_2(x))dx + \int_{0}^{\infty} (1 - F_2(x))dx -$$ $$ - \bigg( \int_{-\infty}^{0} (- F_1(x))dx + \int_{0}^{\infty} (1 - F_1(x))dx \bigg)= $$ $$= \mathbf{E}\xi_2 - \mathbf{E}\xi_1 = \mu_2 - \mu_1.$$ It follows that the answer is $|\mu_2 - \mu_1|.$