Normalized Wasserstein distance The wasserstein_distance will be smaller the longer u_values and v_values are.
from scipy.stats import wasserstein_distance

def wassersteindist(n):    
    a = np.random.randn(n)
    b = np.random.randn(n)
    w = wasserstein_distance(a,b)
    return w
     
np.mean([wassersteindist(100) for r in range(1000)])
0.1786
np.mean([wassersteindist(1000) for r in range(1000)])
0.0579
np.mean([wassersteindist(10000) for r in range(1000)])
0.0180

Is there a way to calculate a normalized wasserstein distance with scipy?
EDIT:
Let's say I 'm interested in comparing the distances from different individuals that happened to have a different amount of time points in their time series.
id1_a = np.random.randn(100)
id1_b = np.random.randn(100)

id2_a = np.random.randn(1000)
id2_b = np.random.randn(1000)

id1_dist = wasserstein_distance(id1_a, id1_b)
0.3539204677483332

id2_dist = wasserstein_distance(id2_a, id2_b)
0.0685546301855615

id1_dist is larger than id2_dist only because the vectors for id1 are shorter than for id2.
EDIT2:
With correlations I don't have the problem that they are consistently lower/higher for longer time series:
def corrr(n):    
    a = np.random.randn(n)
    b = np.random.randn(n)
    c = np.corrcoef(a,b)[0][1]
    return c

np.mean([corrr(100) for r in range(1000)])
0.0004

np.mean([corrr(1000) for r in range(1000)])
0.0012

np.mean([corrr(100) for r in range(1000)])
-0.0001

np.mean([corrr(1000) for r in range(1000)])
-0.0008

 A: The fact that this distance is going down in your case is a good thing. In general, for two samples with empirical distributions $\mathbb P_n$ and $\mathbb Q_m$, we have have
$$
W(\mathbb P_n, \mathbb Q_m) \to W(\mathbb P, \mathbb Q)
$$
as $m, n \to \infty$. Here, $\mathbb P$ and $\mathbb Q$ are the population versions of the two distributions. In your simulations, the two samples are are coming from the same population $\mathbb P = \mathbb Q = N(0,1)$, hence $W(\mathbb P, \mathbb Q) = 0$. As $n,m \to \infty$, you should expect $W(\mathbb P_n, \mathbb Q_m)$ to approach zero.
For you time series, if they are coming from two different populations, $W(\mathbb P, \mathbb Q)$ will be nonzero, so you will converge to something nonzero for large samples.
In other words, you want $W(\mathbb P_n, \mathbb Q_m)$ to vary with $m$ and $n$ to get more accurate for larger samples. In general (in the 1-D case),
$$\mathbb E[ W(\mathbb P_n, \mathbb Q_m) ] = W(\mathbb P,\mathbb Q) + O\Bigr(\frac1{\sqrt{n \wedge m}}\Bigl).$$
Normalizing $W(\mathbb P_n, \mathbb Q_m)$ to be invariant to the sample size does not make much sense, especially when $\mathbb P = \mathbb Q$.

In case you are curious where that rate comes from, first note that by triangle inequality
$$W(\mathbb P_n, \mathbb Q_m) \le W(\mathbb P_n, \mathbb P) + W(\mathbb P, \mathbb Q) + W(\mathbb Q, \mathbb Q_m).$$
Then, use a result like that of Fournier and Guillin on the first and third terms.

EDIT1: The same problem is there with the absolute value of Pearson correlation coefficient:
import numpy as np

np.random.seed(1337)

def corrr(n):    
    a = np.random.randn(n)
    b = np.random.randn(n)
    c = np.corrcoef(a,b)[0][1]
    return c

for n in [10, 100, 1000, 10000]:
    print(np.mean([abs(corrr(n)) for r in range(1000)]))


which produces the following output:
0.2709428042954994
0.07715058968038786
0.023744887030041146
0.008094465791386103

