Similarity/Distance Measures for x = {0,1,0,1}, y = {1,0,1,0}

Setup

$$\mathbf{x} = (0, 1, 0, 1), \quad \mathbf{y} = (1, 0, 1, 0)$$

1. Cosine Similarity

$$\cos(\mathbf{x}, \mathbf{y}) = \frac{\mathbf{x} \cdot \mathbf{y}}{\|\mathbf{x}\|\|\mathbf{y}\|}$$

• Dot product: $(0)(1)+(1)(0)+(0)(1)+(1)(0) = 0$
• $\|\mathbf{x}\| = \sqrt{0^2+1^2+0^2+1^2} = \sqrt{2}$
• $\|\mathbf{y}\| = \sqrt{1^2+0^2+1^2+0^2} = \sqrt{2}$

$$\cos(\mathbf{x},\mathbf{y}) = \frac{0}{\sqrt{2}\cdot\sqrt{2}} = \boxed{0}$$

2. Correlation

Mean: $\bar{x} = \bar{y} = 0.5$

i	$x_i-\bar{x}$	$y_i-\bar{y}$	product	$(x_i-\bar{x})^2$	$(y_i-\bar{y})^2$
1	-0.5	0.5	-0.25	0.25	0.25
2	0.5	-0.5	-0.25	0.25	0.25
3	-0.5	0.5	-0.25	0.25	0.25
4	0.5	-0.5	-0.25	0.25	0.25
Σ			-1.0	1.0	1.0

$$r = \frac{-1.0}{\sqrt{1.0 \times 1.0}} = \boxed{-1}$$
(makes sense: y is exactly the complement of x, so they're perfectly negatively correlated)

3. Euclidean Distance

$$d(\mathbf{x},\mathbf{y}) = \sqrt{(0-1)^2+(1-0)^2+(0-1)^2+(1-0)^2} = \sqrt{1+1+1+1} = \sqrt{4} = \boxed{2}$$

4. Jaccard Similarity

Binary co-occurrence counts across the 4 positions:

Position	x	y	Match type
1	0	1	f01
2	1	0	f10
3	0	1	f01
4	1	0	f10

$f_{11}=0,\ f_{01}=2,\ f_{10}=2,\ f_{00}=0$

$$J(\mathbf{x},\mathbf{y}) = \frac{f_{11}}{f_{01}+f_{10}+f_{11}} = \frac{0}{2+2+0} = \boxed{0}$$

Summary

Measure	Value	Interpretation
Cosine similarity	0	Vectors are orthogonal
Correlation	-1	Perfectly negatively correlated
Euclidean distance	2	Maximum possible distance for these binary vectors
Jaccard similarity	0	No 1-1 overlap at all — completely dissimilar on "presence"

All four measures agree that x and y are maximally dissimilar — they're exact bitwise complements of each other.