Resulting biased distribution when conditionally choosing a RV from uniform distributions

Question

I was reading a paper by Degorre et al. (https://doi.org/10.1103/PhysRevA.72.062314). They present the following protocol, where $\it U_\Lambda$ stands for the uniform distribution on the set $\Lambda$ and $\vec a$ is an arbitrary vector from $\mathbb{R^2}$ (can also be restricted to the unit sphere $S_2$):

1. Alice picks $\vec \lambda_1$ from $\it U_{S_{2}}$
2. Alice picks $\vec \lambda_2$ from $\it U_{S_{2}}$
3. Alice then sets $\vec \lambda_s = \begin{cases} \vec \lambda_1, & \text{if } \lvert \vec a \cdot \vec \lambda_1 \rvert \ge \lvert \vec a \cdot \vec \lambda_2 \rvert\\ \vec \lambda_2, & \text{else} \end{cases} $

They claim that the resulting density of $\vec \lambda_s$ is $\rho(\vec \lambda_s \vert \vec a)=\lvert \vec a \cdot \vec \lambda_s \rvert / 2\pi$.

I do not understand their proof at all and was not able to find one myself because I cannot wrap my head around what integrals I have to evaluate.

I felt that the integral over the cap of the sphere $$\int_{\lvert \vec a \cdot \vec \lambda \rvert \ge \lvert \vec a \cdot \vec \lambda_0 \rvert\\}{d\vec\lambda} = 4\pi [1-\cos(\angle\vec a,\vec \lambda_0)] = 4\pi (1-\vec a \cdot \vec \lambda_0)$$ with $\vec a \in S_2$ and $\vec \lambda_0$ from $\mathcal{U}_{S_2}$ was closely related to the problem, but I could not make it work.

It would be a great help to me if you could help me find the relevant integrals that I need to evaluate to get the resulting distribution.

Answer 1

Not a full answer, but since you asked which integrals to evaluate:

Conditionally on $(\lambda_1,\lambda_2)$ we have

$$ \lambda_s \vert \lambda_1,\lambda_2 = \mathbb{I}\left[\vert \lambda_1^T a \vert \geq \vert \lambda_2^Ta \vert \right] \lambda_1 + (1-\mathbb{I}\left[\vert \lambda_1^T a \vert \geq \vert \lambda_2^Ta \vert \right]) \lambda_2$$ where $\mathbb{I}[\cdots]$ is the indicator function.

Thus, let us write the conditional density suggestively as

$$p(\lambda_s \vert \lambda_1,\lambda_2) = \mathbb{I}\left[\vert \lambda_1^T a \vert \geq \vert \lambda_2^Ta \vert \right] \delta_{\lambda_1}(\lambda_s) + (1-\mathbb{I}\left[\vert \lambda_1^T a \vert \geq \vert \lambda_2^Ta \vert \right]) \delta_{\lambda_2}(\lambda_s).$$

Now since the $\lambda_1,\lambda_2$ are indepedent, uniformly distributed on the unit sphere, the joint probability is just the conditional times $\frac{1}{4\pi^2}$.

To get the marginal probability $p(\lambda_s)$, we now have to evaluate

$$\int \int p(\lambda_s,\lambda_1,\lambda_2) d\lambda_1 d\lambda_2 \propto \int \int \mathbb{I}\left[\vert \lambda_1^T a \vert \geq \vert \lambda_2^Ta \vert \right] \delta_{\lambda_1}(\lambda_s) + (1-\mathbb{I}\left[\vert \lambda_1^T a \vert \geq \vert \lambda_2^Ta \vert \right]) \delta_{\lambda_2}(\lambda_s) d\lambda_1 d\lambda_2 = \\ \int \int \mathbb{I}\left[\vert \lambda_1^T a \vert \geq \vert \lambda_2^Ta \vert \right] \delta_{\lambda_s}(\lambda_1) d \lambda_1 d\lambda_2+ \int \int (1-\mathbb{I}\left[\vert \lambda_1^T a \vert \geq \vert \lambda_2^Ta \vert \right]) \delta_{\lambda_s}(\lambda_2) d\lambda_2 d\lambda_1 \\ = \int_{\vert \lambda_s^T a \vert \geq \vert \lambda_2^Ta \vert} d \lambda_2 + \int_{\vert \lambda_s^T a \vert \geq \vert \lambda_1^Ta \vert} d \lambda_1 \\ = 2 \int_{\vert \lambda_s^T a \vert \geq \vert \lambda^Ta \vert} d \lambda $$

(I suppose there is a more elegant way to get the solution directly, also you should note the radial symmetry in $a$).

Answer 2

To help you visualise what is happening, you are looking at the curved surface area of the grey caps of this sphere or rather at their complement of the pink surface area.

That will give you something proportional to a conditional probability for each point of another randomly selected point being in a sense further from $\mathbf a$, and this can be turned in a density by ensuring that its integral over the whole sphere is $1$.