TITLE: Probability of two vectors lying in the same orthant
QUESTION [19 upvotes]: Let $S^{d-1} = \{x \in \mathbb R^d: \|x\| = 1\}$ denote the unit sphere in $\mathbb R^d$. Let $v$, $w$ be drawn uniformly at random from $S^{d-1}$, conditioned on their inner product being equal to $\langle v, w \rangle = \cos \theta$. In other words, $v$ and $w$ have a fixed common angle $\theta$. I am interested in the probability that these two vectors lie in the same orthant, say the all-positive orthant, averaged over all possible $v$ and $w$:
$$f_d(\theta) = \Pr(v > 0 \mid w > 0, \ \langle v, w \rangle = \cos \theta).$$
Here $v > 0$ means that all $d$ coordinates of $v$ are positive. 
In particular, I am interested in the asymptotics of large $d \to \infty$ of $f_d(\theta)$ for $\theta \in (0, \frac{1}{2} \pi)$. Equivalently, as $f_d$ scales exponentially in $d$, I'm interested in the function
$$g(\theta) = \lim_{d \to \infty} f_d(\theta)^{1/d}.$$
Note that an approximation to the above probabilities can be obtained by replacing the uniform distribution over the sphere by a multivariate Gaussian distribution, where each coordinate is independently drawn from a Gaussian $\mathcal{N}(0, \frac{1}{d})$. For large $d$, with overwhelming probability such a random vector will have norm $1 \pm o(1)$, and with overwhelming probability two such random vectors will have inner product $o(1)$. If we ignore the fact that the norms of such vectors may not exactly be equal to $1$ and that two random vectors may not be exactly orthogonal (which is why this is only an approximation), then two vectors $v$ and $w$ from the sphere with angle $\theta$ can be generated by taking $v = n_1$ and $w = (\cos \theta) n_1 + (\sin \theta) n_2$ for two independent random Gaussian vectors $n_1, n_2 \sim \mathcal{N}(0, \frac{1}{d})^d$. These lie on the sphere exactly if $n_1, n_2$ have norm $1$, and their angle is then equal to $\theta$ if and only if $n_1$ and $n_2$ are orthogonal. 
With this approximation, probabilities can be computed quite easily, as different coordinates are independent and probabilities multiply. However, I'm looking for more precise estimates than using this Gaussian approximation of the uniform distribution on the sphere.
So far I've tried sharing this problem with a few others in the department, and rewriting the probability to computing the expected volume of the intersection of the sphere with $d$ orthogonal hyperplanes, but so far nothing led anywhere. Any pointers on how to solve this would be greatly appreciated!

Update: To verify/compare different approaches, simulations for $\theta = \arccos 0.9$ in dimensions $d \in \{10, \dots, 40\}$ show the following trends for $\ln f$:

The points are the simulation results (using 100.000-500.000 experiments each, so that the number of successes was at least a few hundred) and the line is the linear fit $-0.0133857 - 0.170074 d$, or equivalently $f(\arccos 0.9) \approx C \cdot 0.8436^d$ for a constant $C \approx 1$. This suggests that $g(\arccos 0.9) \approx 0.8436$. The answers given so far say:

Carlo's answer: $0.66$
Other answer (main): $1.08$
Other answer (alternative): $0.71$

So perhaps all answers so far are still far off when $\theta$ is small and the inner product between $v$ and $w$ is large.

REPLY [3 votes]: Thanks to Ofer Zeitouni, unless I made some serious mistakes somewhere in the calculations, I've been able to find the exact answer using large deviations theory. The derivation of the final result is a rather tedious affair, so I will just state the result here. Below $\mathcal{S}^{d-1}$ denotes (the uniform distribution over) the unit ball in $\mathbb{R}^d$. The proof can be found in the appendix here and is about 5 pages long.

Theorem:
Let $\mathbf{X}, \mathbf{Y} \sim \mathcal{S}^{d-1}$, let $\phi(\mathbf{X}, \mathbf{Y})$ denote the angle between $\mathbf{X}$ and $\mathbf{Y}$, and let
\begin{align}
f_d(\theta) = \mathbb{P}(\mathbf{X} > 0 \ | \ \mathbf{Y} > 0, \ \phi(\mathbf{X}, \mathbf{Y}) = \theta).
\end{align}
For $\theta \in (0, \arccos \frac{2}{\pi})$ (respectively $\theta \in (\arccos \frac{2}{\pi}, \frac{\pi}{3})$), let $\beta_0 \in (1, \infty)$ (respectively $\beta_1 \in (1, \infty)$) be the unique solution to:
\begin{align}
\arccos\left(\frac{-1}{\beta_0}\right) = \frac{(\beta_0 - \cos \theta) \sqrt{\beta_0^2 - 1}}{\beta_0 (\beta_0 \cos \theta - 1)} \, , \qquad \arccos\left(\frac{1}{\beta_1}\right) = \frac{(\beta_1 + \cos \theta) \sqrt{\beta_1^2-1}}{\beta_1 (\beta_1 \cos \theta + 1)} \, . \label{eq:beta}
\end{align}
Then, for large $d$,
\begin{align}
f_d(\theta) = \begin{cases}
\left(\displaystyle\frac{(\beta_0 - \cos \theta)^2 }{\pi \beta_0 (\beta_0 \cos \theta - 1) \sin \theta}\right)^{d + o(d)}, & \qquad \text{if } \theta \in [0, \arccos \tfrac{2}{\pi}]; \\[3ex]
\left(\displaystyle\frac{(\beta_1 + \cos \theta)^2}{\pi \beta_1 (\beta_1 \cos \theta + 1) \sin \theta}\right)^{d + o(d)}, & \qquad \text{if } \theta \in [\arccos \tfrac{2}{\pi}, \tfrac{\pi}{3}]; \\[3ex]
\left(\displaystyle\frac{1 + \cos \theta}{\pi \sin \theta}\right)^{d + o(d)}, & \qquad \text{if } \theta \in [\tfrac{\pi}{3}, \tfrac{\pi}{2}); \\[3ex]
0, & \qquad \text{if } \theta \in [\tfrac{\pi}{2}, \pi].
\end{cases}
\end{align}

To illustrate, the following figure shows $g(\theta) = \lim_{d\to \infty} f_d(\theta)^{1/d}$ against $\theta$. The constant $\alpha$ on the y-axis is $\alpha = \pi / (2 \sqrt{\pi^2 - 4}) \approx 0.648$. The dashed lines indicate the boundaries of the piece-wise parts in the theorem.

As stated in the OP, experiments suggested that $g(\arccos 0.9) \approx 0.8436$, and indeed from this theorem we find $g(\arccos 0.9)$ to be approximately $0.843$. This is quite different from Carlo's approximate solution based on Gaussians - if $\theta$ is close to $0$, the Gaussian approximation seems to be further off.
For the case $\theta = \frac{\pi}{3}$, the exact solution is $g(\frac{\pi}{3}) = \frac{\sqrt{3}}{\pi} \approx 0.5513$. Other answers based on different approximations all suggested $g(\frac{\pi}{3}) \in [0.55, 0.57]$ so here the Gaussian approximations are pretty accurate.
And as $\theta$ approaches $\frac{\pi}{2}$ from below, we curiously get $g(\theta) \to \frac{1}{\pi}$. In other words, two almost-orthogonal vectors in high dimensions are in the same orthant with probability proportional to $(\frac{1}{\pi})^{d + o(d)}$. I'm not sure if there is a nice intuitive explanation for this.