Finding the probability that an inner product is positive Let $x$ be a fixed vector in $\mathbb{R}^n$.
Let $a=[a_1,\cdots,a_n]^T$ be a random vector whose entries are iid random variables, say, $a_i \sim P$.
I would like to compute 
$$
P(\langle a, x \rangle > 0) 
$$
Here is my attempt.
Without loss of generality, let $x_1 > 0$. 
Then 
\begin{align*}
P(\langle a, x \rangle > 0) = \int_\mathbb{R} \dots \int_{\mathbb{R}} \int_{a_1 > -\frac{\sum_{j=2}^na_jx_j}{x_1}} p(a_1)\cdots p(a_n)da_1\cdots da_n.
\end{align*}
If $P$ is an uniform distribution on $[-M,M]$, the above becomes
\begin{align*}
P(\langle a, x \rangle > 0) &= \frac{1}{(2M)^n}\int_{-M}^M \dots \int_{-M}^M \int_{a_1 > -\frac{\sum_{j=2}^na_jx_j}{x_1}} da_1\cdots da_n \\
&=\frac{1}{(2M)^n}\int_{-M}^M \dots \int_{-M}^M \left(M + \frac{\sum_{j=2}^na_jx_j}{x_1}\right) da_2\cdots da_n \\
&= \frac{1}{2}.
\end{align*}
If $a_i \sim \mathcal{N}(0,\sigma^2)$, since $\langle a, x \rangle = \sum_{i=1}^na_ix_i \sim \mathcal{N}(0,\|x\|^2\sigma^2)$,
one can easily conclude $P(\langle a,x \rangle > 0) = 0.5$.
Question In what class of distribution $P$, can we derive 
$$P(\langle a,x \rangle > 0) = 0.5?$$
I thought the symmetric distribution around 0 could result in the same result, however, it is unclear to me. 
Any comments/answers will very be appreciated.
 A: Partial answer.
To answer your "I thought... it is unclear to me",
symmetric distributions which give weight $0$ to $0$, ie a distribution $\mu$ on $\mathbb R$ with $\mu(0) = 0$, certainly do. Why? Let me go into detail.
Note that the map $a \mapsto \langle a, x \rangle$ is symmetric in the sense that $\langle -a, x \rangle = - \langle a, x \rangle$. If your distribution is symmetric, then $a$ and $-a$ have the same distribution, and hence so do $\langle a, x \rangle$ and $\langle -a, x \rangle = - \langle a, x \rangle$. Hence
$$ P( \langle a, x \rangle > 0 )
= P( -\langle a, x \rangle > 0 )
= P( \langle a, x \rangle < 0).
$$
Now, assuming that $x$ is not the $0$-vector (ie $x_i = 0$ for all $i$), in which case the probability is always $0$, we have either $\langle a,x \rangle > 0$ or $\langle a, x \rangle < 0$ for all $a$ which are not the $0$-vector.
Now, we have required that our distribution give no weight to $0$, and so $a_i \neq 0$ for all $i$, and so certainly it is not the case that $a$ is the $0$-vector. Hence we must have
$$ P( \langle a, x \rangle > 0 ) + P( \langle a, x \rangle < 0 ) = 1, $$
but since the two terms on the left-hand side are equal, by the previous deiplay, they must both equal $\tfrac12$, ie
$$ P( \langle a, x \rangle > 0 ) = P( \langle a, x \rangle < 0 ) = \tfrac12. $$
