Hardness of LPN problem with small secret The Learning Parity with Noise (LPN) assumption states that, for a fixed secret $s$ chosen uniformly from $\{0,1\}^n$, then the distribution that outputs $(a,a\cdot s+e)$, where $a$ is uniform in $\{0,1\}^n$ and $e$ is sampled according to a Bernoulli distribution $\mathsf{Ber}_\tau$ (with parameter $\tau\in[0,1/2]$), is pseudorandom.

My question is: Does the LPN assumption holds even if $s$ is chosen according to the same distribution as the error, that is, $\mathsf{Ber}_\tau^n$? 

In other words, given $s$ from $\mathsf{Ber}^n_\tau$, is the distribution $(a,a\cdot s+e)$ pseudorandom?
There is a simple trick (known in the LWE literature as the Hermite normal form of the problem) that takes an existing LPN problem and transforms it into a problem in which the secret has the same Bernoulli distribution as the error. This trick is proved in Lemma 2 of Applebaum et al. for a more general case, or on the (Ring-)LPN attacks of Kirchner (Section 4.3.2) or Bernstein and Lange (Section 3).

The idea is that you start with a number of $n$ samples $(\mathbf{a}_i, c_i)$, where $c_i = \mathbf{a}_i \cdot \mathbf{s}$. Suppose, now, that some subset of the $\mathbf{a}_i$ vectors is linearly independent (you do not need much more than $n$ randomly-sampled $\mathbf{a}_i$ for this to be the case), and let them form the matrix $M = \{ \mathbf{a}_{i_1}, \dots, \mathbf{a}_{i_k}\}$. Then we have $$\mathbf{c} = M^T\mathbf{s} + \mathbf{e},$$ where $\mathbf{c} = \{ c_{i_1}, \dots, c_{i_k}\}$ and $\mathbf{e} = \{ e_{i_1}, \dots, e_{i_k}\}$ are the corresponding vectors to the chosen $\mathbf{a}_i$. But we also have $$\mathbf{s} = {M^T}^{-1}(\mathbf{c} + \mathbf{e}).$$

Now that we have this matrix $M$, query the LPN oracle with a new sample $\mathbf{u}$, receiving $(\mathbf{u}, v = \mathbf{s}\cdot \mathbf{u} + e)$. But you can convert this into a  second oracle that returns $(M^{-1}\mathbf{u}, \mathbf{c}\cdot M^{-1}\mathbf{u} + v)$.
Obviously, $M(M^{-1}\mathbf{u}) = \mathbf{u}$. Further, 
$$
\begin{align}
\phantom{=}  &amp;\mathbf{e}\cdot M^{-1}\mathbf{u} + e \\
= &amp;(M^T\mathbf{s} + \mathbf{c})\cdot(M^{-1}\mathbf{u}) + e  \\ 
= &amp;M^T\cdot\mathbf{s}\cdot M^{-1}\mathbf{u} + \mathbf{c}\cdot  M^{-1}\mathbf{u} + e \\
= &amp;\mathbf{s}\cdot M M^{-1}\cdot u + \mathbf{c}\cdot M^{-1}\mathbf{u} + e \\
= &amp;\mathbf{c} \cdot M^{-1}\mathbf{u} + v.
\end{align}
$$ So this is now an instance of LPN with Bernoulli-distributed secret $\mathbf{e}$, from which you can extract the original secret as $\mathbf{s} = {M^T}^{-1}(\mathbf{c} + \mathbf{e})$.

In other words, given some algorithm that solves LPN with a Bernoulli-distributed secret, you can also solve regular LPN with the same error distribution.