# Hybrid argument without efficient samplability

Let's say I have $$k$$ distributions, where $$k$$ is polynomially large, $$D_1, D_2, \ldots, D_k$$ such that each $$D_i$$ is computationally indistinguishable from the uniform distribution.

Is it true that the distribution $$D_1 D_2 \ldots D_k$$ is also computationally indistinguishable from $$k$$ copies of the uniform distribution?

This trivially holds if each $$D_i$$ is efficiently samplable. But let's say they are not.

Does the fact still remain true, by some clever way to bypass the samplability requirement?

• Just a clarification. This paper talks about two distributions, and the setting when we are given $k$ samples from any one distribution out of the two. But, here, we are either given one sample each from $k$ different distributions (each computationally indistinguishable from the uniform), or we are given $k$ samples from the uniform distribution. Do you think the techniques that work for the first setting (that of the paper) also work for the second setting (that of my question)? – BlackHat18 Jul 22 at 13:58