that is, we have $$\begin{aligned}
\mathbb P  \bigl( \lVert Z (\theta )\rVert_ \infty >  C \sqrt k 2 ^{- k \frac {1- \alpha  } 2  } \bigr)
\lesssim 2 ^{k} 
\sup _{\theta } 
\mathbb P  \bigl( \lvert Z (\theta )\rvert  >  C \sqrt k 2 ^{- k \frac {1- \alpha  } 2  } \bigr), \end{aligned}$$ where we have simply used the union bound.

Now, $ Z (\theta )$ is the sum of independent, mean zero random variables, which are bounded by one, and have standard deviation bounded by $ c 2 ^{ - k  \frac {1- \alpha  } 2}$. So by, for instance, the Bernstein inequality, it follows that $$\mathbb P  ( \lvert Z (\theta )\rvert  >  C \sqrt k 2 ^{- k \frac {1- \alpha  } 2  } ) \lesssim 2 ^{-2k},$$ for appropriate $ C$. This completes the proof.

From the previous Lemma, we have the Corollary below. It with the sparse bound for the Hilbert transform completes the proof of Theorem \[t:R\], for the random Hilbert transform. The case for maximal averages is entirely similar.

\[c:almost\] Almost surely, for $  {1+ \alpha } < r < 2$, there is a $ \eta >0$ so that for all 