Suppose a coin has probability $p$ for heads and $(1-p)$ for tail. Let $P_{k,p}$ be the probability that in $N$ flips there is a sequence of consecutive heads of length greater than or equal to $k$. $N$ is some fixed number greater than $k$.
Does it follow that $P_{k,p}< P_{k,q}$ if $p$ < $q$?
I would greatly appreciate any hint/suggestion/solution! Thank you very much!
A simple proof uses the beautiful idea of coupling. Roughly speaking, one realizes the results of the $p$ coins and of the $q$ coins on the same probability space, using the same randomness.
More precisely, consider $N$ i.i.d. random variables $U_k$, uniform on $(0,1)$, and decide that the result of flip $k$ is heads if and only if $U_k\leqslant p$. Then, replacing $p$ by $q\gt p$ increases the number of heads, pointwise, hence $P_{k,p}\leqslant P_{k,q}$.
Finally, the inequality is strict since, when $p\lt U_k\lt q$ for every $k$, the $q$ event is realized while the $p$ is not, and this has positive probability. (The argument shows that $P_{k,q}\geqslant P_{k,p}+(q-p)^N$.)
In the limit cases $p=0$ or $q=1$, the statement is true. $P_{k,p}$ seems continuous in $p$ and if it is continuous, the statement is true also for $p$ close to zero and $q\in(0,1)$.
