$Y_i$ is a i.i.d. random variable following Bernoulli distribution with success probability of $\alpha$.
Then, does $p\left(\frac{Y_1+...Y_k}{k}>x-\frac{a}{k}\right)$ increase with $k$ for any given constant $x>0$? Note that $a>0$.
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$\begingroup$ what is $x$ here? $\endgroup$– user12075Commented Sep 30, 2018 at 3:42
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$\begingroup$ @user12075 Sorry for the confusion. I have updated the question. $x$ is a given constant. $\endgroup$– user3509199Commented Sep 30, 2018 at 3:45
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$\begingroup$ Hint: use the Central Limit Theorem to estimate the probability. If you want a more elementary approach, still use the CLT to guess the answer and then prove it using, say, Chebyshev's Inequality. $\endgroup$– whuber ♦Commented Sep 30, 2018 at 15:21
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No. It converges to either $0$ or $1$.
By the WLLN, the Left hand term converges in probability to $E[Y_i] = \alpha$.
On the other hand, the right hand side clearly converges to $x$ since $a/k \to 0$.
Hence if $x < \alpha$ the probability converges to $1$, but if $x > \alpha$ it converges to $0$.
So if $x < \alpha$ then it increases with $k$, but if $x > \alpha$ it decreases.
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1$\begingroup$ What if $x$ happens to be $E(Y_i)$? $\endgroup$– Glen_bCommented Sep 30, 2018 at 10:00
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$\begingroup$ Use the central limit theorem, the probability will converge to a constant. $\endgroup$ Commented Dec 22, 2020 at 22:53