How to calculate lower bound on $P \left[|Y| > \frac{|\lambda|}{2} \right]$? Let $Y$ be a random variable such that $E[Y] = \lambda$,  $\lambda \in \mathbb{R}$ and $E[Y^2]<\infty$. The  problem is to find a lower bound on the probability
$$
P \left[|Y| > \frac{|\lambda|}{2} \right].
$$
Any leads would be appreciated.
 A: Edit : This answer applies to the original question, that was : "for ANY random variable Y, what is the lower bound of (formula)"
The lower bound is 0.
Let's take $(Y_{n})$ a series following a Bernoulli law such as : $P(Y_{n} = 1) =1/n$.
Then, ${E(Y_{n})= \lambda = 1/n}$.
$P(\lvert Y_{n} \rvert > \frac{\lvert \lambda \rvert}{2}) = P(\lvert Y_{n} \rvert > \frac{1}{2n}) = P(Y_{n} =1) = \frac{1}{n} <  \frac{2}{n}\underset{n\to +\infty}{\longrightarrow} 0 $
Then, $0 < P(\lvert Y_{n} \rvert > \frac{\lvert \lambda \rvert}{2})  < \frac{2}{n}$ , so asymptotically, the lower bound is 0.
A: Given $\lambda,$ there must be a universal lower bound $p(\lambda)$ (because $0$ will certainly work.)  The question is whether there are any $\lambda$ where this bound exceeds $0.$  Regardless, being a lower bound means that for any random variable $Y$ with $E[Y^2]\lt \infty$ and $E[Y]=\lambda,$ $$\Pr(|Y| \gt |\lambda|/2) \ge p(\lambda).\tag{*}$$
Consider $Y=\lambda X/q$ where $X$ is a Bernoulli$(q)$ variable and $\lambda \ne 0.$  Because $E[X]=q,$ $E[Y] = \lambda (q)/q = \lambda.$ $Y$ also has finite second moment because it is bounded.  Moreover, since $q \lt 2,$ the event "$|Y|\gt|\lambda|/2$" equals the event "$Y=\lambda/q.$"  We may therefore conclude from $(*)$ that
$$q = \Pr(X=1) \ge \Pr(Y=\lambda/q) = \Pr(|Y| \gt |\lambda|/2) \ge p(\lambda).$$
If we choose $q = p(\lambda)/2,$ this says
$$p(\lambda)/2 \ge  p(\lambda),$$
which is true only for non-positive numbers $p(\lambda).$ (You can check that this example still works when $\lambda=0,$ for then $Y$ reduces to the atom at $0$ and $\Pr(|Y|\gt |\lambda|/2) = \Pr(Y \ne 0) = 0.$) Therefore $0$ is the only universal lower bound, no matter what the value of $\lambda$ might be.
