Equality in Cantelli's inequality ($\mathbb{P}(X=\mu +a)={\sigma }^2/({\sigma }^2+a^2)$)

For $a>0$ and ${\sigma }^2>0$, place probabilitiesTheir weighted deviations cancel, giving expected value $\mu$; their weighted squared deviations give variance ${\sigma }^2$. Only the upper point belongs to $X-\mu \ge a$, so its tail probability attains Cantelli's inequality. With $a=2$, ${\sigma }^2=1$, $\mu =0$, the locations are two and $-1/2$, with masses $1/5$ and $4/5$. The optimized shift makes the lower point zero before squaring, explaining equality in the Markov's inequality step.

The threshold convention matters: this example has $\mathbb{P}(X-\mu >a)=0$. For a strict threshold, placing the upper point at $a+\epsilon$ instead gives probabilities approaching the same bound as $\epsilon \downarrow 0$, with the prescribed mean and variance retained. Extra restrictions on the support of a probability distribution can change which extremal models are allowed.