Consider the random variables $X_1$ and $X_2$, both describing the experiment of tossing an unbaised coin as follows:
For $X_1$,
$X_1 = -1$ means tails and $X_1 = 1$ means heads. Hence $$E(X_1) = 0.5 \times (-1) + 0.5 \times (1) = 0.$$
Now I want to calculate $P(X_1 \geq 1)$ (i.e. probability of heads) through Chebychev inequality i.e. $$ P(X_1 \geq 1) \leq E(X_1)/1 = 0.$$ Since Probability is always positive, we have $P(X_1 \geq 1) = 0$ but the direct calculation shows $P(X_1 \geq 1) = P(X_1 = 1) = 1/2$ - a contradiction to Chebychev inequality.
For $X_2$,
$X_2 = 0$ means tails and $X_2 = 1$ means heads. Hence $$E(X_2) = 0.5 \times (0) + 0.5 \times (1) = 1/2.$$
Now I want to calculate $P(X_2 \geq 1)$ (i.e. probability of heads) through Chebychev inequality i.e. $$ P(X_2 \geq 1) \leq E(X_2)/1 = (1/2)/1 = 1/2.$$ Also a direct calculation shows $P(X_2 \geq 1) = P(X_2 = 1) = 1/2$ - a verification to Chebychev inequality.
So my question is, does the chebyshev inequality is only true for 1 among many random variable available to model a given expirement. If yes how do I recognise which one is it?