Discussion
Let X be a r.v. such that variance of X is \(\dfrac{1}{2}\). Then, an upper bound for \(P[|X - FX| \gt 1]\) as given by the Chebychev's inequality is :
Answer: B
Chebyshev's inequality gives the bound \(\dfrac{1}{2}\).
Here FX stands for the expectation \(E(X)\). Chebyshev's inequality states that for any \(k \gt 0\),
\[ P[|X - E(X)| \ge k] \le \frac{\text{Var}(X)}{k^{2}} \]
With \(\text{Var}(X) = 1/2\) and \(k = 1\), the bound is \((1/2)/1^{2} = 1/2\). The event with strict inequality, \(|X - E(X)| \gt 1\), is contained in the one with \(\ge\), so the same bound applies.
1/4 is what you get by squaring the variance instead of dividing by \(k^{2}\). 1 is the trivial bound every probability satisfies and uses none of the information given. 3/4 does not follow from the inequality with these numbers at all; the complementary form gives \(P[|X - E(X)| \lt 1] \ge 1 - 1/2 = 1/2\).
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