Discussion
Let X be a discrete r.v. with the following pmf :
\[ P[X = x] = \begin{cases} k; & x = 0, \pm j; \ j = 1, 2, \ldots n \\ 0; & \text{otherwise} \end{cases} \]
Let \(Y = X^{2}\), then, \(P[Y = 4]\) is :
Answer: D
\(P[Y = 4] = \)\(\dfrac{2}{2n + 1}\).
X takes the \(2n + 1\) values \(0, \pm 1, \pm 2, \ldots, \pm n\), each with the same probability \(k\). The probabilities must add to 1, so \((2n + 1)k = 1\) and \(k = 1/(2n + 1)\).
\(Y = X^{2} = 4\) happens exactly when \(X = 2\) or \(X = -2\) (both values exist once \(n \ge 2\)), so
\[ P[Y = 4] = P[X = 2] + P[X = -2] = \frac{2}{2n + 1} \]
\(1/(2n + 1)\) counts only one of the two values. \(2/(n + 1)\) comes from miscounting the support as \(n + 1\) points, and \(n/(2n + 1)\) is roughly the probability that X is positive, not that \(X^{2} = 4\).
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