If $f(x) = x/2$ for $-2 < x < 2$ (and 0 elsewhere), is it a valid p.d.f.?
ACannot decide without more info
BYes — $\int_{-2}^2 (x/2)\,dx = 1$
CNo — $f(x)$ takes negative values on $-2 < x < 0$
DYes — by symmetry
Answer & Solution
Correct answer: C. No — $f(x)$ takes negative values on $-2 < x < 0$
p.d.f. requires $f \geq 0$. Here $f(x) = x/2 < 0$ for $x \in (-2, 0)$, violating the non-negativity condition. So $f$ is NOT a valid p.d.f.
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