Discussion
Let X be a random variable with distribution function :
\(F_{\alpha}(t) = 1 - \exp(-(\lambda t)^{\alpha})\), \(\lambda, \alpha \gt 0\), \(t \ge 0\)
Then :
Answer: D
\(F_{\alpha}\) is IFR if \(1 \le \alpha\).
This is the Weibull distribution. Its hazard rate is
\[ \begin{aligned} h(t) &= \frac{f(t)}{1 - F(t)} \\ &= \alpha\lambda^{\alpha}t^{\alpha - 1} \end{aligned} \]
which is increasing in \(t\) when \(\alpha \gt 1\), constant when \(\alpha = 1\) (the exponential case, which counts as both IFR and DFR), and decreasing when \(\alpha \lt 1\). So the distribution is IFR for \(\alpha \ge 1\) and DFR for \(\alpha \le 1\).
It is not IFR or DFR for every \(\alpha\), so A and B fail, and C has the range reversed.
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