= Solution
Choose an individual uniformly from the population. The <law of total probability> gives \b[the population density and survivor function] as mixtures:
$$
\boxed{\overline f(t)=\frac1n\sum_{i=1}^n f_i(t),\qquad\overline F(t)=\frac1n\sum_{i=1}^n F_i(t).}
$$
A <hazard function> is a ratio of density to survival, so averaging the densities does not average those ratios. Instead,
$$
\boxed{\overline h(t)=\frac{\sum_iF_i(t)h_i(t)}{\sum_iF_i(t)}=\sum_iw_i(t)h_i(t),\qquad w_i(t)=\frac{F_i(t)}{\sum_jF_j(t)}.}
$$
These are the proportions of the original individual types among those still event-free at $t$, rather than their equal starting proportions. For example, a mixture of different <exponential distributions> progressively favours the lower-rate individuals. This <population hazard of a survival mixture> generally differs from $n^{-1}\sum_i h_i(t)$.
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