h (t ∣ z) = c (z) h_{0} (t),

c (z)

is positive and independent of time. Equivalently, its cumulative hazard functions satisfy

H (t ∣ z) = c (z) H_{0} (t)

S (t ∣ z) = S_{0} (t)^{c (z)}

i

Solution

Writing the two functions in the question as survivor functions,

S_{2} (t) = S_{1} (t)^{λ}

. Since

H_{k} (t) = - lo g S_{k} (t)

H_{2} (t) = - lo g S_{2} (t) = - λ lo g S_{1} (t) = λ H_{1} (t) .

Differentiating at times where the hazard functions exist gives

h_{2} (t) = λ h_{1} (t)

. Their hazard ratio is therefore the constant

λ

The transformation is

T_{k} = a_{k} U_{k}^{b}

. Because

U_{k}

S_{k} (t) = P (U_{k} > (t / a_{k})^{1/ b}) = exp [- (t / a_{k})^{1/ b}] .

Thus

T_{k}

H_{k} (t) = (t / a_{k})^{1/ b}, h_{k} (t) = \frac{1}{b} a_{k}^{- 1/ b} t^{1/ b - 1} .

Consequently

h_{2} (t) / h_{1} (t) = (a_{1} / a_{2})^{1/ b}

is constant, proving proportional hazards.

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