Survival Analysis

Let T ≥ 0 have density function f(t) and distribution function F (t). The survival function S(t) is

S(t) = 1 − F (t) = P (T > t)

and the hazard rate or hazard function λ(t) is

λ(t) = \frac{f(t)}{1 − F (t)}

interpreted as

λ(t) ≈ P (t < T < t + dt | T > t) = P (expiring in the interval (t, t + dt) | survived past time t). Integrating λ(t) gives

∫_{0}^{t} λ(u)du = − log S(t)
⇒ S(t) = e^{− ∫_{0}^{t} λ(u)du}

Censoring What distinguishes survival analysis from other fields of statistics is censoring. Vaguely speaking, a censored observation contains only partial information about the random variable of interest.

Let T_{1}, T_{2}, …, T_{n} ∼^{iid} F (t).

Type I Let t_{c} be a preassigned fixed number which we call the fixed censoring time. We observe Y_{1}, …, Y_{n}

Y_{i} = \{ \begin{matrix}T_{i} & if T_{i} ≤ t_{c}, \\ t_{c} & if T_{i} > t_{c}. \end{matrix}

Type II Let r < n be fixed, and let T_{(1)} < T_{(2)} < … < T_{(n)} be the order statistics of T_{1}, T_{2}, …, T_{n}. Observation ceases after the r^{th} failure, so we can observe T_{(1)}, T_{(2)}, …, T_{(r)}. The full observed sample is

Y_{(1)} = T_{(1)}
Y_{(2)} = T_{(2)}
⋮
Y_{(r)} = T_{(r)}
Y_{(r+1)} = T_{(r)}
⋮
Y_{(n)} = T_{(r)}

Type III (Random Censoring)

Let C_{1}, C_{2}, …, C_{n} be iid each with distribution function G. Each C_{i} is the censoring time associated with T_{i}. We can only observe (Y_{1}, δ_{1}), (Y_{2}, δ_{2}), …, (Y_{n}, δ_{n}) where

Y_{i} = min(T_{i}, C_{i}) = T_{i} ∧ C_{i}

δ_{i} = I(T_{i} ≤ C_{i}) = \{ \begin{matrix}1 & if T_{i} ≤ c_{i} (not censored), \\ 0 & if T_{i} > c_{i} (censored). \end{matrix}

notice Y_{1}, …, Y_{n} are iid with some distribution function H.

Example (p.185)

From February 1972 to February 1975, 29 severe vival hepatitis patients satisfied the admission criteria for a 16 week study of the effects of steroid therapy at the Stanford, Veterans Administration, and Santa Clara Valley Hospitals. These patients were randomized into either the steroid or control group. The survival times (in weeks) of the 14 patients in the steroid group were 1, 1, 1, 1+, 4+, 5, 7, 8, 10, 10+, 12+, 16+, 16+, 16+.

Assume an exponential distribution S(t) = e^{−λt}

a) Estimate λ by maximum likelihood and construct an approximate 95% CI for λ.

b) Estimate S(16) and construct an approximate 95% CI for S(16).

c) Estimate the median survival time and construct an approximate 95% CI for the median.

Statistics 6502, Spring 2007 3 Solutions:

a) Assume random censoring, the pair (y_{i}, δ) has likelihood

L(y_{i}, δ_{i}) = \{ \begin{matrix}f(y_{i}) & if δ_{i} = 1 (uncensored) \\ S(y_{i}) & if δ_{i} = 0 (censored) \end{matrix}

= f(y_{i})^{δ_{i}}S(y_{i})^{1−δ_{i}}.

The likelihood for the full sample

L = L(y_{1}, …, y_{n}; δ_{1}, …, δ_{n})

= ∏_{i=1}^{n} L(y_{i}, δ_{i})

= [∏_{u} f(y_{i})] [∏_{c} S(y_{i})]

= [λ^{n_{u}}e^{−λ ∑_{u} t_{i}}] [e^{−λ ∑_{c} c_{i}}] = λ^{n_{u}}e^{−λ ∑_{i=1}^{n} y_{i}}

log L = n_{u} log λ − λ ∑_{i=1}^{n} y_{i}
\frac{∂}{∂λ} log L = \frac{n_{u}}{λ} − ∑_{i=1}^{n} y_{i}
Setting \frac{∂}{∂λ} log L = 0 gives λ̂ = \frac{n_{u}}{∑_{i=1}^{n} y_{i}}
\frac{∂^{2}}{∂λ^{2}} log L = −\frac{n_{u}}{λ^{2}}
I(λ) = \frac{n_{u}}{λ^{2}}

So

\frac{λ̂ − λ}{\sqrt{\frac{λ^{2}}{n_{u}}}} ⟶^{d} N(0, 1)

Note that the normality approximation can be improved by transforming the estimate. By the δ-method,

λ̂ ⟶^{d} N (λ, \frac{λ^{2}}{n_{u}} )

then

log λ̂ ⟶^{d} N (log λ, \frac{1}{n_{u}} )

Notice that 1/n_{u}, the asymptotic variance of log λ̂, does not depend on the unknown parameter λ. It is an empirical fact that transforming an estimate to remove the dependence of the variance on the unknown parameter tends to improve the convergence to normality by reducing the skewness.

log λ̂ ⟶^{d} N (log λ, \frac{1}{n_{u}} )

The 95% CI becomes

e^{log λ̂±z_{α}/\sqrt{n_{u}}}
(λ̂e^{−z_{α}/\sqrt{n_{u}}}, λ̂e^{z_{α}/\sqrt{n_{u}}})
(0.031, 0.136)

b) { \atop S(16)} = e^{−λ̂(16)}

95% CI for S(16) is

\begin{matrix}(e^{−0.136(16)}, e^{−0.031(16)}) \\ (0.113, 0.609) \end{matrix}

c) t̂_{med} = log 2/λ̂ = 10.69

95% CI for the median is

\begin{matrix}(\frac{log 2}{0.136}, \frac{log 2}{0.031}) \\ (5.097, 22.36) \end{matrix}

Reference:

Miller Jr., Rupert G., Survival Analysis, Wiley, 1981.