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New kernel estimators of the hazard ratio and their asymptotic properties

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Abstract

We propose a kernel estimator of a hazard ratio that is based on a modification of Ćwik and Mielniczuk (Commun Stat-Theory Methods 18(8):3057–3069, 1989)’s method. A naive nonparametric estimator is Watson and Leadbetter (Sankhyā: Indian J Stat Ser A 26(1):101–116, 1964)’s one, which is naturally given by the kernel density estimator and the empirical distribution estimator. We compare the asymptotic mean squared error (AMSE) of the hazard estimators, and then, it is shown that the asymptotic variance of the new estimator is usually smaller than that of the naive one. We also discuss bias reduction of the proposed estimator and derived some modified estimators. While the modified estimators do not lose nonnegativity, their AMSE is small both theoretically and numerically.

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References

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Acknowledgements

The authors would like to appreciate the editor’s and referees’ valuable comments that helped us to improve this manuscript significantly. The authors gratefully acknowledge JSPS KAKENHI Grant Number JP15K11995 and JP16H02790.

Author information

Correspondence to Taku Moriyama.

Appendices: Proofs of Theorems

Appendices: Proofs of Theorems

Proof of Theorem 1

For simplicity, we will use the following notation,

$$\begin{aligned}&\eta (z) = \int _{-\infty }^z F(u) \text {d}u ,\quad \gamma (z) = \int _{-\infty }^z \eta (u) \text {d}u, \\&M(z) = z -\eta (z) \quad \mathrm{and}\quad m(z) =M'(z) =1-F(z). \end{aligned}$$

To begin with, we consider the following stochastic expansion of the direct estimator:

$$\begin{aligned}&\widehat{H}(x_0) \\&\quad =\frac{1}{h} \int K\left( \frac{M(w) -M(x_0) }{h} \right) \text {d}F_n(w) \\&\qquad +\, \frac{1}{h^2} \int K'\left( \frac{M(w) -M(x_0) }{h} \right) \bigg \{[\eta (w) -\eta _n(w)] -[\eta (x_0) -\eta _n(x_0)] \bigg \} \text {d}F_n(w) \\&\qquad +\, \frac{1}{h^3} \int K''\left( \frac{M(w) -M(x_0) }{h} \right) \bigg \{[\eta (w) -\eta _n(w)] -[\eta (x_0) -\eta _n(x_0)] \bigg \}^2 \text {d}F_n(w) \\&\qquad +\, \frac{1}{h^4} \int K^{(3)}\left( \frac{M_n^*(w) -M(x_0) }{h} \right) \bigg \{[\eta (w) -\eta _n(w)] -[\eta (x_0) -\eta _n(x_0)] \bigg \}^3 \text {d}F_n(w) \\&\quad = J_1 + J_2 + J_3 + J_4^* \quad (\mathrm{say}), \end{aligned}$$

where

$$\begin{aligned} \eta _n (w)= \int _{-\infty }^w F_n(u) \text {d}u = \frac{1}{n}\sum _{i=1}^n (w- X_i)_+ \end{aligned}$$

and \(M_n^*(w)\) is a r.v. between \(M_n(w)\) and M(w) with probability 1.

The main term of the expectation of \(\widehat{H}(x_0)\) is given by \(J_1\), as we will show the later. Since \(J_1\) is a sum of i.i.d. random variables, the expectation can be obtained directly:

$$\begin{aligned} E[J_1]= & {} E \left[ \frac{1}{h} \int K\left( \frac{M(w) -M(x_0) }{h} \right) \text {d}F_n(w) \right] \\= & {} \frac{1}{h} \int K\left( \frac{M(w) -M(x_0) }{h} \right) f(w) \text {d}w \\= & {} \int K(u) H(M^{-1} (M(x_0) +h u)) \text {d}u \\= & {} H(x_0) +\frac{h^2}{2} A_{1,2} \left[ \frac{(1 -F)\{(1-F)f'' +4f f' \} +3f^3}{(1 -F)^5}\right] (x_0) + O(h^4). \end{aligned}$$

Combining the following second moment,

$$\begin{aligned}&\frac{1}{n h^2} \int K^2\left( \frac{M(w) - M(x_0) }{h} \right) f(w) \text {d}w \\&\quad = \frac{1}{n h}\int K^2(u) H(M^{-1} (M(x_0) +h u)) \text {d}u \\&\quad = \frac{A_{2,0}}{n h} H(x_0) +O(n^{-1}), \end{aligned}$$

we get the variance,

$$\begin{aligned} V[J_1] = \frac{1}{n h} H(x_0) A_{2,0} +O(n^{-1}). \end{aligned}$$

Next, we consider the following representation of \(J_2\)

$$\begin{aligned} J_2 = \frac{1}{n^2 h^2} \sum _{i=1}^n \sum _{j=1}^n K'\left( \frac{M(X_i) -M(x_0) }{h} \right) Q(X_i,X_j), \end{aligned}$$

where

$$\begin{aligned} Q(x_i,x_j) =[\eta (x_i) -(x_i -x_j)_{+}] -[\eta (x_0) -(x_0 -x_j)_{+}]. \end{aligned}$$

Using the conditional expectation, we get the following equation:

$$\begin{aligned} E[J_2]= & {} \frac{1}{n h^2} \sum _{j=1}^n E\left[ K'\left( \frac{M(X_i) -M(x_0) }{h} \right) Q(X_i,X_j) \right] \\= & {} \frac{1}{n h^2} E\left[ K'\left( \frac{M(X_i) -M(x_0) }{h} \right) E\left[ \sum _{j=1}^n Q(X_i,X_j) \Bigm | X_i \right] \right] \\= & {} \frac{1}{n h^2} E\left[ K'\left( \frac{M(X_i) -M(x_0) }{h} \right) \left\{ \eta (X_i) -\left[ \eta (x_0) -(x_0-X_i)_{+} \right] \right\} \right] \\= & {} \frac{1}{n h} \int K'\left( u \right) \left\{ \eta (M^{-1} (M(x_0) +h u)) -\eta (x_0) +(x_0-M^{-1} (M(x_0) +h u))_{+} \right\} \\&\quad \times \, H(M^{-1} (M(x_0) +h u)) \text {d}u\\= & {} \frac{1}{n h} \int K'\left( u \right) O(h u) H(x_0) \text {d}u = O\left( \frac{1}{n}\right) . \end{aligned}$$

Next, we have

$$\begin{aligned} J_2^2= & {} \frac{1}{n^4 h^4} \sum _{i=1}^n \sum _{j=1}^n \sum _{s=1}^n \sum _{t=1}^n K'\left( \frac{M(X_i) -M(x_0) }{h} \right) K'\left( \frac{M(X_s) -M(x_0) }{h} \right) \\&\times \, Q(X_i,X_j) Q(X_s,X_t) \\= & {} \frac{1}{n^4 h^4} \sum _{i=1}^n \sum _{j=1}^n \sum _{s=1}^n \sum _{t=1}^n \varXi (i,j,s,t)\quad (\mathrm{say}). \end{aligned}$$

After taking the conditional expectation, we find that if all of the (ijst) are different,

$$\begin{aligned} E[\varXi (i,j,s,t)] = E\left[ E\left\{ \varXi (i,j,s,t) | X_i,X_s\right\} \right] = 0, \end{aligned}$$

and

$$\begin{aligned}&E[\varXi (i,j,s,t)]=0 \quad (\text {if }i=j\text { and all of }(i,s,t)\text { are different}),\\&E[\varXi (i,j,s,t)]=0 \quad (\text {if }i=s\text { and all of }(i,j,t)\text { are different}),\\&E[\varXi (i,j,s,t)]=0 \quad (\text {if }i=t\text { and all of }(i,j,s)\text { are different}), \end{aligned}$$

the term in which \(j=t\) and all of the (ijs) are different is the main term of \(E[J_2^2]\). If \(j=t\) and all of the (ijs) are different, we have

$$\begin{aligned}&E[\varXi (i,j,s,t)] \\&\quad = \frac{n(n-1)(n-2)}{n^4 h^4} E\biggl [ K'\left( \frac{M(X_i) -M(x_0) }{h} \right) K'\left( \frac{M(X_s) -M(x_0) }{h} \right) \\&\qquad \times \, Q(X_i,X_j) Q(X_s,X_j) \biggl ]. \end{aligned}$$

Using the conditional expectation of \(Q(X_i,X_j) Q(X_s,X_j)\) given \(X_i\) and \(X_s\), we find that

$$\begin{aligned}&E\Bigl [ E\Bigl \{ Q(X_i,X_j) Q(X_s,X_j) \Bigm | X_i, X_s\Bigl \} \Bigl ] \\&\quad =E\Bigl [ \eta (X_i)\eta (x_0) +\eta (X_s)\eta (x_0) -\eta ^2(x_0) +2\gamma (x_0) -\eta (X_i)\eta (X_s) \\&\qquad -\,(x +X_i -2\min (x ,X_i))\eta (\min (x ,X_i)) -2\gamma (\min (x ,X_i)) \\&\qquad -\,(x +X_s -2\min (x ,X_s))\eta (\min (x ,X_s)) -2\gamma (\min (x ,X_s)) \\&\qquad +\,(X_i +X_s -2\min (X_i ,X_s))\eta (\min (X_i ,X_s)) +2\gamma (\min (X_i ,X_s)) \Bigl ]. \end{aligned}$$

Therefore, the entire expectation of the last row is

$$\begin{aligned}&E\biggl [ K'\left( \frac{M(X_i) -M(x_0) }{h} \right) K'\left( \frac{M(X_s) -M(x_0) }{h} \right) \\&\qquad \times \, (X_i +X_s -2\min (X_i ,X_s))\eta (\min (X_i ,X_s)) +2\gamma (\min (X_i ,X_s)) \Bigl ]\\&\quad =\int \Biggl [ \int _{\infty }^w K'\left( \frac{M(z) -M(x_0) }{h} \right) K'\left( \frac{M(w) -M(x_0) }{h} \right) \\&\qquad \times \, \bigg \{ (-z+w)\eta (z) + 2\gamma (z) \bigg \} f(w) \text {d}z \\&\qquad +\int _{w}^{\infty } K'\left( \frac{M(z) -M(x_0) }{h} \right) K'\left( \frac{M(w) -M(x_0) }{h}\right) \\&\qquad \times \,\left\{ (z-w)\eta (w) + 2\gamma (w) \right\} f(z) dz \Biggr ] f(w) \text {d}w. \\ \end{aligned}$$

Finally, we get

$$\begin{aligned}&E\biggl [ K'\left( \frac{M(X_i) -M(x_0) }{h} \right) K'\left( \frac{M(X_s) -M(x_0) }{h} \right) \\&\qquad \times \, \{X_i +X_s -2\min (X_i ,X_s)\}\eta (\min (X_i ,X_s)) +2\gamma (\min (X_i ,X_s)) \Bigl ]\\&\quad = -h^2 \int K'\left( \frac{M(w) -M(x_0) }{h} \right) f(w) \text {d}w\\&\qquad \times \, \Biggl ( \Biggm [W\left( \frac{M(w) -M(x_0) }{h}\right) \biggl ( \left\{ \eta (x_0) +(-x_0 +w)F(x_0) \right\} \left[ \frac{f}{m^2}\right] (x_0) \\&\qquad +\,\{(-x_0 +w) \eta (x_0) +2\gamma (x_0)\} \left[ \frac{f'm -fm'}{m^3}\right] (x_0) \biggl ) +O(h)\Biggm ] \\&\qquad +\,\Biggm [\left( 1 -W\left( \frac{M(w) -M(x_0) }{h} \right) \right) \\&\qquad \times \,\left( \eta (w) \frac{f}{m^2}(x_0) +\{(x_0-w) \eta (w)+ 2\gamma (w)\}\left[ \frac{f'm -fm'}{m^3}\right] (x_0) \right) +O(h) \Biggm ] \Biggl )\\&\quad = h^4 \left[ \frac{Ff^2}{m^4}(x_0) +\left\{ \frac{f'm -fm'}{m^3} \left( 2\eta \frac{f}{m^2} +2\gamma \frac{f'm -fm'}{m^3} \right) \right] (x_0) \right\} +O(h^5). \end{aligned}$$

After similar calculations of the other terms, we find that if \(j=t\) and all of the (ijs) are different,

$$\begin{aligned} E[\varXi (i,j,s,t)] = O\left( \frac{1}{n}\right) . \end{aligned}$$

In addition, it is easy to see that the expectations of the other combinations of (ijst) are \(o(n^{-1})\). As a result, we have

$$\begin{aligned} E[J_2^2] = O(n^{-1}) \quad \text {and}\quad V[J_2] = O(n^{-1}). \end{aligned}$$

The moments of \(J_3\) can be obtained in a similar manner; we find that

$$\begin{aligned} E[J_3] = O(n^{-1}) \quad \text {and}\quad V[J_3] = O(n^{-2}). \end{aligned}$$

Finally, we will obtain upper bounds of \(|E[J_4^*]|\) and \(E[{(J_4^*)}^2]\). From the assumption of Theorem 1, we can see

$$\begin{aligned}&\left| E \left[ \frac{1}{h^4} K^{(3)}\left( \frac{M_n^*(X_i) -M(x_0) }{h} \right) \{[\eta (X_i) -\eta _n(X_i)] -[\eta (x_0) -\eta _n(x_0)] \}^3 \right] \right| \\&\quad \le \frac{\max _{u} |K^{(3)}(u)|}{h^4} E \left[ \{[\eta (X_i) -\eta _n(X_i)] -[\eta (x_0) -\eta _n(x_0)] \}^3 \right] \\&\quad = O\left( \frac{1}{n^2 h^4}\right) . \end{aligned}$$

Similarly, it follows that

$$\begin{aligned}&E[{(J_4^*)}^2] \\&\quad = \frac{1}{h^8} E \left[ \left( \int K^{(3)}\left( \frac{M_n^*(w) -M(x_0) }{h} \right) \{[\eta (w) -\eta _n(w)] -[\eta (x_0) -\eta _n(x_0)] \}^3 \text {d}F_n(w) \right) ^2\right] \\&\quad \le \frac{\max _{u} (K^{(3)}(u))^2}{n^2 h^8} \sum _{i=1}^n \sum _{j=1}^n E \Bigm [\{[\eta (X_i) -\eta _n(X_i)] -[\eta (x_0) -\eta _n(x_0)] \}^3 \\&\qquad \times \, \bigg \{[\eta (X_j) -\eta _n(X_j)] -[\eta (x_0) -\eta _n(x_0)] \bigg \}^3 \Bigm ] \\&\quad = O\left( \frac{1}{n^4 h^8}\right) . \end{aligned}$$

To sum up, we conclude that \(J_2 + J_3 + J_4^*\) is asymptotically negligible for fixed \(x_0\). The main bias of \(\widehat{H}(x_0)\) comes from \(J_1\). From the Cauchy–Schwarz inequality, we find that the main term of the variance coincides with \(V[J_1]\). Now, we can get the AMSE of the direct estimator and prove Theorem 1. \(\square \)

Proof of Theorems 2 and 3

It follows from the above discussion that

$$\begin{aligned}&\sqrt{n h} \left\{ \widehat{H}(x_0) -H(x_0) \right\} \\&\quad = \sqrt{n h} \big \{ J_1 -H(x_0) \big \} +o_P(1) \\&\quad = (\sqrt{n h} h^2) B_1 + \sqrt{n h} \left\{ J_1 -H(x_0)-h^2 B_1 \right\} +o_P(1). \end{aligned}$$

Since \(J_1\) is a sum of i.i.d. random variables and the expectation of the second term is o(1), asymptotic normality holds for Theorem 2. \(\square \)

Furthermore, we can show that

$$\begin{aligned} E \left[ \widehat{H}(x_0) \right]= & {} E[J_1] +O(n^{-1}) \\= & {} \int K(u) H(M^{-1} (M(x_0) +h u)) \text {d}u +O(n^{-1}), \end{aligned}$$

and we can directly get Theorem 3 by taking a Taylor expansion. \(\square \)

Proof of Theorem 4

We follow the proof of Jones et al. (1995). \(\widehat{H}_{N}(x_0)\) is written as

$$\begin{aligned} \widehat{H}_{N}(x_0) = \widehat{H}(x_0)\widehat{\alpha }(x_0) = H(x_0)\left\{ 1+\frac{\widehat{H}(x_0) -H(x_0)}{H(x_0)} \right\} \{1+(\widehat{\alpha }(x_0)-1)\}. \end{aligned}$$

It follows from the asymptotic expansion that

$$\begin{aligned} \widehat{\alpha }(x_0)\approx & {} \frac{1}{n} \sum _{i=1}^n \frac{1}{h{H}(X_i)} K\left( \frac{M_n(x_0) - M_n(X_i) }{h} \right) \\&\times \,\left[ 1 - \frac{\widehat{H}(X_i) -H(X_i)}{H(X_i)} + \left\{ \frac{\widehat{H}(X_i) -H(X_i)}{H(X_i)} \right\} ^2 \right] . \end{aligned}$$

By taking the expectation of the ith term in this sum conditional on \(X_i\), we have

$$\begin{aligned}&E\Biggm [\frac{1}{h{H}(X_i)} K\left( \frac{M_n(x_0) - M_n(X_i) }{h} \right) \\&\qquad \times \, \left[ 1 - \frac{\widehat{H}(X_i) -H(X_i)}{H(X_i)} + \left\{ \frac{\widehat{H}(X_i) -H(X_i)}{H(X_i)} \right\} ^2 \right] \Biggm | X_i \Biggm ]\\&\quad =\frac{1}{h{H}(X_i)} K\left( \frac{M(x_0) - M(X_i) }{h} \right) \left[ 1 - \frac{h^2 B_1(X_i) + h^4 B_2(X_i)}{H(X_i)} + \left( \frac{h^2 B_1(X_i)}{H(X_i)}\right) ^2\right] \\&\qquad +\, O_P((nh)^{-1}) + o_P(h^4). \end{aligned}$$

Thus, we have

$$\begin{aligned} E[\widehat{\alpha }(x_0)]= & {} 1 - h^2 \frac{B_1}{H}(M^{-1}(M(x_0)+ hu)) \\&+\, h^4 \biggm [ -\,\frac{B_2(x_0)}{H(x_0)} + \left( \frac{B_1(x_0)}{H(x_0)}\right) ^2 \biggm ] + O((nh)^{-1}) + o(h^4) \\= & {} 1 - h^2 {C_1}(x_0) \\&+\, h^4 \biggm [ -\,\frac{B_2(x_0)}{H(x_0)} + \left( \frac{B_1(x_0)}{H(x_0)}\right) ^2 + \frac{C_1''}{2 m^3}(x_0) - \frac{C_1 m''}{2 m^4}(x_0) \\&-\,3\frac{C_1'm'}{2m^4}(x_0) +3\frac{C_1(m')^2}{2m^5}(x_0) \biggm ] +\, O((nh)^{-1}) + o(h^4). \end{aligned}$$

It follows that

$$\begin{aligned}&E[\widehat{H}_{N}(x_0) -H(x_0)] \\&\quad = h^2 B_1(x_0) + h^4 [ B_2(x_0) - B_1(x_0) C_1(x_0)] - h^2 H(x_0) {C_1}(x_0) \\&\qquad +\, h^4 H(x_0) \biggm [ -\frac{B_2(x_0)}{H(x_0)} + \left( \frac{B_1(x_0)}{H(x_0)}\right) ^2 + \frac{C_1''}{2 m^3}(x_0) - \frac{C_1 m''}{2 m^4}(x_0) \\&\qquad -\,3\frac{C_1'm'}{2m^4}(x_0) +3\frac{C_1(m')^2}{2m^5}(x_0) \biggm ] + O((nh)^{-1}) + o(h^4) \\&\quad = h^4 H(x_0) \biggm [ \frac{C_1''}{2 m^3}(x_0) - \frac{C_1 m''}{2 m^4}(x_0) -3\frac{C_1'm'}{2m^4}(x_0) +3\frac{C_1(m')^2}{2m^5}(x_0) \biggm ] \\&\qquad +\, O((nh)^{-1}) + o(h^4). \end{aligned}$$

From the proof of Theorem 1, we can see that the following approximation holds:

$$\begin{aligned} \widehat{H}_{N}(x_0)= & {} \frac{1}{n^2h}\sum _{i=1}^n \sum _{j=1}^n K\left( \frac{M_n(x_0) - M_n(X_i) }{h} \right) \frac{K\left( \frac{M_n(x_0) - M_n(X_j) }{h} \right) }{\sum _{\ell =1}^n K\left( \frac{M_n(X_j) - M_n(X_{\ell }) }{h} \right) } \\\approx & {} \frac{1}{n^2h} \sum _{i=1}^n \sum _{j=1}^n K\left( \frac{M(x_0) - M(X_i) }{h} \right) \frac{K\left( \frac{M(x_0) - M(X_j) }{h} \right) }{\sum _{\ell =1}^n K\left( \frac{M(X_j) - M(X_{\ell }) }{h} \right) } \\=: & {} H_N(x_0) \end{aligned}$$

\(H_N(x_0)\) can be seen as a Jones et al. (1995)’s density estimate for the random variable M(X) at the point \(M(x_0)\). Then, the asymptotic variance is given by

$$\begin{aligned} V[\widehat{H}_{N}(x_0)] \approx \frac{H(x_0)}{nh} \int \bigg \{ 2K(u) - K*K(u) \bigg \}^2 \text {d}u \end{aligned}$$

(see Jones et al. 1995). \(\square \)

Proof of Theorem 5

To obtain the asymptotic variance of the modified estimator \(\widehat{H}^{\dagger }\), we need to calculate the covariance term \(Cov[\widehat{H}_h(x_0), \widehat{H}_{2h}(x_0)]\) as shown in Sect. 4. From the proof of Theorem 1, it is easy to see

$$\begin{aligned}&Cov \left[ \widehat{H}_h(x_0), \widehat{H}_{2h}(x_0) \right] \\&\quad = E\left[ \widehat{H}_h(x_0)\widehat{H}_{2h}(x_0) \right] - E\left[ \widehat{H}_h(x_0) \right] E\left[ \widehat{H}_{2h}(x_0) \right] \\&\quad = \frac{1}{2n^2h^2} E\left[ \sum _{i=1}^{n} \sum _{j=1}^{n} K\left( \frac{M(X_i) -M(x_0)}{h}\right) K\left( \frac{M(X_j) -M(x_0)}{2h}\right) \right] \\&\qquad - \left\{ H^2(x_0) + 5 h^2 H(x_0) B_1(x_0) + O(h^4 + n^{-1})\right\} + O\left( \frac{1}{n h^{1/2}} \right) . \end{aligned}$$

Consequently, we have

$$\begin{aligned}&Cov \left[ \widehat{H}_h(x_0), \widehat{H}_{2h}(x_0) \right] \\&\quad = \frac{1}{2n h^2} E\left[ K\left( \frac{M(X_1) -M(x_0)}{h}\right) K\left( \frac{M(X_1) -M(x_0)}{2h}\right) \right] \\&\qquad -\,\frac{1}{n} \left\{ H^2(x_0) + 5 h^2 H(x_0) B_1(x_0) + O(h^4 + n^{-1})\right\} + O\left( \frac{1}{n h^{1/2}} \right) \\&\quad = \frac{1}{nh} H(x_0) \int K(2u) K(u) \text {d}u + O\left( \frac{1}{n h^{1/2}} \right) \end{aligned}$$

and the desired result. \(\square \)

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Moriyama, T., Maesono, Y. New kernel estimators of the hazard ratio and their asymptotic properties. Ann Inst Stat Math 72, 187–211 (2020). https://doi.org/10.1007/s10463-018-0685-6

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Keywords

  • Kernel estimator
  • Hazard ratio
  • Nonparametric estimator
  • Mean squared error