Kernel based estimation of the distribution function for length biased data

Abstract

Empirical and kernel estimators are considered for the distribution of positive length biased data. Their asymptotic bias, variance and limiting distribution are obtained. For the kernel estimator, the asymptotically optimal bandwidth is calculated and rule of thumb bandwidths are proposed. At any point below the median, the asymptotic mean squared error of the kernel estimator is smaller than that of the empirical estimator. A suitably truncated kernel estimator is positive and we prove the strong uniform, and $L_2$ consistency of this estimator. Simulations reveal the improved performance of the truncated kernel estimator in estimating tail probabilities based on length biased data.

Appendix

Proof

We can write

$$\begin{aligned} \frac{N}{D}=\frac{N}{E(D)(1-r)},\quad \text {where}\ 1-r=\frac{D}{E(D)}. \end{aligned}$$

Under the stated assumptions $P(D=E(D))=0$, and therefore $P(r\ne 1)=1$. We know that for $r\ne 1$,

$$\begin{aligned} \frac{1}{1-r}=1+r+\frac{r^2}{(1-r)}.\end{aligned}$$

(6.1)

It is easy to verify that

$$\begin{aligned} E\left[ \left( 1+r+\frac{r^2}{1-r}\right) \frac{N}{E(D)}\right]= & {} \frac{E(N)}{E(D)}-\frac{1}{E^2(D)}Cov(N,D)\nonumber \\&+\frac{1}{E^2(D)}E\left[ \frac{N}{D}(D-E(D))^2\right] .\end{aligned}$$

(6.2)

Multiplying both sides of (6.1) by $\frac{N}{E(D)}$, taking expectation and using (6.2) we get the stated inequality. $\square$

Proof of Lemma 2

To prove Lemma 2 it is enough to see that

$$\begin{aligned}&E\left[ \frac{1}{Y^2_1}I(Y_1\le y)\right] =\frac{1}{\mu }\int ^y_0 \frac{f(z)}{z}dz,\ \ E\left[ \frac{1}{Y_1}I(Y_1\le y)\right] \nonumber \\&\quad =\frac{1}{\mu }\int ^y_0f(z)dz=\frac{F(y)}{\mu }. \end{aligned}$$

(6.3)

$$\begin{aligned} \text {and}&\ \ E\left[ \frac{1}{Y^2_1}I(Y_1\le y)\right] =\frac{1}{\mu }\int ^y_0 \frac{f(z)}{z}dz.\end{aligned}$$

(6.4)

Using (6.3) and (6.4), it is straight forward to verify the expressions of $Var\left[ \frac{1}{Y_1}I(Y_1\le y)\right] ,\ Cov\left[ \frac{1}{Y_1}I(Y_1\le y),\frac{1}{Y_1}\right]$ in Lemma 2. $\square$

Proof of Lemma 3

$$\begin{aligned}&E\left[ \frac{1}{Y^2_1}K^2\left( \frac{y-Y_1}{h}\right) \right] \nonumber \\&\quad = \frac{1}{\mu }\int ^\infty _0\nonumber K^2\left( \frac{y-z}{h}\right) \frac{f(z)}{z}dz=\frac{h}{\mu }\int ^{\frac{y}{h}}_{-\infty }K^2(v)\frac{f(y-vh)}{y-vh}dv\nonumber \\&\quad =\frac{h}{y\mu }\int ^{\frac{y}{h}}_{-\infty }K^2(v)f\left( y-vh\right) dv+ \frac{h^2}{\mu }\int ^{y/h}_{-\infty }vK^2(v)\frac{f(y-vh)}{y(y-vh)}dv\nonumber \\&\quad = \frac{h}{y\mu }\int ^{\frac{y}{h}}_{-\infty }K^2(v)f\left( y-vh\right) dv+O\left( h^2\right) \ (\text {Assumptions (A1)-(A3)})\nonumber \\&\quad =-\frac{1}{y\mu }\int ^{\frac{y}{h}}_{-\infty }K^2(v)dF\left( y-vh\right) +O\left( h^2\right) .\end{aligned}$$

(6.5)

Using integration by parts we get

$$\begin{aligned}&\int ^{\frac{y}{h}}_{-\infty }K^2(v)dF\left( y-vh\right) +\int ^{\frac{y}{h}}_{-\infty }F(y-hv)dK^2(v)\nonumber \\&\quad = F(0)K^2(y/h)-F(\infty )K^2(-\infty ).\nonumber \\&\quad \Rightarrow -\int ^{\frac{y}{h}}_{-\infty }K^2(v)dF\left( y-vh\right) \\&\quad = \int ^{\frac{y}{h}}_{-\infty }F(y-hv)dK^2(v)\ (\text {as}\ K(-\infty )=0 \ \text {and}\ F(0)=0) \end{aligned}$$

Therefore, as $n\rightarrow \infty$

$$\begin{aligned} -\int ^{\frac{y}{h}}_{-\infty }K^2(v)dF\left( y-vh\right)= & {} F(y)K^2\left( \frac{y}{h}\right) -hf(y)\int ^{\frac{y}{h}}_{-\infty }vdK^2(v) + O(h^2),\nonumber \\= & {} F(y)-hf(y)\int ^1_{-1}vdK^2(v)+ O(h^2) (\text {as}\ K(1)=1).\nonumber \\ \end{aligned}$$

(6.6)

From (6.5) and (6.6) we get

$$\begin{aligned} E\left[ \frac{1}{Y^2_1}K^2\left( \frac{y-Y_1}{h}\right) \right] =\frac{1}{y\mu }[F(y)-hf(y)\int ^1_{-1}vdK^2(v)]+O\left( h^2\right) \ \text {as}\ n\rightarrow \infty .\end{aligned}$$

(6.7)

Further under Assumptions (A1) to (A3) we get

$$\begin{aligned} E\left[ \frac{1}{Y_1}K\left( \frac{y-Y_1}{h}\right) \right] =-\frac{1}{\mu }\int ^{\frac{y}{h}}_{-\infty }K(v)dF(y-vh)=\frac{F(y)}{\mu }+O\left( h^2\right) \text {as}\ n\rightarrow \infty .\end{aligned}$$

(6.8)

Equations (6.7) and (6.8) imply that

$$\begin{aligned} Var(N)= & {} \frac{1}{n}Var\left[ \frac{1}{Y_1}K\left( \frac{y-Y_1}{h}\right) \right] \\= & {} \frac{1}{n\mu }\left[ \frac{F(y)}{y}-\frac{hf(y)}{y}\int ^1_{-1}vdK^2(v)-\frac{F^2(y)}{\mu }\right] \\&+O\left( \frac{h^2}{n}\right) \text {as}\ n\rightarrow \infty .\end{aligned}$$

$\square$

Using similar arguments we get

$$\begin{aligned} E\left[ \frac{1}{Y^2_1}K\left( \frac{y-Y_1}{h}\right) \right]= & {} \frac{1}{y\mu }[F(y)-hf(y)\int ^1_{-1}vdK(v)]+O\left( h^2\right) \nonumber \\= & {} \frac{F(y)}{y\mu }+O\left( h^2\right) \quad \text {as}\ n\rightarrow \infty .\end{aligned}$$

(6.9)

Equations (6.8) and (6.9) imply that

$$\begin{aligned} Cov(N,D)= & {} \frac{1}{n}Cov\left( \frac{1}{Y_1}K\left( \frac{y-Y_1}{h}\right) ,\frac{1}{Y_1}\right) \\= & {} \frac{F(y)}{n\mu }\left[ \frac{1}{y}-\frac{1}{\mu }\right] +O\left( \frac{h^2}{n}\right) \ \text {as}\ n\rightarrow \infty . \end{aligned}$$

$\square$

Proof of Lemma 4

$$\begin{aligned} \hat{\sigma }^2_E= & {} \frac{1}{\frac{1}{n}\left( \sum ^n_{i=1}\frac{1}{Y_i}\right) ^2} \sum ^n_{i=1}\frac{1}{Y^2_i}\left[ I(Y_i\le y)-\hat{F}_n(y)\right] ^2\\= & {} \frac{1}{\left( \frac{1}{n}\sum ^n_{i=1}\frac{1}{Y_i}\right) ^2}\left[ \frac{1}{n}\sum ^n_{i=1}\frac{I(Y_i\le y)}{Y^2_i}+[\hat{F}_n(y)]^2\frac{1}{n}\sum ^n_{i=1}\frac{1}{Y^2_i} -2\hat{F}_n(y)\sum ^n_{i=1}\frac{I(Y_i\le y)}{Y^2_i}\right] \end{aligned}$$

As $n\rightarrow \infty$, under Assumption (A2),

$$\begin{aligned}&\frac{1}{n}\sum ^n_{i=1}\frac{1}{Y_i}\rightarrow E\left( \frac{1}{Y_1}\right) =\frac{1}{\mu },\ \frac{1}{n}\sum ^n_{i=1}\frac{I(Y_i\le y)}{Y^2_i}\rightarrow \frac{1}{\mu }\int ^y_0 \frac{f(z)}{z}dz, \ \text {almost surely} \end{aligned}$$

(6.10)

$$\begin{aligned}&\frac{1}{n}\sum ^n_{i=1}\frac{1}{Y^2_i}\rightarrow \frac{1}{\mu }\int ^\infty _0 \frac{f(z)}{z}dz\ \text {and}\ \hat{F}_n(y)\rightarrow F(y),\ \text {almost surely}. \end{aligned}$$

(6.11)

Therefore under Assumption (A2), from (6.10) and (6.11) we see that as $n\rightarrow \infty$,

$$\begin{aligned} \hat{\sigma }^2_E \rightarrow&\mu \left[ \int ^y_0 \frac{f(z)}{z}dz + [F(y)]^2\int ^\infty _0 \frac{f(z)}{z}dz-2F(y)\int ^y_0 \frac{f(z)}{z}dz\right] , \text {almost surely}\\&= \mu \int ^\infty _0 \frac{f(z)}{z}\left[ I(z\le y)-F(y)\right] ^2= {\sigma }^2_E.\end{aligned}$$

Hence, $\frac{\hat{\sigma }^2_E}{{\sigma }^2_E}\rightarrow 1$, almost surely, as $n\rightarrow \infty$.

Let $\hat{f}_n(y)$ be a strongly consistent estimator of f(y). Using (6.10) and (6.11), we see that under the stated conditions (A1) to (A3) as $n\rightarrow \infty$

$$\begin{aligned} \hat{\sigma }^2_K= & {} \hat{\mu }\left[ \frac{\hat{F}_n(y)}{y}+\hat{F}^2_n(y)\frac{\hat{\mu }}{n}\sum ^n_{i=1}\frac{1}{Y^2_i}-\frac{2\hat{F}^2_n(y)}{y} -\frac{h\hat{f}_n(y)}{y}\int ^1_{-1}vdK^2(v)\right] \\\rightarrow & {} \mu \left[ \frac{F(y)}{y}+F^2(y)\int ^\infty _0 \frac{f(z)}{z}dz-\frac{2F^2(y)}{y}-\frac{hf(y)}{y}\int ^1_{-1}vdK^2(v)\right] ,\ \text {almost surely}.\end{aligned}$$

From Theorem (3) we recall that as $n\rightarrow \infty$

$$\begin{aligned} \sigma ^2_K=\mu \left[ \frac{F(y)}{y}+F^2(y)\int ^\infty _0 \frac{f(z)}{z}dz-\frac{2F^2(y)}{y}-\frac{hf(y)}{y}\int ^1_{-1}vdK^2(v)\right] +O(h^2). \end{aligned}$$

Therefore under the stated assumptions $\frac{\hat{\sigma }^2_K}{\sigma ^2_K}\rightarrow 1,$ almost surely, as $n\rightarrow \infty$. $\square$

Proof of Lemma 5

The asmyptotic mean squared error AMSE(h) of $\hat{F}_h(y)$ is sum of the asymptotic variance and square of the asymptotic bias. The expressions for the asymptotic bias and variance are obtained from the leading terms in Theorem 1 (ii) and $\sigma ^2_K$ in Theorem 3.

Proof of Lemma 7

Since $||\hat{F}_{C,\ \hat{h}}-F||\le ||\hat{F}_{C,\ \hat{h}}-\hat{F}_{\hat{h}}||+||\hat{F}_{\hat{h}}-F||$ and under the stated conditions $||\hat{F}_{\hat{h}}-F||\rightarrow 0$, almost surely, as $n\rightarrow \infty$ (see Lemma 6), therefore it is enough to prove $||\hat{F}_{C,\ \hat{h}}-\hat{F}_{\hat{h}}||\rightarrow 0$, almost surely, as $n\rightarrow \infty$.

$$\begin{aligned} \hat{F}_{C,\ \hat{h}}(y)-\hat{F}_{\hat{h}}(y)={\left\{ \begin{array}{ll}-\hat{F}_{\hat{h}}(0)\frac{(1-\hat{F}_{\hat{h}}(y))}{1-\hat{F}_{\hat{h}}(0)},\ y>0\\ -\hat{F}_{\hat{h}}(y),\ y\le 0.\end{array}\right. }\end{aligned}$$

Therefore

$$\begin{aligned} ||\hat{F}_{C,\ \hat{h}}-\hat{F}_{\hat{h}}||\le \hat{F}_{\hat{h}}(0). \end{aligned}$$

But, $\hat{F}_{\hat{h}}(0)\rightarrow F(0)=0$ almost surely, as $n\rightarrow \infty$ (see Lemma 6). Consequently, $||\hat{F}_{C,\ \hat{h}}-\hat{F}_{\hat{h}}||\rightarrow 0$, almost surely, as $n\rightarrow \infty$. This completes the proof of the first part.

Since $\hat{F}_{C,\ \hat{h}}$ and F are both dfs, $||\hat{F}_{\hat{h}}-F||\le 1$ almost surely. Therefore using the almost sure convergence of $||\hat{F}_{C,\ \hat{h}}-F||$ and the dominated convergence theorem we see that $E||\hat{F}_{C,\ \hat{h}}-F||^2\rightarrow 0$ as $n\rightarrow \infty$. This completes the proof. $\square$