Goodness of fit test for Rayleigh distribution with censored observations

Abstract

We develop new goodness of fit tests for Rayleigh distribution based on fixed point characterization. We use U-Statistic theory to derive the test statistics. First we develop a test for complete data and then discuss, how the right censored observations can be incorporated in the testing procedure. The asymptotic properties of the test statistic in both uncensored and censored cases are studied in detail. Extensive Monte Carlo simulation studies are carried out to validate the performance of the proposed tests. We illustrate the procedures using real data sets. We also provide, a goodness of fit test for the standard Rayleigh distribution based on jackknife empirical likelihood.

Appendices

Appendix

Simplification of $\Delta (F)$

Consider

$$\begin{aligned} \Delta (F)= \,& {} \int _{0}^{\infty }E\left[ \left( \frac{x^2-\sigma ^2}{x\sigma ^2}\right) min(X,t)\right] dF(t)-\int _{0}^{\infty }F(t)dF(t).\nonumber \\=\, & {} \int _{0}^{\infty }\int _{0}^{\infty }\left( \frac{x^2-\sigma ^2}{x\sigma ^2}\right) min(x,t)dF(x)dF(t)- 1/2\nonumber \\=\, & {} \frac{1}{\sigma ^2}\int _{0}^{\infty }\int _{0}^{t}\left( {x^2-\sigma ^2}\right) dF(x)dF(t)\nonumber \\{} & {} \qquad +\frac{1}{\sigma ^2}\int _{0}^{\infty }\int _{t}^{\infty }\left( {x^2-\sigma ^2}\right) \frac{t}{x}dF(x)dF(t)- 1/2\nonumber \\= \,\,& {} \frac{1}{\sigma ^2}\int _{0}^{\infty }\int _{0}^{t}x^2dF(x)dF(t)-\frac{1}{2}\int _{0}^{\infty }\int _{0}^{t}2dF(x)dF(t)\nonumber \\ {}{} & {} \qquad +\frac{1}{\sigma ^2}\int _{0}^{\infty }\int _{t}^{\infty }\left( {x^2-\sigma ^2}\right) \frac{t}{x}dF(x)dF(t)- \frac{1}{2}\nonumber \\=\,\, & {} \frac{1}{\sigma ^2}(\Delta _1(F)+\Delta _2(F))-1 \,\,(say). \end{aligned}$$

(14)

Now, changing the order of integration we have

$$\begin{aligned} \Delta _1(F)= & {} \int _{0}^{\infty }\int _{0}^{t}x^2dF(x)dF(t)\nonumber \\= & {} \int _{0}^{\infty }{x^2}\int _{x}^{\infty }dF(t)dF(x)\nonumber \\= & {} \int _{0}^{\infty }{x^2}{\bar{F}}(x)dF(x)\nonumber \\=\, & {} \frac{1}{2}E\left( \min (X_{1},X_{2})^2\right) \end{aligned}$$

(15)

where $\bar{F}(x) = 1 - F(x)$. The last identity follows from the fact that $2\bar{F}(x)dF(x)$ is the density function of $\min (X_1,X_2)$. Again,

$$\begin{aligned} \Delta _2(F)= & {} \int _{0}^{\infty }\int _{t}^{\infty }\left( {x^2-\sigma ^2}\right) \frac{t}{x}dF(x)dF(t)\nonumber \\= & {} \int _{0}^{\infty }\int _{t}^{\infty }{t}{x}dF(x)dF(t)-\sigma ^2\int _{0}^{\infty }\int _{t}^{\infty }\frac{t}{x}dF(x)dF(t)\nonumber \\=\, & {} E\left( (X_1X_2)I(X_2<X_1)\right) -\sigma ^2E\left( \frac{X_2}{X_1}I(X_2<X_1)\right) . \end{aligned}$$

(16)

Substituting (15) and (16) in (14), we obtain

$$\begin{aligned} \Delta (F)=\frac{1}{\sigma ^2}E\left( \min (X_{1},X_{2})^2+X_1X_2I(X_2<X_1)\right) -E\left( \frac{X_2}{X_1}I(X_2<X_1)\right) -1. \end{aligned}$$

(17)

Proof of Theorem 3:

Define

$$\begin{aligned} {\widehat{\Delta }}^{*}(F) =\frac{U_{1}}{\sigma ^2}-U_{2}. \end{aligned}$$

Since ${{\widehat{\sigma }}}^2$ is a consistent estimators of $\sigma ^2$, by Slutsky’s theorem, the asymptotic distribution of $\sqrt{n}({\widehat{\Delta }}-\Delta (F))$ and $\sqrt{n}({\widehat{\Delta }}^{*}-E({\widehat{\Delta }}^{*}))$ are same. Now we observe that ${\widehat{\Delta }}^*$ is a U statistic with symmetric kernel,

$$\begin{aligned} h(X_1,X_2) = \frac{1}{2}\left( \frac{2\min (X_{1},X_{2})^2}{\sigma ^2}+\frac{X_1X_2}{{\sigma ^2}}- \frac{X_{1}}{X_{2}}I(X_{1}<X_{2})-\frac{X_{2}}{X_{1}}I(X_{2}<X_{1})\right) . \end{aligned}$$

Hence using the central limit theorem for U-statistics we have the asymptotic normality of ${\widehat{\Delta }}^*$. The asymptotic variance is $4\sigma _1^2$ where $\sigma _1^2$ is given by Lee (2019)

$$\begin{aligned} \sigma _1^2= Var\left[ E\left( h(X_{1},X_{2})|X_{1}\right) \right] . \end{aligned}$$

(18)

Consider

$$\begin{aligned} E[2\min (x,X_{2})^2+xX_2]=\, & {} 2E[x^2 I(x<X_{2})+X_2^2I(X_{2}<x)+xX_2]\nonumber \\=\, & {} 2x^2P(x<X_2)+2\int _{0}^{\infty }y^2I(y<x)dF(y)+x\mu \nonumber \\=\, & {} 2x^2{\bar{F}}(x)+2\int _{0}^{x}y^2dF(y)+x\mu . \end{aligned}$$

(19)

Also

$$\begin{aligned} E[\frac{x}{X_2}I(x<X_2]+E[\frac{X_2}{x}I(X_2<x)]= &\, {} xE[\frac{1}{X_2}I(x<X_2]+\frac{1}{x}E[X_2I(X_2<x)]\nonumber \\= &\, {} x\int _{0}^{\infty }\frac{1}{y}I(x<y)dF(y)+\frac{1}{x}\int _{0}^{x}ydF(y)\nonumber \\= &\, {} x\int _{x}^{\infty }\frac{1}{y}dF(y)+\frac{1}{x}\int _{0}^{x}ydF(y). \end{aligned}$$

(20)

Substituting equations (19) and (20) in equation (18) we obtain the variance expression as specified in the theorem. $\square$