# The two sample scale problem where the underlying distributions are one-sided

M. Raghavachari

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## To cite this version:

M. Raghavachari. The two sample scale problem where the underlying distributions are one-sided. Annales de l'ISUP, 1970, XIX (3), pp.191-196. hal-04083418

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# THE TWO SAMPLE SCALE PROBLEM WHERE THE UNDERLYING DISTRIBUTIONS ARE ONE-SIDED

## M. RAGHAVACHARI *

### SUMMARY

Let $X_1, X_2, ..., X_m$ be independent and identically distributed random variables with absolutely continuous cdf $P\{X \le x\} = F(x - \mu)$ where $F(0) = 0$. Let $Y_1, Y_2, ..., Y_n$ be independent and identically distributed random variables with cdf $F\{(x - \eta) \Theta\}$. One may wish to test the hypothesis $H: \Theta = 1$ against one of the alternatives $K: \Theta > 1$ or $K': \Theta < 1$. In what follows we shall be concerned with the case where the location parameters $\mu$ and $\eta$ are unknown but equal. This means that the two distributions are allowed to differ only in the scale parameter. For this problem, M. A. H. Taha [10] proposed a rank test based on $L = \sum_{i=1}^{n} R^2(Y_i)$ where $R(Y_i)$ is the rank of $Y_i$ in the combined sample of the X's and Y's. I. R. Savage [8] proposed a rank test (S-test) using the exponential order scores and one can hope that this test would perform well even when the underlying distributions are not exponential. In this paper the S-test is compared with L-test and also a few comparisons are made when the competitors are the classical F-test and the Normal scores test. See Klotz [5].

### THE S-TEST

Define $Z_{N,i} = 1$ if the $i^{th}$ ordered statistic from the combined sample of the X's and Y's is a X and $= 0$ otherwise. Let $E_{N,i}$ be the expected value of the $i^{th}$ order statistic from a sample of size $N = m + n$ drawn from the exponential distribution with density $p(x) = \exp(-x)$; $x \ge 0$.

<footnote>* Carnegie-Mellon University and Indian Institute of Management, Ahmedabad.</footnote>
Define

$$mT_N = \sum_{i=1}^{N} E_{N,i} Z_{N,i} \quad (1)$$

The S-test is based on the test statistic $T_N$. This test is the locally most powerful rank test when the underlying distribution is $F(x)=1-\exp(-x)$. See [2]. If $E_{N,i}$ in (1) is the expected value of the $i^{th}$ order statistic from a sample of size N from a standard normal distribution the resulting test is called the Normal scores test. The asymptotic normality of the $T_N$-statistic was shown by Chernoff-Savage [3]. Efficiency evaluation: The measure of efficiency is the Pitman-efficiency. Under some regularity conditions, the Pitman-efficiency is given by the ratio of the efficacies (See Klotz [5]) of the two tests. If F satisfies the regularity conditions of Noether's theorem [6], the efficacy of the S-test is

$$\frac{mn}{N} \left( \int_{0}^{\infty} \frac{x f^2(x)}{1-F(x)} dx \right)^2$$

where f is the density of F. See Basu & Woodworth [1] for a derivation. It appears there is a typographical error in their expression for efficacy where $\psi(x)$ should be replaced by $\{1 - F(x)\}$. The efficacy of the L-test was shown by Taha [10] to be

$$\frac{45mn}{N} \left[ \int_{0}^{\infty} F(x) \cdot x f^2(x) dx \right]^2$$

The Pitman efficiency of S-test relative to L-test $e(S,L)$ is given by

$$e_{S,L} = \left\{ \int_{0}^{\infty} \frac{x f^2(x)}{1-F(x)} dx \right\}^2 / 45 \left\{ \int_{0}^{\infty} F(x) \cdot x f^2(x) dx \right\}^2 \quad (2)$$

Comparison of S-test and L-test: Since $F(x)$ ($1-F(x)$) $\le 1/4$ for all $x$, $e_{S,L}$ is clearly $\ge 16/45$. This shows $e_{S,L}$ is bounded away from zero. When $F(x) = 1-\exp(-x^K)$, ($K > 0$), $e_{S,L}$ is independent of K and equals
1.16. For the underlying distribution $F(x) = 1-(1-x)^p, p > 1$; $e_{S,L} = 2.22$ for $p=2$, and as $p \to \infty$, $e_{S,L} \to 1296/1125 = 1.16$. When $p \to 1$, $e_{S,L} \to \infty$. This shows that where the underlying distribution is uniform over $(0,1)$, the efficiency $e_{S,L}$ is infinite. Comparison of S-test with other tests: There are many tests proposed for the general two-sample scale problem. We shall in this section compare S-test with the F-test and the Normal scores test. The expressions for efficacies of the F-test and the Normal scores test (NS-test) have been given by Sukhatme [9] and Klotz [5] respectively. From these expressions we can write

$$e_{S,F} = \frac{(\beta_2 - 1)}{4} \left[ \int_0^\infty \frac{x f^2(x)}{1-F(x)} dx \right]^2 \quad (3)$$

$$e_{NS,S} = \frac{2 \left[ \int_0^\infty \frac{\Phi^{-1}\{F(x)\}}{\Phi[\Phi^{-1}\{F(x)\}]} \times f^2(x) dx \right]^2}{\left[ \int_0^\infty \frac{x f^2(x)}{1-F(x)} dx \right]^2} \quad (4)$$

where $\beta_2$ is the kurtosis of $f(x)$ and is given by the formula

$$\beta_2 = E\{X - E(X)\}^4 / \{E\{X - E(X)\}^2\}^2.$$

It was shown in [7] that the efficiency of the Normal scores test relative to F-test may be made arbitrarily small for a class of distributions. It is shown below that for the class of distributions given by the density

$$f(x) = 2cx^{-\alpha} (1-x)^{\alpha}; \qquad \begin{cases} x \ge 0 \\ 0 < \alpha < 1 \end{cases} \qquad (5)$$

where $c = 1/2 \Gamma(1-\alpha) \Gamma(1+\alpha)$, $e_{S,F} \to 0$ as $\alpha \to 1$. We can check that $(\beta_2 - 1)(1-\alpha) \to 9/5$ as $\alpha \to 1$. Now

$$1 - F(x) = 2c \int_x^1 t^{-\alpha} (1-t)^{\alpha-1} dt$$
$$ 
\begin{aligned}
&= 2c \left[ \frac{t^{1-\alpha} (1-t)^{\alpha}}{1-\alpha} \right]_{\substack{t=1 \\ t=x}} + 2c (\alpha/1-\alpha) \int_x^1 t^{1-\alpha} (1-t)^{\alpha-1} dt \\
&\ge 2c (\alpha/1-\alpha) \int_x^1 t^{1-\alpha} (1-t)^{\alpha-1} dt \\
&\ge 2c (\alpha/1-\alpha) (x/1-x)^{1-\alpha} (1-x)
\end{aligned}
$$

since $t/(1-t)$ is monotone increasing in the interval $(x, 1)$. Now the right hand side equals $\alpha x^{1-\alpha} (1-x)^{\alpha/\Gamma(2-\alpha)\Gamma(1+\alpha)}$. We have therefore

$$ \int_0^1 \frac{x f^2(x)}{1-F(x)} dx \le \frac{\Gamma(2-\alpha) \Gamma(1+\alpha)}{\alpha[\Gamma(1-\alpha) \Gamma(1+\alpha)]^2} \int_0^1 x^{-\alpha} (1-x)^{\alpha} dx = (1-\alpha)/\alpha $$ 
Hence $e_{S,F} \le \frac{(\beta_2-1)}{4} \frac{(1-\alpha)^2}{\alpha^2} \to 0$ as $\alpha \to 1$.

**Remark:** For the same class of distributions given by (5) $e_{L,F} \to 0$ as $\alpha \to 1$. This follows since $e_{L,F} = e_{L,S}$, $e_{S,F}$ and $e_{L,S} \ll 1/0.36$. On the other hand one can make $e_{S,F}$ as large as we please by considering the class of distributions given by c.d.f. $F(x) = 1 - (1-x)^p$, $p > 1$ and taking $p$ close to 1. For in this case the efficacy of the S-test tends to $\infty$ as $p \to 1$ and $\beta_2 \to$ a finite limit. For the comparison of the S-test with the NS-test, consider the expression of the efficiency $e_{NS,S}$ given by (4). When $F(x) = 1 - \exp(-x)$, the expression $e_{NS,S}$ has been known to be equal to 0.177. See [1]. When the underlying distributions are rectangular over (0,1) the expressions for the efficacies of the S-test and the NS-test give infinite values. However a limiting argument shows that $e_{S,NS}$ (Rectangular) = 2. More explicitly consider a sequence of distributions $F_p(x) = 1 - (1-x)^p$, $p > 1$. We shall show that $e_{NS,S}(F_p) \to 2$ as $p \to 1$. Observe that for any $\delta > 0$ (we shall take $\delta > 1/2$)
$$p^2 \int_0^\delta \frac{|\Phi^{-1}\{1-(1-x)^p\}|}{\phi[\Phi^{-1}(1-(1-x)^p)]} x (1-x)^{2p-2} dx \quad (6)$$

$$= p^2 \int_0^\delta \frac{|\Phi^{-1}\{1-(1-x)^p\}|}{\phi[\Phi^{-1}\{1-(1-x)^p\}]} \{1-(1-x)^p\} \frac{x}{1-(1-x)^p} (1-x)^{2p-2} dx$$

$$\le p^2 k \int_0^\delta (1-x)^{2p-2} dx, \quad \text{since} \quad \frac{|\Phi^{-1}\{1-(1-x)^p\}|}{\phi[\Phi^{-1}\{1-(1-x)^p\}]} \{1-(1-x)^p\}$$

is bounded as $x \to 0$ and $p \to 1$ so that the expression (6) divided by

$$\int_0^1 x (1-x)^{p-2} dx \text{ tends to zero as } p \to 1. \text{ Since } (p-1) \int_0^1 x (1-x)^{p-2} dx$$

tends to 1 as $p \to 1$, it is enough to show

$$(p-1) \int_{\delta}^{1} \frac{\Phi^{-1}\{1-(1-x)^p\}}{\phi[\Phi^{-1}\{1-(1-x)^p\}]} (1-x)^p x (1-x)^{p-2} dx \to 1$$

This follows by noting that

$$\frac{\Phi^{-1}\{1-(1-x)^p\}}{\phi[\Phi^{-1}\{1-(1-x)^p\}]} (1-x)^p$$

converges to 1 uniformly in $p$ as $x \to 1$. This completes the argument to show $e_{NS,S}(F_p) \to 2$ as $p \to 1$. For these classes of distributions with $p$ close to unity the Normal Scores test is more efficient than the S-test and infinitely more efficient than the L-test.

Throughout the present paper we have assumed that the unknown location parameters $\mu$ and $\eta$ are equal. The author hopes to report on the modifications of these tests when $\mu$ and $\eta$ are completely unknown.
REFERENCES

[1] BASU (A.P.) and WOODWORTH (G.) (1967). A note on nonparametric tests for scale. *Ann. Math. Statist.*, **38**, 274-277.

[2] CAPON (J.) (1961). Asymptotic efficiency of certain locally most powerful rank tests. *Ann. Math. Statist.*, **32**, 88-100.

[3] CHERNOFF (H.) and SAVAGE (I.R.) (1958). Asymptotic normality and efficiency of certain nonparametric test statistics. *Ann. Math. Statist.*, **29**, 972-994.

[4] FELLER (W.). *An Introduction to Probability Theory and Its Applications*, Vol. I. John Wiley & Sons, Inc., New York.

[5] KLOTS (J.) (1962). Nonparametric tests for scale. *Ann. Math. Statist.*, **33**, 498-512.

[6] NOETHER (G.) (1955). On a theorem of Pitman. *Ann. Math. Statist.*, **26**, 64-68.

[7] RAGHAVACHARI (M.) (1965). On the efficiency of the Normal Scores test relative to the F-test. *Ann. Math. Statist.*, **36**, 1306-1307.

[8] SAVAGE (I.R.) (1956). Contributions to the theory of rank order statistics - The two sample case. *Ann. Math. Statist.*, **27**, 590-614.

[9] SUKHATME (B.V.) (1957). On certain two-sample nonparametric tests for variances. *Ann. Math. Statist.*, **29**, 188-194.

[10] TAHA (M. A. H.). *Publication of Institute of Statistics. University of Paris., Vol. 13 (1964)*, 169-180.

Reçu en Mars 1970

Carnegie-Mellon University

Department of Statistics

Schenley Park

PITTSBURGH,

Pennsylvania 15213

U.S.A.

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