Question

Optimal signed-rank based methods also exist for the one-sample problem. In this exercise, we briefly discuss these methods. Let \(X_{1}, X_{2}, \ldots, X_{n}\) follow the location model $$ X_{i}=\theta+e_{i}, \quad(10.5 .39) $$ where \(e_{1}, e_{2}, \ldots, e_{n}\) are iid with pdf \(f(x)\), which is symmetric about \(0 ;\) i.e., \(f(-x)=\) \(f(x)\) (a) Show that under symmetry the optimal two-sample score function \((10.5 .26)\) satisfies $$ \varphi_{f}(1-u)=-\varphi_{f}(u), \quad 00 $$ Our decision rule for the statistic \(W_{\varphi^{+}}\) is to reject \(H_{0}\) in favor of \(H_{1}\) if \(W_{\varphi^{+}} \geq\) \(k\), for some \(k\). Write \(W_{\varphi^{+}}\) in terms of the anti-ranks, \((10.3 .5) .\) Show that \(W_{\varphi^{+}}\) is distribution-free under \(H_{0}\). (f) Determine the mean and variance of \(W_{\varphi^{+}}\) under \(H_{0}\). (g) Assuming that, when properly standardized, the null distribution is asymptotically normal, determine the asymptotic test.

Short answer

The signed-rank method is a non-parametric statistical hypothesis test used when the distribution of the data is unknown. It's used in the exercise to prove that the two-sample score function is an odd function. This odd function and its derivative are then used to define the score and then prove the common tests that can in the terms of this score statistic. Finally, distribution property of the score statistic is verified, and its mean, variance and asymptotic distribution is calculated.

Step-by-step solution

Proof of the optimal two-sample score function to be an odd function

We have to prove that under symmetry, the optimal two-sample score function satisfies \(\varphi_{f}(1-u)=-\varphi_{f}(u), 0<u<1\). From the understanding of symmetric distributions we know that if \(f(x)\) is a density function of a symmetric distribution, \(F(x)\) the associated cumulative distribution function and \(\varphi(u)\) is the associated score function then \(F(x) = 1- F(-x)\) and \(\varphi(u) = F^{-1}(1-u)\). Hence when we substitute \(1-u\) in place of \(u\) for the score function we get \(\varphi_{f}(1-u)=F^{-1}(1-(1-u)) = F^{-1}(u)\). Now by using the property of a symmetric distribution \(\varphi_{f}(u)\=-F^{-1}(1-u)\). Hence proved. Also, for a symmetric distribution, the value at \(u=\frac{1}{2}\) is 0 as both halves of the distribution mirror each other.

Proof of the non-negativity of the two-sample score function

We need to prove that if \(\varphi(u)\) is a nondecreasing odd function about \(\frac{1}{2}\), then \(\varphi^{+}(u)=\varphi[(u+1) / 2] \geq 0\). For this, we need to see that since \(\varphi(u)\) is nondecreasing and an odd function about \(\frac{1}{2}\), for any \(u \leq \frac{1}{2}\), \(\varphi(u) <= 0\). And for \(u \geq \frac{1}{2}\), we have \(\varphi(u) >= 0\). Thus, for \(\varphi^{+}(u)=\varphi[(u+1) / 2]\) the value of \((u+1) / 2\) lies between \(0.5\) and \(1\), hence \(\varphi^+(u)\geq0\).

Proof of converting the score to two well known test statistics

We have to show that \(W_{\varphi+}\) reduces to a linear function of the signed-rank test statistic and the sign test statistic under two different conditions. First, let's assume that \(\varphi(u) = 2u - 1\). Writing out the expression for \(W_{\varphi+}\), it becomes \(\sum_{i=1}^{n} \operatorname{sgn}\left(X_{i}\right) a^{+}\left(R\left|X_{i}\right|\right)\), where \(a^{+}(i) = \varphi^+(i / (n+1)) = 2(i/(n+1)) - 1\). Now the signed rank statistic is given as \(\sum_{i=1}^{n} \mathrm{sgn}(\xi_i)R_i\) (where \(R_i\) is the rank of \(|\xi_i|\)). Hence it can be seen that \(W_{\varphi+}\) is a linear function of signed rank statistic for this case. Now consider \(\varphi(u)=\operatorname{sgn}(2u - 1)\). In this case \(a^{+}(i) = \varphi^{+}(i / (n+1)) = \mathrm{sgn}((2i / (n+1)) - 1)\). The sign test statistic is given as \(\sum_{i=1}^{n} \mathrm{sgn}(\xi_i)\). Hence it can be also seen that \(W_{\varphi+}\) is a linear function of sign test statistic for this case.

Writing the score in terms of anti-ranks and proof of distribution

The anti-rank \(S_i\) is given by \(n + 1 - R_i\). We know that \(W_{\varphi+} = \sum_{i=1}^{n}\mathrm{sgn}(X_i) a^{+}(R_{|X_i|})\). It is also stated that we can write \(W_{\varphi+}\) in terms of \(S_i\), or anti-ranks, so that we can say \(W_{\varphi+} = \sum_{i=1}^{n}\mathrm{sgn}(X_i) a^{+}(S_{|X_i|})\). Now we need to prove that \(W_{\varphi+}\) is distribution-free under \(H_0\). This can be proved by recognizing the fact that when \(\theta = 0\), \(W_{\varphi+}\) distribution only depends on the ranks of \(|X_i|\) or \(S_{|X_i|}\), and is unaffected by the true distribution of \(X_i\), making it distribution-free.

Finding the mean and variance of the score statistic under null hypothesis

Mean and variance of \(W_{\varphi+}\) under \(H_0\) where \(\theta=0\) have standard formulas given by \(\mu_{W_{\varphi+}} = n(\mu_{\varphi+})\) \(Var_{W_{\varphi+}} = n(\sigma^2_{\varphi+}-(\mu_{\varphi+})^2+nCov_{\varphi+,\varphi+})\). Here, \(\mu_{\varphi+}\), \(\sigma_{\varphi+}\) and \(Cov_{\varphi+,\varphi+}\) are the mean, standard deviation and covariance of the score function.

Finding the asymptotic test

The asymptotic test can be found by using the Central Limit Theorem (CLT) which states that if you have a sum of many independent random variables, under certain conditions, the distribution of that sum will approximate a normal distribution. In this case, under \(H_0\), if we take \(W_{\varphi+}\) as the sum of many independent random variables, when standardized it will follow an asymptotically normal distribution, z~\(N(0,1)\).