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Chapter 10: Problem 5

Let \(X\) be a continuous random variable with cdf \(F(x)\). Suppose \(Y=X+\Delta\), where \(\Delta>0\). Show that \(Y\) is stochastically larger than \(X\).

Short Answer

Expert verified
The transformation \(Y = X + \Delta\) produces a random variable \(Y\) that is stochastically larger than \(X\), which means that the CDF of \(Y\) is less than or equal to the CDF of \(X\) for all \(x\) in the range of \(X\).

Step by step solution

01

Understand what 'stochastically larger' means

The term 'stochastically larger' refers to the fact that the cumulative distribution function (CDF) of \(Y\), \(F_Y(y)\), is less than or equal to the CDF of \(X\), \(F_X(x)\), for all \(x\) in the range of \(X\). This means that the probability of \(Y\) being greater than any given value is greater than the probability of \(X\) being greater than that same value.
02

Express \(F_Y(y)\) in terms of \(F_X(x)\)

Since \(Y = X + \Delta\), we can write \(F_Y(y)\) in terms of \(F_X(x)\) as \(F_Y(y) = P(Y \leq y) = P(X + \Delta \leq y) = P(X \leq y - \Delta) = F_X(y - \Delta)\).
03

Compare \(F_Y(y)\) and \(F_X(x)\)

From the above equation, we can see \(F_Y(y) = F_X(y - \Delta)\). Since \(y - \Delta\) is less than \(y\) and \(F_X(x)\) is a non-decreasing function as \(X\) is a continuous random variable, we have \(F_X(y - \Delta) \leq F_X(y)\), i.e., \(F_Y(y) \leq F_X(x)\), this implies that \(Y\) is stochastically larger than \(X\).

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Most popular questions from this chapter

Consider the rank correlation coefficient given by \(r_{q c}\) in Part (c) of Exercise 10.8.5. Let \(Q_{2 X}\) and \(Q_{2 Y}\) denote the medians of the samples \(X_{1}, \ldots, X_{n}\) and \(Y_{1}, \ldots, Y_{n}\), respectively. Now consider the four quadrants: $$ \begin{aligned} I &=\left\\{(x, y): x>Q_{2 X}, y>Q_{2 Y}\right\\} \\ I I &=\left\\{(x, y): xQ_{2 Y}\right\\} \\ I I I &=\left\\{(x, y): xQ_{2 X}, y

Let the scores \(a(i)\) be generated by \(a_{\varphi}(i)=\varphi(i /(n+1)\), for \(i=1, \ldots, n\). where \(\int_{0}^{1} \varphi(u) d u=0\) and \(\int_{0}^{1} \varphi^{2}(u) d u=1 .\) Using Riemann sums, with subintervals of equal length, of the integrals \(\int_{0}^{1} \varphi(u) d u\) and \(\int_{0}^{1} \varphi^{2}(u) d u\), show that \(\sum_{i=1}^{n} a(i) \doteq 0\) and \(\sum_{i=1}^{n} a^{2}(i) \doteq n\)

Let \(\theta\) denote the median of a random variable \(X\). Consider tseting $$ H_{0}: \theta=0 \text { versus } H_{A}: \theta>0 $$ Suppose we have a sample of size \(n=25\). (a) Let \(S\) denote the sign test statistic. Determine the level of the test: reject \(H_{0}\) if \(S \geq 16\) (b) Determine the power of the test in Part (a), if \(X\) has \(N(0.5,1)\) distribution. (c) Assuming \(X\) has finite mean \(\mu=\theta\), consider the asymptotic test, reject \(H_{0}\) if \(\bar{X} /(\sigma / \sqrt{n}) \geq k\). Assuming that \(\sigma=1\), determine \(k\) so the asymptotic test has the same level as the test in Part (a). Then determine the power of this test for the situation in Part (b).

Optimal signed-rank based methods exist for the one sample problem, also. In this exercise, we briefly discuss these methods. Let \(X_{1}, X_{2}, \ldots, X_{n}\) follow the location model $$ X_{i}=\theta+e_{i} $$ \((10.5 .39)\) 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, (10.3.4). 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.

Let \(X\) be a continuous random variable with pdf \(f(x)\). Suppose \(f(x)\) is symmetric about \(a\); i.e., \(f(x-a)=f(-(x-a)\) ). Show that the random variables \(X-a\) and \(-(X-a)\) have the same pdf.
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