likelihood ratio test for shifted exponential distribution

The likelihood ratio function \( L: S \to (0, \infty) \) is defined by \[ L(\bs{x}) = \frac{f_0(\bs{x})}{f_1(\bs{x})}, \quad \bs{x} \in S \] The statistic \(L(\bs{X})\) is the likelihood ratio statistic. But, looking at the domain (support) of $f$ we see that $X\ge L$. for $x\ge L$. Understanding simple LRT test asymptotic using Taylor expansion? as the parameter of the exponential distribution is positive, regardless if it is rate or scale. If \( g_j \) denotes the PDF when \( p = p_j \) for \( j \in \{0, 1\} \) then \[ \frac{g_0(x)}{g_1(x)} = \frac{p_0^x (1 - p_0)^{1-x}}{p_1^x (1 - p_1^{1-x}} = \left(\frac{p_0}{p_1}\right)^x \left(\frac{1 - p_0}{1 - p_1}\right)^{1 - x} = \left(\frac{1 - p_0}{1 - p_1}\right) \left[\frac{p_0 (1 - p_1)}{p_1 (1 - p_0)}\right]^x, \quad x \in \{0, 1\} \] Hence the likelihood ratio function is \[ L(x_1, x_2, \ldots, x_n) = \prod_{i=1}^n \frac{g_0(x_i)}{g_1(x_i)} = \left(\frac{1 - p_0}{1 - p_1}\right)^n \left[\frac{p_0 (1 - p_1)}{p_1 (1 - p_0)}\right]^y, \quad (x_1, x_2, \ldots, x_n) \in \{0, 1\}^n \] where \( y = \sum_{i=1}^n x_i \). The UMP test of size for testing = 0 against 0 for a sample Y 1, , Y n from U ( 0, ) distribution has the form. uoW=5)D1c2(favRw `(lTr$%H3yy7Dm7(x#,nnN]GNWVV8>~\u\&W`}~= , where $\hat\lambda$ is the unrestricted MLE of $\lambda$. Understand now! p_5M1g(eR=R'W.ef1HxfNB7(sMDM=C*B9qA]I($m4!rWXF n6W-&*8 0 The likelihood ratio statistic is L = (b1 b0)n exp[( 1 b1 1 b0)Y] Proof The following tests are most powerful test at the level Suppose that b1 > b0. We now extend this result to a class of parametric problems in which the likelihood functions have a special . endstream I made a careless mistake! The following tests are most powerful test at the \(\alpha\) level. Learn more about Stack Overflow the company, and our products. Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site If the size of \(R\) is at least as large as the size of \(A\) then the test with rejection region \(R\) is more powerful than the test with rejection region \(A\). }, \quad x \in \N \] Hence the likelihood ratio function is \[ L(x_1, x_2, \ldots, x_n) = \prod_{i=1}^n \frac{g_0(x_i)}{g_1(x_i)} = 2^n e^{-n} \frac{2^y}{u}, \quad (x_1, x_2, \ldots, x_n) \in \N^n \] where \( y = \sum_{i=1}^n x_i \) and \( u = \prod_{i=1}^n x_i!