Question

Let the random variable \(X\) have the pdf \(f(x ; \theta)=(1 / \theta) e^{-x / \theta}, 0

Short answer

Yes, the hypothesis test \(H_{0}\) against \(H_{1}\) can be performed using the statistic \(X_{1}+X_{2}\) according to the Neyman-Pearson lemma which suggests that the optimal test depends on the statistic \(X_{1}+X_{2}\).

Step-by-step solution

Establish The Likelihood Ratio

Start by defining the likelihood ratio, \(λ(x)\), for the given hypotheses. According to the Neyman-Pearson lemma, this ratio is the basis for the most powerful test of \(H_{0}\) against \(H_{1}\). The likelihood ratio is defined as \[ λ(x) = \frac{L(θ'=2|x)}{L(θ''=4|x)} \]where \(L(θ'|x)\) and \(L(θ''|x)\) are the likelihoods under \(H_{0}\) and \(H_{1}\) respectively, and 'x' is the observed data.

Calculate the Likelihoods

Calculate the likelihoods for both \(H_{0}\) and \(H_{1}\). Given a random sample \(X_{1}, X_{2}\), for \(H_{0}\):\[ L(θ'|x) = f(X_{1};2)f(X_{2};2) \]and for \(H_{1}\):\[ L(θ''|x) = f(X_{1};4)f(X_{2};4) \]Plug in the given pdf \(f(x;θ) = (1/θ)e^{-x/θ}\),For \(H_{0}\):\[ L(θ'|x) = (1/2)e^{-X_{1}/2} * (1/2)e^{-X_{2}/2} = (1/4)e^{-(X_{1}+ X_{2})/2} \]And for \(H_{1}\):\[ L(θ''|x) = (1/4)e^{-X_{1}/4} * (1/4)e^{-X_{2}/4} = (1/16)e^{-(X_{1}+ X_{2})/4} \]

Insert The Likelihoods Into The Likelihood Ratio

Substitute the found likelihoods for \(H_{0}\) and \(H_{1}\) into the formula for the likelihood ratio \(λ(x)\), \[ λ(x) = \frac{L(θ'=2|x)}{L(θ''=4|x)} = \frac{ (1/4)e^{-(X_{1}+ X_{2})/2} }{ (1/16)e^{-(X_{1}+ X_{2})/4} } \]which simplifies to\[ λ(x) = 4e^{(X_{1} + X_{2})/4} \]So, as per the Neyman-Pearson lemma, the testing rule depends on the statistic \(X_{1}+X_{2}\). We hence conclude that the optimal test of \(H_{0}\) against \(H_{1}\) can indeed be performed using the statistic \(X_{1}+X_{2}\).