Uniformly most powerful test for $H_0: \lambda \ge \lambda_0$

Question

I cannot find anywhere answer to this question. Let's take exponential distribution i.e.

$$f(x) = \lambda e^{-\lambda x}\mathbb{1}_{(0, +\infty)}(x)$$

I want to derive uniformly most powerful test for hypothesis: $H_0:\lambda\ge\lambda_0$ and $H_1:\lambda<\lambda_0$.

How I would start with this problem is to take two $\lambda_1 \ge\lambda_0$ and $\lambda_2 < \lambda_0$ and then start to write:

$$\frac{L(x, \lambda_2)}{L(x, \lambda_1)} = \frac{\lambda_2^ne^{-\lambda_2(\sum_{i=1}^nx_i)}}{\lambda_1^ne^{-\lambda_1(\sum_{i=1}^nx_i)}}$$

Did I start to write correct likelihood proportion for my problem?

Answer 1

I would be interested to see a technique for solving this problem

the system to be verified is the following

$$\begin{cases} H_0: \theta \ge \theta_0 \\ H_1: \theta < \theta_0 \end{cases} $$

now what we have to do first, is to check if our model has a monotonic likelihood ratio. To do that, lets set aribtrarily

$$\theta_1<\theta_2$$

and write down LR (note that here LR is not the generalized likelihood ratio but it is only the ratio of two likelihood functions)

$$LR=\frac{\theta_1^n e^{-\theta_1\Sigma_i x_i}}{\theta_2^n e^{-\theta_2\Sigma_i x_i}}=\left( \frac{\theta_2}{\theta_1}\right)^n e^{(\theta_2-\theta_1)\Sigma_i x_i}$$

which is clearly an increasing funciont in $Y=\Sigma_i x_i$

Now let's apply the following theorem, part (i), taken from Mood Graybill Boes

Using this theorem, the critical region is

$$\mathbb{P}[Y>k|\theta_0]=\alpha$$

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Example

Let's verify the following system

$$\begin{cases} H_0: \theta \ge 1\\ H_1: \theta < 1 \end{cases} $$

Using the following random sample $\mathbf{x}=\{2.1;3.4;2.7\}$ and a significance level $\alpha=5\%$

Solution

First, remind that this system is equivalent to

$$\begin{cases} H_0: \theta = 1\\ H_1: \theta < 1 \end{cases} $$

Our critical region is

$$\mathbb{P}[Y>k|\theta=1]=0.05$$

now observe that, under $H_0$, $Y=\Sigma_i x_i \sim\text{Gamma}[3;1]$ that is $2Y\sim \chi_{(6)}^2$ thus the critical value, read in a chi square table with 6 d.o.f., is $k=12.6$. Being in our observations $2Y=16.4$ we reject $H_0$ in favour of the alternative hypothesis that the mean is greater than 1

Answer 2

This is not an complete answer.

Take $\Lambda_0=[\lambda_0,\infty)$ and $\Lambda_1=[0,\lambda_0)$. If $\lambda_0\geq \frac{1}{\bar{x}}$ we have that $$\mathcal{L}(x_1,...,x_n)=\frac{\sup_{\lambda \in \Lambda_0}\lambda^ne^{-\lambda n \bar{x}}}{\sup_{\lambda \in \Lambda_1} \lambda^ne^{-\lambda n \bar{x}}}=(\lambda_0 \bar{x})^ne^{n(1-\lambda_0 \bar{x})}$$ Now if $\lambda_0 <\frac{1}{\bar{x}},$ $$\mathcal{L}(x_1,...,x_n)=\frac{\sup_{\lambda \in \Lambda_0}\lambda^ne^{-\lambda n \bar{x}}}{\sup_{\lambda \in \Lambda_1} \lambda^ne^{-\lambda n \bar{x}}}=(\lambda_0 \bar{x})^{-n}e^{-n(1- \lambda_0 \bar{x})}$$ You need to find $c$ such that $\mathbb{P}(\mathcal{L}(X_1,...,X_n)<c)\leq{\alpha}$. Well $$\begin{eqnarray*}\mathbb{P}(\mathcal{L}(X_1,...,X_n)<c) &=& \mathbb{P}\Big((\lambda_0 \bar{X})^ne^{n(1-\lambda_0 \bar{X})}<c,\bar{X}\geq {1 \over \lambda_0}\Big)+\mathbb{P}\Big((\lambda_0 \bar{X})^{-n}e^{-n(1-\lambda_0 \bar{X})}<c,\bar{X}< {1 \over \lambda_0}\Big) \\ &=& \mathbb{P}\Big(S^n e^{-\lambda_0 S}<c\big({n \over \lambda _0 e}\big)^n,S \geq {n \over \lambda_0}\Big) + \mathbb{P}\Big(S^n e^{-\lambda_0 S}>\frac{1}{c}\big({n \over \lambda _0 e}\big)^n,S < {n \over \lambda_0}\Big)\end{eqnarray*}$$ where $S\sim \text{Erlang}(\lambda,n)$. This is where I'm stuck. If we assume $\lambda \in \Lambda_0$ I believe we can bound the last probability above by $$\mathbb{P}\Big(S^n_0 e^{-\lambda_0 S_0}<c\big({n \over \lambda _0 e}\big)^n,S_0 \geq {n \over \lambda_0}\Big) + \mathbb{P}\Big(S^n_0 e^{-\lambda_0 S_0}>\frac{1}{c}\big({n \over \lambda _0 e}\big)^n,S_0 < {n \over \lambda_0}\Big) $$ where $S_0 \sim \text{Erlang}(\lambda_0,n)$ but even if we could I still wouldn't know how to manipulate these inequalities to obtain the value of $c$ we're looking for. I would be interested to see a technique for solving this problem. Maybe this is an indication that the LRT shouldn't be applied to find the most powerful test.