What is the NP?

Suppose $$X_1, X_2, X_3,\ldots, X_n$$ are i.i.d. variables Poisson $$(\lambda)$$

and $$g(λ)=\lambda(c - e^{-cλ})$$
c:constant

What is the NP for $$H_0:g(λ)=c1$$ vs $$H_1:g(λ)=c2$$ ??

My thought:

Step 1: We prove (monotone likelihood ratio property) $$\forall \lambda_2>\lambda_1\:\: \frac{g(x\mid g(λ_2))}{g(x\mid g(λ_1))}$$ and (If I have not done any mistake) I proved $$\frac{g(x\mid g(λ_2))}{g(x\mid g(λ_1))}$$ is non-decreasing (increasing) in $$T(X)=\sum(X_i)$$

Step 2 : after this point I am not sure how to continue the function hypothesis "$$g(λ)$$ confuses me .

• You're comparing two simple hypotheses, and Neyman-Pearson (en.wikipedia.org/wiki/Neyman–Pearson_lemma) tells us that in that case the likelihood ratio test is UMP. Apr 13 '19 at 16:43
• $g(\lambda)=1$ is the same as $\lambda\approx1.349976$ and and $g(\lambda=2)$ is the same as $\lambda \approx 2.238646.$ So you're testing $\lambda\approx1.349976$ against $\lambda \approx 2.238646.$ After that, the function $g$ no longer matters. Apr 13 '19 at 20:47
• @MichaelHardy Could you please elaborate on how you found λ1 and λ2 values? for example how would I calculate those only by hand.
– GAGA
Apr 14 '19 at 6:21
• @GAGA : I did it numerically, using software, by the secant method. Apr 14 '19 at 18:30
• @GAGA Please be careful of making edits that end up making your questions less informative. Your new title is ambiguous. Please make it clearer. Aug 20 '19 at 23:03

Let's start by observing that $$g(\lambda)$$ is a monotonically increasing function of $$\lambda$$. This implies that for each value of $$g(\lambda)$$, there will be a unique value of $$\lambda$$; we can find a unique $$\lambda$$ such that $$g(\lambda)=1$$ and another such that $$g(\lambda)=2$$. If we find these $$\lambda$$s, our problem reduces to finding the UMP test for the two point hypotheses associated with those two $$\lambda$$ values, and the Neyman-Pearson lemma assures us that such a test exists.

Unfortunately, there is no closed-form expression for the inverse function of $$g(\lambda) = \lambda(1-e^{-\lambda})$$, but we can find it numerically easily enough through any of a number of univariate root-finding algorithms, e.g., interval halving:

$$\lambda_1 = g^{-1}(1) \approx 1.35$$ $$\lambda_2 = g^{-1}(2) \approx 2.239$$

Defining $$T(X) = \sum X_i$$, as you have done, allows us to write the likelihood ratio for $$H_1/H_0$$ as:

$$\mathcal{L} = 2^{T(X)}e^{-1} \propto 2^{T(X)}$$

Taking the log of this last expression simplifies things, and is permissible as it is a monotonic transform of the likelihood ratio. Our task then becomes finding the critical value $$c$$ for $$T(X)$$, where, under the null hypothesis, $$T(X) \sim \text{Poisson}(g^{-1}(1)n) \approx \text{Poisson}(1.35n)$$. If we do not care whether our test is exact, this amounts to finding $$\min_c : P(c | n) \geq 1-\alpha$$, e.g, if $$n=10$$ and $$\alpha = 0.05$$, then $$c=20$$, as $$P(20|13.5)=0.965$$ but $$P(19|13.5) = 0.942$$.

Unfortunately, as $$T(X)$$ takes on only integer values, there may be no $$c$$ such that the probability that $$T(X) > c$$ under $$H_0$$ is exactly equal to our chosen test level $$\alpha$$. If we wish an exact test, we will have to randomize.

Following our example, this randomization results in the following criteria:

\begin{align} &\text{If}\,\, T(x) > 20, \text{reject} \,H_0 \\ &\text{If}\,\, T(x) < 20, \text{don't reject}\,H_0 \\ &\text{If}\,\, T(x) = 20, \text{reject}\, H_0\,\text{with probability}={0.95-0.942 \over 0.965-0.942} \approx 0.348 \end{align}

where the randomization could occur by generating a $$U(0,1)$$ random number and comparing to the stated probability.

• Could you please elaborate on how you found λ1 and λ2 values? for example how would I calculate those only by hand.
– GAGA
Apr 14 '19 at 6:21
• You can use a simple root-finding routine to do so, such as interval halving. Apr 14 '19 at 15:31
• Is it an easier way to just say that the UMP exists without calculating λ1 , λ2 (for example by using the facts 1) g(λ) monotone (non decreasing ) and 2) $H_0:g(λ)\leq1$ vs $H_0:g(λ)>1$
– GAGA
Apr 29 '19 at 21:39
• Yes, for existence all you really need in this case is that $g(1) \neq g(2)$ and the Neyman-Pearson lemma. If they aren't equal, then you can find which one maps back to $\lambda = 1$ and which to $\lambda = 2$ (somehow) and Neyman-Pearson gets you the rest of the way there. Apr 29 '19 at 22:03
• I think you mean $H_0: g(\lambda) = c_1 \dots$, right? If so, yes, $g(\lambda)$ increasing does imply that $\lambda_2 > \lambda_1$ if $c_2 > c_1$. Apr 29 '19 at 23:04