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

and $g(λ)=\lambda(c - e^{-cλ})$

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 .

  • 2
    $\begingroup$ 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. $\endgroup$
    – jbowman
    Apr 13 '19 at 16:43
  • 1
    $\begingroup$ $g(\lambda)=1$ 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. $\endgroup$ Apr 13 '19 at 20:47
  • $\begingroup$ @MichaelHardy Could you please elaborate on how you found λ1 and λ2 values? for example how would I calculate those only by hand. $\endgroup$
    – GAGA
    Apr 14 '19 at 6:21
  • $\begingroup$ @GAGA : I did it numerically, using software, by the secant method. $\endgroup$ Apr 14 '19 at 18:30
  • $\begingroup$ @GAGA Please be careful of making edits that end up making your questions less informative. Your new title is ambiguous. Please make it clearer. $\endgroup$
    – Glen_b
    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.

  • $\begingroup$ Could you please elaborate on how you found λ1 and λ2 values? for example how would I calculate those only by hand. $\endgroup$
    – GAGA
    Apr 14 '19 at 6:21
  • $\begingroup$ You can use a simple root-finding routine to do so, such as interval halving. $\endgroup$
    – jbowman
    Apr 14 '19 at 15:31
  • $\begingroup$ 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 $H_0:g(λ)>1$ $\endgroup$
    – GAGA
    Apr 29 '19 at 21:39
  • 1
    $\begingroup$ 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. $\endgroup$
    – jbowman
    Apr 29 '19 at 22:03
  • 1
    $\begingroup$ 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 $c_2 > c_1$. $\endgroup$
    – jbowman
    Apr 29 '19 at 23:04

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