# Likelihood Ratio Test for Comparing Two Poisson Distributions

I am currently studying statistical hypothesis testing, specifically the Likelihood Ratio Test (LRT), and I've come across a problem that I'm struggling to solve. I would greatly appreciate any guidance or help.

The problem involves two independent random variables, $$X_1$$ and $$X_2$$, each following a Poisson distribution with parameters $$\lambda_1$$ and $$\lambda_2$$ respectively, where $$\lambda_1, \lambda_2 > 0$$.

The null hypothesis ($$H_0$$) is that $$\lambda_1 > \lambda_2$$ I need to formulate a likelihood ratio test for this hypothesis at a given significance level $$\alpha$$.

I understand the basic concept of the LRT, which involves comparing the likelihood of the observed data under the null hypothesis to the likelihood of the observed data under the alternative hypothesis. However, I'm having trouble applying this concept to the specific case of Poisson distributions.

Any help in understanding how to formulate this test would be greatly appreciated. Thank you in advance!

• welcome to cv! The likelihood in a Poisson distribution is the probability of your observatiins, since Poisson distribution is discrete. Do you know how to write that down? How far did you get on your own with the question, what have you tried?
– Ute
Jul 17 at 8:12
• Here's what I've tried so far: I started by writing down the likelihood functions for $X_1$ and $X_2$ under the null and alternative hypotheses. I then tried to formulate the likelihood ratio, which can be written as: $\lambda = \frac{\sup_{\theta \in H} L(X_1,X_2; \theta)}{\sup_{\theta \in \Theta } L(X_1,X_2; \theta)}$ and that in the denominator we obtain the supremum at the values $\lambda_1 = X_1$ and $\lambda_2=X_2$. However, I do not know how to determine the supremum of the denominator. I also do not know if we can then apply Wilson's theorem here, as we have a small $n$ . @Ute Jul 17 at 17:42
• Very good so far! For the numerator, you have the restriction $(\lambda_1,\lambda_2)\in H$. Can you write down $H$? Btw, did you mean Wilks' theorem? That one does not apply here, because the problem wants you to specify the test statistic used in the LRT, that is, a function of $X_1$ and $X_2$.
– Ute
Jul 17 at 18:03
• I am actually wondering what you would prefer: slow guidance to working yourself towards the solution, or just a quick solution. If you want guidance, check out the self-study tag.
– Ute
Jul 17 at 18:16
• In reference to Wilk's theorem, it is known that $-2 \log{\lambda(X_1,…,X_n )} \rightarrow_D \chi_{\text{dim }Θ-\text{dim }H}^2$. Given this, if $-2 \log{\lambda(X_1,…,X_n )}$ exceeds $\chi_{\text{dim }Θ-\text{dim }H}^2$, we reject our null hypothesis $H$, where $H$ is $H: \lambda_1 > \lambda_2$. @Ute Jul 17 at 18:25

As you said, it is difficult to find the maximum likelihood under the compound hypothesis $$H:\lambda_1>\lambda_2$$. Generally, finding the maximum is often not possible in the standard way for compound hypotheses that cover only a part of the parameter space.

It can sometimes help to reparametrize - also here, as you will see later. We have the likelihood $$\mathcal{L}(\lambda_1, \lambda_2; x_1, x_2)= \frac{\mathrm{e}^{-\lambda_1} \lambda_1^{x_1}}{x_1!}\cdot \frac{\mathrm{e}^{-\lambda_2} \lambda_2^{x_2}}{x_2!}$$ but for convenience, we maximize the log likelihood instead, $$\log\mathcal{L}(\lambda_1, \lambda_2; x_1, x_2)= -(\lambda_1+\lambda_2) + x_1\log\lambda_1 + x_2\log\lambda_2 - \log(x_1!x_2!).$$

By standard calculus, we find that this function is maximized over the unconstrainted parameter space, $$(\lambda_1,\lambda_2)\in \mathbb{R}^+\times\mathbb{R}^+$$, for $$\hat{\lambda}_1=x_1$$ and $$\hat{\lambda}_2=x_2.$$

The figure below shows a contour plot of the log likelihood for $$x_1=3, x_2=5$$. The black dot is the maximum likelihood estimator for the unconstrainted case. Under the null hypothesis, we require $$\lambda_1>\lambda_2$$: this is the region under the identity line. The maximizer for this region is depicted as a circle with a cross.

##### How do we find that maximizer?

Notation becomes a bit less cluttered, if we write $$\lambda := \lambda_1$$ and $$\rho := \lambda_2/\lambda_1$$, that is, $$\lambda_2=\rho\lambda$$. Since the term $$\log(x_1!x_2!)$$ does not change the maximizer, we will then find the values $$\rho, \lambda)$$ that maximize $$h(\lambda, \rho) = - \lambda(1+\rho) + x_1\log \lambda + x_2\log(\lambda\rho)$$ We find the root of $$\frac{\partial h(\lambda,\rho)}{\partial\lambda}=-(1+\rho)+\frac{x_1+x_2}{\lambda}=0$$ as $$\hat\lambda = \frac{x_1+x_2}{1+\rho},$$ and double checking the second partial derivative of $$h$$, we convince ourselves that this is actually a maximizer. Plugging it in to $$h$$, we get $$h(\hat\lambda,\rho)=x_2\log\rho - (x_1+x_2)\log(1+\rho)-(x_1+x_2)+(x_1+x_2)\log(x_1+x_2),$$ and from there $$\frac{\partial h(\hat\lambda,\rho)}{\partial \rho}=\frac{x_2}{\rho} -\frac{x_1+x_2}{1+\rho}=\frac{x_2-\rho x_1}{\rho(1+\rho)}.$$ The function $$\rho\to h(\hat\lambda,\rho)$$ takes one local maximum in $$\hat\rho = x_2/x_1.$$

Under the constraint $$\lambda_1 > \lambda_2$$, only values $$\rho < 1$$ are admitted.

• If $$x_1>x_2$$, then $$\hat\rho = x_2/x_1$$ is the maximizer also under the constraint, and we get $$\hat\lambda_1 = x_1\quad \text{and}\quad \hat\lambda_2 = x_2.$$
• If $$x_1\leq x_2$$, then the local maximizer $$x_2/x_1 \geq 1$$. As $$h$$ is increasing in $$\rho$$ for $$\rho\leq 1 \leq x_2/x_1$$, choose $$\hat\rho$$ as large as allowed, so here: $$\hat\rho = 1$$, which gives $$\hat\lambda_1 = \hat\lambda_2 = (x_1 + x_2)/2.$$
##### Remarks on the Likelihood ratio itself
• The alternative to $$H_0:\lambda_1>\lambda_2$$ is, strictly spoken $$H_A: \lambda_1 \leq \lambda_2$$. When you choose the complete parameter space instead, you would get likelihood ratio equal to 1 whenever $$x_1\geq x_2$$.

• For one sided null hypotheses $$\lambda_1 > \lambda_2$$ or $$\lambda_2 > \lambda_1$$ and opposite alternative, nominator and denominator swap their place in the two cases $$x_1>x_2$$ and $$x_1 < x_2$$. When $$x_1=x_2$$, the likelihood ratio is 1.

• You can easily adapt the calculations to nulhypoteses of the form $$\lambda_1> 10 \lambda_2$$ or $$\lambda_1 < 1.5 \lambda_2$$.

##### Code for the figure (R)
# countour plot of Poisson log likelihood

loglik <- function(lambda1, lambda2, x = c(3,5))
dpois(x[1], lambda1, log = TRUE)+ dpois(x[2], lambda2, log = TRUE)

la1 <- seq(1, 7, 0.01)
la2 <- la1

ll <- outer(la1, la2, loglik)

# draw contour map of log likelihood function
contour(la1, la2, ll, method = "edge",
xlab = expression(lambda[1]),
ylab = expression(lambda[2]),
main="log likelihood, x=(3,5)")

# unconditional maximum likelihood estimate at x
points (3, 5, pch = 16)

# identity line separating Omega_0 and Omega_1
abline(0, 1)
# conditional maximum likelihood: local maximum at (4,4)
points(4, 4, pch = 10)