what is Likelihood ratio test for one observation $X\sim\mathrm{Pois}(μ+c)$? We have one observation $X\sim\mathrm{Pois}(\mu+c)$ where $c$ is a known constant. We wish to test $H_0: \mu=0$ vs $H_a: \mu>0$. Derive the likelihood ratio test for this problem. Do the test at the 5% significance level if $c=7$ and we observe $x=15$.
I am really lost here, I guess I have to find the MLE, so $f(x\mid \mu+c) = \frac{(\mu+c)^xe^{-(\mu+c)}}{x!}$ so $\ell(\mu+c\mid x)=\log f(x\mid \mu+c)=x\log(\mu+c)-\mu-c-\log(x!)$. But I get lost here, any advice?
 A: Since $c$ is known, you need
$$
\ell(\mu\mid x) = x\log(\mu+c)-\mu-c+\text{constant}.
$$
Then
$$
\ell'(\mu\mid x) = \frac x {\mu+c} - 1 = \frac{(x-c)-\mu}{\mu+c}\qquad\begin{cases} >0 & \text{if } 0<\mu<x-c, \\ =0 & \text{if }\mu=x-c, \\ <0 & \text{if }\mu>x-c. \end{cases}
$$
In this context the statement that $c$ is "known" means that the family of probability distributions is parametrized only by $\mu$ which $c$ remains the same regardless of the value of $\mu$.  And the word "constant" above means not depending on $\mu$.
This tells you that $x-c$ is the MLE if we don't assume $H_0$.  If we do assume $H_0$ then the MLE is $0$.  The maximum value of the likelihood if we don't assume $H_0$ is then
$$
\frac{(\hat\mu+c)^x e^{-(\hat\mu+c)}}{x!} = \frac{x^x e^{-x}}{x!}.
$$
The maximized value under $H_0$ is $\dfrac{c^x e^{-c}}{x!}$.  The likelihood ratio is therefore
$$
\frac{\left(\dfrac{x^x e^{-x}}{x!}\right)}{\left( \dfrac{c^x e^{-c}}{x!} \right)} = \left(\frac x c\right)^x e^{-(x-c)}.
$$
This is an increasing function of
$$
x\log x - x\log c - x
$$
so you reject $H_0$ if that is too big.  Notice that this statistic is negative until $x$ gets above some point, actually bigger than $c$, and as $x$ grows, it decreases at first and then increases.
