Discussion
Let \(X_{1}, X_{2}, \ldots X_{n}\) be a random sample of size n from Poisson distribution with mean \(\lambda\). The log likelihood function \(L(\lambda \mid x_{1}, x_{2}, \ldots x_{n})\) is :
Answer: C
The log likelihood is a concave function of \(\lambda\).
For a Poisson sample the log likelihood is
\[ \begin{aligned} L(\lambda) &= -n\lambda + \Big(\sum x_{i}\Big)\log\lambda \\ &\quad - \sum \log(x_{i}!) \end{aligned} \]
Its second derivative is \(L''(\lambda) = -\dfrac{\sum x_{i}}{\lambda^{2}} \le 0\), so L is concave on \(\lambda \gt 0\) (strictly so whenever some \(x_{i} \gt 0\)). This is why the maximum likelihood estimate \(\hat{\lambda} = \bar{x}\), where \(L'(\lambda) = 0\), is the unique maximum.
L is not constant, since it depends on \(\lambda\). It is not monotonic in general: it rises up to \(\bar{x}\) and falls after it. And it is not convex, because the second derivative is negative, not positive.
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