Poisson distribution of random variable $X$ with parameter $\lambda$ is defined as $p(x;\lambda)=\frac{e^{-\lambda} \lambda^x}{x!}$.
Question: How to show analytically that $p(\lambda)$ is not concave but log-concave in parameter $\lambda$ for a fixed value of $x$?
$p(\lambda)$ looks like quasi-concave shown below:
$\operatorname{Log}p(\lambda)$ looks like log-concave shown below:
My Work: The sign of the second derivative would determine the concavity/convexity of $p(x)$ in $\lambda$ but its second derivative is not that easy to examine:
$p''(\lambda)= \frac{e^{-\lambda} (x-1) x \lambda^{x-2}-2 e^{-\lambda} x \lambda^{x-1}+e^{-\lambda} \lambda^x}{x!}$.
Whereas second derivative of $\operatorname{Log}(p(\lambda))$ is:
$\operatorname{Log}(p(\lambda))^{''}=-e^{\lambda } x x! \lambda ^{-x-1} \left(\frac{e^{-\lambda } x \lambda ^{x-1}}{x!}-\frac{e^{-\lambda } \lambda ^x}{x!}\right)+e^{\lambda } x! \lambda ^{-x} \left(\frac{e^{-\lambda } (x-1) x \lambda ^{x-2}}{x!}-\frac{2 e^{-\lambda } x \lambda ^{x-1}}{x!}+\frac{e^{-\lambda } \lambda ^x}{x!}\right)+e^{\lambda } x! \lambda ^{-x} \left(\frac{e^{-\lambda } x \lambda ^{x-1}}{x!}-\frac{e^{-\lambda } \lambda ^x}{x!}\right).$