Assume that $X$ is a random variable following Poisson distribution with parameter $\lambda $. Note that $X$ is non-negative, i.e., $X>0$. I want to calculate the expectation of ${{\log }_{2}}X$. What I have done is as follows:\begin{align} \mathbb{E}\left\{ {{\log }_{2}}X \right\}&=\sum\limits_{k=1}^{\infty }{{{\log }_{2}}k\frac{{{\lambda }^{k}}}{k!}{{e}^{-\lambda }}} \\ & =\sum\limits_{k=0}^{\infty }{{{\log }_{2}}\left( k+1 \right)\frac{{{\lambda }^{k+1}}}{\left( k+1 \right)!}{{e}^{-\lambda }}} \\ & =\lambda \sum\limits_{k=0}^{\infty }{\frac{{{\log }_{2}}\left( k+1 \right)}{k+1}\frac{{{\lambda }^{k}}}{k!}{{e}^{-\lambda }}} \\ & =\lambda \mathbb{E}\left\{ \frac{{{\log }_{2}}\left( Y+1 \right)}{Y+1} \right\} \end{align}
The problem of calculating the expectation of ${{\log }_{2}}X$ has been converted to calculating the expectation of $\frac{{{\log }_{2}}\left( Y+1 \right)}{Y+1}$ with $Y\sim Poisson\left( \lambda \right)$. The variable $Y$ follows a standard Poisson distribution of mean $\lambda $ without any restriction.
How can I make a further analysis to get the approximation of the expectation or the upper bound or lower bound of $\mathbb{E}\left\{ \frac{{{\log }_{2}}\left( Y+1 \right)}{Y+1} \right\}$. Any help is greatly appreciated.
Another method to get the upper bound with Jensen's inequality is as follows:\begin{align} \mathbb{E}\left\{ {{\log }_{2}}X \right\}& \le {{\log }_{2}}\left( \mathbb{E}\left\{ X \right\} \right) \\ & ={{\log }_{2}}\left( \sum\limits_{k=1}^{\infty }{k\frac{{{\lambda }^{k}}}{k!}{{e}^{-\lambda }}} \right) \\ & ={{\log }_{2}} \lambda \end{align}
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