Consider integers $m$ and $n$, with $m < n$.
Consider a random variable $X$, distributed according to the $\text{Gamma}~(2^{n-m}, 2^{-n})$ distribution. Consider the random variable
$$\mathrm{Z} = \frac{2^{m}}{2}\left|X - \frac{1}{2^{m}}\right|.$$
I am trying to compute the expected value of $\mathrm{Z}$. It is given by the following integral.
$$ \mathbb{E}(\mathrm{Z}) = \frac{2^{m}}{2}\int_{0}^{\infty} \left| x - \frac{1}{2^{m}} \right| \frac{\left(2^{-n}\right)^{-2^{n-m}} e^{-2^n x} x^{2^{n-m}-1}}{\Gamma \left(2^{n-m}\right)}dx. $$
A calculation in Mathematica tells me that the value is \begin{equation} \frac{2^{2^{-m} \left(2^m-2^n\right) (m-n)} e^{-2^{n-m}}}{\Gamma \left(2^{n-m}\right)}. \end{equation}
Now, consider the cases when $m \leq \log(n)$ and $m \leq n - \log(n)$.
I am trying to find an asymptotic upper bound on the expected value, for both cases. For example, is it upper bounded by
\begin{equation} \frac{\text{poly}(n)}{2^{\text{poly}(n)}}, \end{equation} for some polynomial in n (or maybe, some other inverse polynomial or inverse exponential upper bound)? I tried some numerical simulations with trivial examples (like $\frac{1}{2^{n/2}}, \frac{n}{2^{n/4}} \frac{1}{n} )$ etc. These do not work.
It is known that $$ \Gamma (x) \ge \sqrt {2\pi }\, x^{x - 1/2} \mathrm{e}^{ - x} $$ for all $x>0$ (cf. $(5.6.1)$). Because of Stirling's formula, this is a rather sharp lower bound. This gives the simple bound $$ \frac{{2^{2^{ - m} (2^m - 2^n )(m - n)} \mathrm{e}^{ - 2^{n - m} } }}{{\Gamma (2^{n - m} )}} \le \frac{1}{{\sqrt {2\pi } \,2^{(n - m)/2} }}. $$ When $m \le \log n$ or $m \le n - \log n$, we obtain the upper bounds $$ \frac{1}{{\sqrt {2\pi } \,2^{\frac{{n - \log n}}{2}} }},\quad \frac{1}{{\sqrt {2\pi } \,2^{\frac{1}{2}\log n} }} $$ respectively.