What's the expected value of the expression: $x^{random(y, y+1)}$ where $random(a,b)$ is a uniformly random real number between a and b.
Simple testing confirms my suspicion that this is not simply $x^{E[random(y, y+1)]}$, i.e. $x^{y+0.5}$.
(I'm coming up with some exponential backoff retry logic for a system that involves this randomization to prevent "collisions", and I'm trying to analyze its expected cumulative retry time.)
$\endgroup$2 Answers
$\begingroup$The expected value of $f(X)$ for some random variable $X$ is calculated as follows: $\mathbb{E}(f(X)) = \int f(x)p(x)dx$ for $p$ the probability density function associated to $X$.
In your case we have: $\mathbb{E} x^n = \int_y^{y+1} x^n dn = \left[\frac{1}{\log x}x^n\right]^{y+1}_y = \frac{1}{\log x}\left(x^{y+1}-x^y\right)$
$\endgroup$ 5 $\begingroup$If $Z$ is a uniform random variable on $(y,y+1)$, then it has density function $f_Z$ given by $f_Z (s) = 1$ if $y < s < y+1$, and $f_Z (s) = 0$ otherwise. Thus, $$ {\rm E}[x^Z ] = \int {x^s f_Z (s)ds} = \int_y^{y + 1} {x^s ds} = \ldots . $$
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