I am recording observed speeds in a histogram with fixed bin size. The histogram also keeps track of the variance and delay. Now given a fixed distance $d$ I need to derive a histogram of delays from this (how long it will possibly take to go the distance). So the bins of the new histogram are determined by $\frac{d}{v_i}$ (with $v_i$ the speed recorded in the $i$th bin).
I could of course recalculate the mean and variance for the new histogram, but I was wondering if it is possible to calculate these from the original histogram's mean and variance?
Right now I'm recording the inverse of the speed $1/v$ instead of $v$ as this allows me to use $\mu_t = \mu_{1/v} * d$ and $\sigma_t^2 = \sigma_{1/v}^2 * d^2$, but I'm afraid that $v^{-1} = \frac{t}{d}$ will explode because $d$ could be arbitrarily small.
Without further assumptions or info about the underlying distribution, I don't think you can get $E[1 / X]$ from $E[X]$ and $E[X^2]$.
Here $X$ is the random variable you call `speed'. If $X$ is sort of continuous then what you want is
$$ E[\frac{1}X] = \int \frac{1}u\cdot f_X(u)\,\mathrm{d}u$$
integrated over the relevant domain. Equivalently, denote a new random variable that is the speed-inverse as $Y \equiv 1/X$
$$ E[Y] = \int u\cdot f_Y(u)\,\mathrm{d}u$$
which is the same thing as the integral above with $~f_X(u)$. This is just the variable transformation in calculus.
Most introductory probability/statistics emphasize that $E[1 / X] \neq 1/E[X]$, and as far as I know, there doesn't exist some fancy expression to obtain the first inverse moment from other moments, esp. with just the first two moments.
One can make assumptions about the shape of $~f_X$ and obtain theoretical bounds for $E[1 / X]$ using $E[X]$ and $E[X^2]$, but I don't see the point in that since you already have the empirical distribution.
Do you have a semi-theoretical model for the speed? Is this some kind of physics experiment?