Why is the energy probability different from its speed probability in the Maxwell-Boltzmann Distribution? I know this has been asked before here and I understand how the formula changes when the Maxwell-Boltzmann Distribution (MBD) is written in terms of speed. But I am trying to understand this more intrinsically. 
I don't understand why the probability for a particle with energy $E \geq E+dE$ is different from the probability of the speed that belongs to that energy range.
The answer that's being said (I think) is because when the MBD is written in terms of energy, a $dE$ covers a different number of particles than a $dv$ does when it is written in terms of speed.
However, I have a struggle with that answer because of the following:
Starting from a certain speed $v_0$, a $dv$ covers the number of particles that have speeds between $v_0 \geq v_0 + dv$. Those particles would have energies between $\frac{1}{2}mv_0^2 \geq \frac{1}{2}m(v_0 + dv)^2$ which 
is exactly what $dE$ is.
Thus, $dE$ covers those same number of particles that have those speeds $v_0 \geq v_0 + dv$. 
How would the number of particles differ?
 A: It is confusing whats the difference between probabilities and probability densities. If you write the the probability distribution expressed with velocities you get $$ 1 = \int_0^\infty f(v)dv.$$ If you want to calculate the probability of a particle having velocity in some finite interval $a < v <b$ you get $$ P(a<v<b) = \int_a^b f(v)dv.$$ If you make a change of variable to energy $E = \frac{1}{2}mv^2$ you get $$ P(a<v<b) = \int_{ma^2/2}^{mb^2/2} f(v(E)) \frac{dv}{dE}dE.$$ Now you can in particular choose $a=v_0$ and $b=v_0+dv$ and the $\textit{probabilities}$ of being in interval $v_0<v<v_0 +dv$ is equal to $mv_0^2/2< E < m(v_0+dv)^2/2$. This just follows from making a variable change in an integral. However, the probability densities of having velocity $v$ and energy $E=\frac{1}{2}mv^2$ are not equal. This is because the probability density of having energy $E$ is not just $f(v(E))$ but you get a contribution from the derivative $\frac{dv}{dE}$.
So, in general the probability densities are not equal $$f(v) \neq f(v(E)) \frac{dv}{dE}.$$ But when you integrate the probability density you get equal probabilities for corresponding velocity and energy intervals. 
A: $$ E = \frac 1 2 m v^2$$
$$ dE = mv dv$$
so:
$$ \frac{dE}{E} = \frac{mv dv}{\frac 1 2 m v^2}$$
$$\frac{dE}{E} = 2 \frac{dv}{v} $$
Or in the terms you presented:
$$ \frac 1 2 m(v_0 + dv)^2 = \frac 1 2 m(v_0 + 2mv_0dv)$$
(where $dv^2 \rightarrow 0$, has been used):
$$ = E_0 + mdv = E_0 + 2dE \ne E_0 + dE$$
