Most probably value of velocity does not coincide with most probable energy for a 2D ideal gas I'm having trouble understanding the physical concept of the following statistical mechanical problem.
I'm given the function
$$ \Phi(v_x) = \Big(\dfrac{m}{2\pi \kappa T}\Big)^{1/2} \cdot e^{-mv_x^2/2\kappa T}.$$
To solve the density of a a speed $v = [v_x, v_y]$
$ \Phi(v_x,v_y) = \Phi(v_x)\Phi(v_y) = \Big(\dfrac{m}{2\pi \kappa T}\Big) \cdot e^{-m(v_x^2+v_y^2)/2\kappa T}$
A coordinate transfer taking $dxdy=r\ dr\ d\theta$
$f(v)dv = \int_0^{2\pi} d\theta \ v\Big(\dfrac{m}{2\pi \kappa T}\Big) \cdot e^{-m(v_x^2+v_y^2)/2\kappa T}$
Which gives
$f(v) = 2\pi v\big(\dfrac{m}{2\pi \kappa T}\big) \cdot e^{-mv^2/2\kappa T}$
The most probable speed is then just
$v_{\text{mpv}} = \sqrt{\kappa T/m}$
I would have expected that the most probable kinetic energy would then be
$\epsilon_{\text{mpv}} = \dfrac{mv_{\text{mpv}}^2}{2}$
However by doing the correct transformations the distribution of $\epsilon$ looks instead like
$f(\epsilon) = \dfrac{e^{-\epsilon/\kappa T}}{\kappa T}$
which has a most probable value of
$\epsilon_{\text{mpv}} = 0$
I think I missed something conceptually as to why the most probable values of $\epsilon$ and $v$ don't coincide. Any help is greatly appreciated.
Edit: I would like to add that this is not a homework problem, but an example problem. I took a stab at it without looking at the answer and am confused by the disconnect with my logic and the answer.
 A: I believe your result is correct. To undestand better this counter intuitive result, let me make the problem simpler by considering a random variable $v$ between $0$ and $1$ with the density $f(v) = 3/2 v^{1/2}$. Since $f(0) = 0$, you will agree that the most probable value of $v$ cannot be in any case $v=0$ (it is in fact $v = 1$). Now if we calculate the interated probability $F(x) = P(v < x)$ we find that $F(x) = x^{3/2}$. If we introduce a new variable $E = v^2$ with values between $0$ and $1$ as well, you can find the integrated probability $G(x)$ associated with $E$ by simply noting that $F(x) = P(v < x) = P(E < x^2) = G(x^2)$. So in the end $G(x) = x^{3/4}$ and the associated density is $g(x) = 3/4 x^{-1/4}$. Suddenly the "most probable" value of $E$ is around $0$ (the density even goes to infinity as $x \to 0$).
By the way, this is also why we can "equate" the two densities if we take into account the differential elements $g(E)dE = f(v)dv$. This simply comes from $G(E(v)) = F(v)$, which becomes $\frac{dE}{dV} g(E(v)) = f(v)$ after derivating with respect to $v$.
A similar behaviour occurs with the thermal velocity distribution in $2 \mathrm{D}$. The difference in most probable velocity/energy simply comes from the differential elements $dv$ and $dE$. In any case, don't attach too much significance to the "most probable" value. In general, the most probable value for $v$ or $E$ differs from the "typical" value, which can be defined either as the median velocity/energy or the mean velocity/energy.
A: I stumbled also into this problem and took me a while to realize why the most probable speed and the most probable energy do not have to correspond.
I want to preface this by saying that it is always the integral of the distribution function that matters, not the value of the distribution function itself. Therefore, asking for the most probable velocity is not really physically meaningful -- there are zero particles with exactly  the most probable velocity! (There are zero particles with any exact velocity value, you can only ask "How many particles are there with velocities in a given interval")
The problem of why the maxima of the distribution functions do not match is because the transformation of variables here is not linear ($E \propto v^2$). The transformation of the distribution function preserves area under the distribution function at each interval:
$$
   f(v) dv = g(E) dE
$$
Now, if you transform your x-axis of the distribution function with $E \propto v^2$, you will have to spread out the number of particles such that the area-preserving relation holds at every portion of the distribution. This will modify the shape of the function and displace your maxima.
Let me show this by an animation. In the following I start with a histogram (approximation of the distribution function) for the speed variable (2D case), then modify the variable shown to be $\xi = v^p / 2$ where I change $p$ to be from $1$ to $3$. Notice that at $p = 2$ this corresponds to the energy distribution.
(The plotting script is here.)

