Effect of expansion of space on CMB Is it true that the expansion of space time cause the CMB to become microwaves from a shorter wavelength. If it is has the amplitude been increased? 
Seeing as the amplitude has decreased; why hasn't it increased (/"stretched") in the same way the wavelength has?
 A: Just to be clear: by "amplitude" you mean the amplitude of a classical electromagnetic wave -- that is, the peak value of the electric field -- right? In that case, the answer is that the amplitude goes down.
For definiteness, let's consider a wave packet of electromagnetic radiation with some fairly well-defined wavelength. At some early time, it has a wavelength $\lambda_1$ and energy $U_1$. (I'm not calling it $E$ because I want to reserve that for the electric field.) After the Universe has expanded for a while, it has a longer wavelength $\lambda_2$ and a smaller energy $U_2$. (Fine print: wavelengths and energies are measured by a comoving observer -- that is, one who's at rest in the natural coordinates to use.) In fact, the ratios are both just the factor by which the Universe has expanded:
$$
{\lambda_2\over\lambda_1}={U_1\over U_2}={a_2\over a_1}\equiv 1+z,
$$
where $a$ is the "scale factor" of the Universe. $1+z$ is the standard notation for this ratio, where $z$ is the redshift. 
The physical extent of the wave packet is also stretched by the same factor. So the energy density in the wave packet goes down by a factor $(1+z)^2$.
What does that mean about the amplitude of the wave? The energy density in the wave packet is proportional to the electric field amplitude squared. So if the energy density has gone down by $(1+z)^2$, the electric field amplitude must have gone down by $(1+z)$. 
Specifically, if the Universe doubles in size, the wavelength of any given wave packet doubles, and the amplitude (peak value of ${\bf E}$) is cut in half.
A: Yes to the first part of your question. This phenomenon is known as cosmological redshift. Also due to the increase in the volume due to cosmlogical expansion the same amount of photons, as were present during the CMB, now have to occupy a much larger region. Consequently the average temperature ("amplitude" in your words) of such a gas drops from a high of $T \sim 150,000$ deg. Kelvin (corresponding to $T=E/k_B$ with $E$ being the ionization energy of hydrogen (13.6 ev) and $k_B$ is Boltzmann's constant $k_B = 8.6 \times 10^{-5}$ eV/Kelvin) to the currently observed $T'=2.7$ deg. Kelvin.
A: It is easier to think in terms of a gas of photons. The photons (in thermal equilibrium) have their wavelength distributed according with the Bose-Einstein (BE) distribution $$n(p) = \frac{2}{\exp(pc/k_BT) - 1}.$$ 
From the null geodesics in a Friedmann metric we have that the photon momentum decreases with $a^{-1}$ (the wavelength is stretched with the scale factor $a$) and applying the Vlasov equation in a Friedmann metric we have that the temperature also decreases with $a^{-1}$. 
Using the BE distribution and the results above one obtain that the total number of photons ($\propto T^3V$) in a volume $V$ is conserved, therefore, the density of photons decreases with $a^{-3}$. This shows that the energy density decreases with $a^{-4}$ ($a^{-1}$ from the stretch of the wavelength and $a^{-3}$ from the decrement of the number density). Other way to obtain that is to use the continuity equation for a gas of relativistic particles in Friedmann i.e, $$\dot{\rho} + 3\frac{\dot{a}}{a}(\rho+p) = 0,$$ and since $p=\rho/3$ we obtain $\rho \propto a^{-4}$.
Finally, since the energy density is proportional to the electric field squared $E^2$, then, the electric field decreases with $E \propto a^{-2}$.
