$\mathbf{\text{PP=RP}}$ consequences

If $\mathbf{\text{PP}}=\mathbf{\text{RP}}$ then you have a collapse to the first level, namely $\mathbf{\text{PH}}=\mathbf{\text{NP}}$. Assume $\mathbf{\text{PP}}=\mathbf{\text{RP}}$, since $\mathbf{\text{PP}}$ is closed under complement we have $\mathbf{\text{RP}}=co\mathbf{\text{RP}}=\mathbf{\text{ZPP}}$, thus $\mathbf{\text{PP}}=\mathbf{\text{ZPP}}$.

Let $L$ be a language in ${\mathbf{\text{PP}}}^{\mathbf{\text{ZPP}}}$, then there exists a probabilistic polynomial time Turing machine with access to a $\mathbf{\text{ZPP}}$ oracle $\mathcal{O}$, call it $M_{\mathcal{O}}$, such that:

$x\in L\Rightarrow Pr\left[M_{\mathcal{O}}\text{ accepts x}\right]>\frac{1}{2}$, and

$x\notin L\Rightarrow Pr\left[M_{\mathcal{O}}\text{ accepts x}\right]\le \frac{1}{2}-\frac{1}{2^{n^c}}$.

Where $n^c$ is the running time of $M_{\mathcal{O}}$ (in the standard definition, the second inequality appears with $\le \frac{1}{2}$, however it can be improved to $<\frac{1}{2}$ and thus to the above).

Suppose that the Turing machine which decides $\mathcal{O}$ has expected runtime $n^d$. Now, let us look at the machine $M$, which replaces each oracle call of $M_{\mathcal{O}}$ by a $t$ steps simulation of the $\mathbf{\text{ZPP}}$ machine for $\mathcal{O}$, and repeats this simulation $k$ times. If we come upon an oracle call, for which this process did not result in an answer (none of the $k$ $t$-steps simulations halted), then $M$ rejects.

Let $H$ denote the event that all of the simulations halted in the given time, then:

$Pr\left[\text{M accepts x}\right]=Pr\left[\text{M accepts x}|H\right]Pr[H]+Pr\left[\text{M accepts x}|\bar{H}\right]Pr\left[\bar{H}\right]=Pr\left[\text{M accepts x}|H\right]Pr[H]=Pr\left[M_{\mathcal{O}}\text{ accepts x}\right]Pr[H]$.

Thus, we have:

$x\in L\Rightarrow Pr\left[\text{M accepts x}\right]>\frac{1}{2}Pr[H]$, and

$x\notin L\Rightarrow Pr\left[\text{M accepts x}\right]\le (\frac{1}{2}-\frac{1}{2^{n^c}})Pr[H]\le \frac{1}{2}-\frac{1}{2^{n^c}}$.

By Markov’s inequality, a single simulation does not halt with probability $\le \frac{n^d}{t}$, so $k$ $t$-steps simulations do not output an answer with probability $\le {\left(\frac{n^d}{t}\right)}^k$. By the union bound we have $Pr[H]\ge 1-n^c{\left(\frac{n^d}{t}\right)}^k\ge 1-\frac{1}{2^{n^c}}$, for $t=2n^d$ and $k=2n^c$. In this case $M$ acheives the following separation:

$x\in L\Rightarrow Pr\left[\text{M accepts x}\right]>\frac{1}{2}(1-2^{-n^c})$, and

$x\notin L\Rightarrow Pr\left[\text{M accepts x}\right]\le \frac{1}{2}-2^{-n^c}<\frac{1}{2}-\frac{1}{2}{2}^{-n^c}$

Which proves $L\in \mathbf{\text{PP}}$, since we can replace the constant $\frac{1}{2}$ with any function $f(x)\in \mathbf{\text{FP}}$.