Is $f(t)R(x;\sigma = 1)$ equivalent to $R(x; \sigma = f(t))$, where $R(x; \sigma)$ is a PDF with a single mode defined by $\sigma$? If I have some function $f(t)$, which is known, and a PDF - for example the Rayleigh distribution:
$$R(x;\sigma) = \frac{x}{\sigma^2}e^{-x^{2}/(2 \sigma^2)}$$
and I use this function, $f(t)$, to define the mode (or shape parameter) of the Rayleigh distribution, such that
$$\sigma = f(t) \text{.}$$
This means that we can have a random variable, $F$ which is now drawn from
$$F(t) \sim \frac{x}{f(t)^2}e^{-x^{2}/(2 f(t)^2)}.$$
This must, "somehow", be connected, or equivalent, with the following
$$G(t) \sim f(t) R(x;\sigma = 1).$$
which could represent some noisy data which follows $f(t)$.
Are these expressions equivalent, or related, and if so how may I show this?

Additional thoughts/rationale
As I see it, assuming one fixes the seed of a random number generator, both $G(t)$ and $F(t)$ must produce the same random number, as:

*

*In the case of $F(t)$ the mode, $\sigma$ is now defined by the function $f(t)$, so the most likely number to be drawn from the distribution $F(t)$ should indeed be $f(t)$, for some value of $t$.

*In the case of $G(t)$ we have the function, f(t), scaled by a random number which is most likely to be $1$, as we have defined the mode of the Rayligh distribution to be $\sigma = 1$
Apart from this "logical" reasoning I can't find a way to prove/disprove this.

I have tried one additional train of thought, whereby if we consider
$$f(t)R(x;,\sigma = 1)$$
to be the "global" picture, and
$$R(x;,\sigma = f(t))$$
to be the local one. If we perform the following integral
$$\int_{0}^{\infty} R(x;\sigma) \ {\rm{d}}\sigma = \sqrt{\frac{\pi}{2}}$$
which is the mean of a Rayleigh distribution when $\sigma = 1$ -- the mean being defined as
$$\mu =  \sqrt{\frac{\pi}{2}} \sigma $$
This may be spurious connection I am making out of desperation.

Illustrative simulation
This may be superfluous, but I thought an illustrative simulation may be helpful. It is written in Mathematica but it should be relatively easy to follow. Consider the function $f(t)$, I have defined it as a Lorentzian peak like function as an arbitrary function
f[w_, t0_, t_] := Abs[w^2 / (w^2 + (t - t0)^2)]

Then consider the two cases, as discussed above:
CaseOne = Table[RandomVariate[RayleighDistribution[f[500, 5000, t]]],{t, 1, 10000}];
CaseTwo = Table[f[500, 5000, t] RandomVariate[RayleighDistribution[1]],{t, 1, 10000}];

If we examine the histograms of CaseOne (red) and CaseTwo (blue) and both overlapped (purple), we can see the distributions are identical.

 A: Let $X$ be a random variable with the standard Rayleigh pdf:
$$R(x;\sigma=1) = xe^{-x^{2}/2}$$
In other words, $X$ is one of the data points you are randomly generating. If we multiply the data point by some constant $c>0$ we get a data point $Y = cX$. You are asking whether the pdf of $Y$ is:
$$R(y;\sigma=c) = \frac{y}{c^2}e^{-y^{2}/(2c^2)}$$

Proof: Denote the probability that $X$ is less than or equal to some real number $x$ by:
$$P(X \leq x)$$
Then, the probability $Y$ is less than or equal to some real number $y$ is:
$$\begin{align}
P(Y \leq y) &= P(cX \leq y) \\
&= P(X \leq y/c)
\end{align}$$
Now, the pdf of $Y$ is defined as the derivative of $P(Y \leq y)$ with respect to $y$:
$$\begin{align}
\frac{d}{dy}P(Y \leq y) &= \frac{d}{dy} P(X \leq y/c) \\
&= \frac{d}{dy} \int_{0}^{y/c} R(x, \sigma = 1) dx
\end{align}$$
By Leibniz's integral rule:
$$= c^{-1}R(y/c, \sigma=1)$$
Plugging $x=y/c$ into the pdf $R(x, \sigma = 1) = xe^{-x^{2}/2}$ gives:
$$\begin{align}
&= c^{-1} \left( \frac{y}{c}e^{-(y/c)^{2}/2} \right) \\
&= \frac{y}{c^2}e^{-y^{2}/(2c^2)}
\end{align}$$
