377. The density function $f(x)$ of a random variable $X$, whose possible values are contained in the interval $(a, b)$, is given. Find the density function of the random variable $Y=3X$.
S o l u t i o n. Since the function $y=3 x$ is differentiable and strictly increasing, the formula

$$
g(y)=f[\psi(y)] \cdot\left|\psi^{\prime}(y)\right|
$$

can be applied, where $\Psi(y)$ is the inverse function of $y=3 x$.

Let's find $\psi(y):$

Let's find $f[\psi(y)]:$

$$
\psi(y)=x=y / 3
$$

$$
f[\psi(y)]=f(y / 3)
$$

Let's find the derivative $\boldsymbol{\psi}^{\prime}(y)$ :

$$
\phi^{\prime}(y)=\left(y^{\prime} 3\right)^{\prime}=1 / 3
$$

It is obvious that

$$
\left|\Psi^{\prime}(y)\right|=1 / 3 \text {. }
$$

Let's find the required density function by substituting (**) and (***) into $\left(^{*}\right): g(y)=(1 / 3) f \mid(y / 3)$.

Since $x$ varies in the interval $(a, b)$ and $y=3 x$, then $3 a<y<3 b$.