Proof that if $Y=2X+3$ then $f_Y(y)=\frac12f_X(\frac{y-3}2)$

Question

I want to prove that if probability density function (PDF) of a random variable $X$ is $f_X(x)$ and another random variable $Y$ is defined as $Y=2X+3$, then the PDF of $Y$ is $f_Y(y)=\frac{1}{2} f_X(\frac{y-3}{2})$.

My Approach:
We Know: $P(X \leq x)= \int_{-\infty}^x f_X(x) \, dx$
similarly,$$P(Y \leq y)= \int_{-\infty}^y f_Y(y) \, dy$$ Now,$\, Y=2X+3$ $$\implies P(Y \leq y)= P((2X+3) \leq y)$$ $$=P(X \leq (\frac{y-3}{2}))$$ $$\implies P(Y \leq y)=\int_{-\infty}^{\frac{y-3}{2}} f_X(y) \, dy$$ $$\implies \int_{-\infty}^y f_Y(y) \, dy=\int_{-\infty}^{\frac{y-3}{2}} f_X(y) \, dy $$ Now how can I proceed further to find a relationship between $f_Y$ and $f_X $?

Answer 1

"We know: $P(X \leq x)= \int_{-\infty}^x f_X(x) \, dx$... "

Observe that on RHS the $x$ is fixed and variable at the same time. That is confusing notation. You should write something like:$$P(X \leq x)= \int_{-\infty}^x f_X(u) \, du$$

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You found: $$P(Y\leq y)=\int_{-\infty}^{\frac{y-3}2}f(u)du=\int_{-\infty}^{+\infty}\mathbf1_{\left(-\infty,\frac{y-3}2\right]}(u)f(u)du$$

(Notice that I introduce $u$ here in order to avoid the confusing notation).

Now substitute $u=\frac{v-3}2$ to get $$\cdots=\int_{-\infty}^{\infty}\mathbf1_{\left(-\infty,\frac{y-3}2\right]}\left(\frac{v-3}2\right)f\left(\frac{v-3}2\right)d\left(\frac{v-3}2\right)=\int^y_{-\infty}\frac12 f\left(\frac{v-3}2\right)dv$$

Answer 2

$$F_Y(y) = F_X\left(\frac{y-3}{2} \right)$$

Now, let's differentiate with respect to $y$ by chain rule

$$\frac{d}{dy}F_Y(y) = \frac{d}{dy}F_X\left(\frac{y-3}{2} \right)=f_X\left(\frac{y-3}{2} \right)\frac{d}{dy}\left(\frac{y-3}{2} \right)$$

that is $$f_Y(y)=\frac12f_X\left(\frac{y-3}{2} \right)$$