Find the probability density function of $Y=3X+2$.

A continuous random variable $X$ has probability density function,
$$f(x)=\begin{cases}\frac{3}{5}e^{-3x/5}, & x>0 \\ 0, & x\leq0\end{cases}$$
Then find the probability density function of $Y=3X+2$.

So I have difficulty in solving this problem. I am confuse about how to relate the density function of $X$ to that of $Y$. So I look up for hints, so the hint says replace $x$ by $\frac{y-2}{3}$ and transform the domain to $Y$. So the question is why this is true? What is the concept behind it. If anybody can explain this to me, it would be helpful to learn. Thanks.

Solutions Collecting From Web of "Find the probability density function of $Y=3X+2$."

The first thing you have to do is to find the support of the transformed
random variable $Y,$ which is $(2, \infty).$ Then read in your text about
transformation methods. Many texts discuss two.

What is often called the CDF method finds the CDF of $Y$ as $$F_Y(y) = P(Y \le y) = P\left(X \le \frac{y-2}{3}\right),$$
for $y > 2.$ Then you need the formula for the CDF of $X,$ plug in $(y-2)/3)$ for $x$ and differentiate the CDF of $Y$ to get its PDF. (The Answer of @RonaldB has correctly started this.)

The second method for monotone transformations (increasing or decreasing) is
to use $f_Y(y) = f_X(h^{-1}(y))|dh^{-1}/dy|,$ where $h^{-1}$ is the inverse
of the transformation $y = h(x).$ (In this problem $\frac{dh^{-1}(y)}{dy} = 1/3.$) By multiplyiing by 3 in $Y = 3X + 2,$ you have ‘stretched out’
the distribution by a factor of 3, thus the constant needs to be adjusted
to maintain the area under the density curve of $Y$ as unity.

You should learn how to use both methods on this simple problem. You should get the PDF of $Y$ to be $f_Y(y) = .2e^{-.2(y-2)},$ for $y > 2$ (and $0$ otherwise).

In this problem $X \sim Exp(rate = 3/5),$ the exponential distribution with
rate $\lambda = 3/5.$ See your text or Wikipedia exponential distribution.

Here is a simulation in R statistical software in which I have generated 100,000 realizations of $X \sim Exp(3/5),$
transformed them to get $Y = 3X + 2,$ made a histgram of the 100,000 $Y$’s and plotted the PDF of $Y$ through the histogram. (Limits on axes are the same in
both panels to illustrate change in rate.)

 x = rexp(10^5, 3/5)
 y = 3*x + 2
 hist(y, prob=T, br = 50, col="wheat")  # righthand panel
 curve(dexp(x-2,.2), 1, 50, n=10001, col="blue", lwd=2, add=T)
 abline(v=2, col="red")

enter image description here

The random variable $Y$ is sometimes said to have a ‘delayed’ exponential

We want $Pr(Y\lt y)$ where small y is arbitrary we’re given Y=3X+2

So we want $Pr(3X+2\lt k)$ and lets isolate: $Pr\left(X\lt\frac{k-2}{3}\right)$

Steps after that should be straight forward I think


This will give you the CDF, differentiate and you’ll get what you’re looking for.

Let $y=3x+2$. Solve it for $x$ as $x=\frac{y-2}3$.

The bound $x=0$ becomes $y=2$. Put $x=\frac{y-2}3$ in the given $f_X(x)$, and update the domain for $y$. Finally, scale the result to get a valid pdf such that $\int_{-\infty}^{\infty} f_Y(y)dy=1$. Since $f_Y(y)=0$ for $y≤2$, you just need to have $\int_2^\infty f_Y(‌​y)dy=1$.

$$f_Y(y)=\begin{cases}\frac{1}{5}e^{-(y-2)/5}, y>2\\ 0,~~~~~~~~~~~~~~~~ y\leq2\end{cases}$$

Clearly $Y$ takes values in $[2,\infty)$. Now note that the odds $a\le Y\le b$ are the same odds as $(a-2)/3\le X\le (b-2)/3$. This means our density function should be $f_Y(y) = {1\over 3}f_X\left(\displaystyle{x-2\over 3}\right)$.

The basic idea is that $Y$ is spread out more than $X$ since the multiplication by $3$ expands the more concentrated probability from a smaller $X$ interval to a less dense one for a $Y$ interval.