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

Question

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.

Answer 1

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).

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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")

The random variable $Y$ is sometimes said to have a 'delayed' exponential distribution.