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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.

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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")
```

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

distribution.

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

EDIT:

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.

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