Find the p.d.f of $Y=4X+3.$

Question

A random variable $X$ has the p.d.f

$$f_X(x)=\begin{cases} 2x & 0<x<1 \\ 0 & \text{ else} \end{cases} $$

Then find the p.d.f of $Y=4X+3.$

My attempt

$y=4x+3\implies x=(y-3)/4.$ Apply in $f_X(x)$. We get $f_Y(y)=3[(y-3)/4], $ for $0<(y-3)/4<1\implies 3<y<7.$ and $f_Y(y)=0$ else. But answer was given was $f_Y(y)=(3/16)(y-3).$ I am learning the basics of statistics. I am facing this type of problem for the first time. I am not sure about the logic. Please help me.

Answer 1

Hint $:$ Let $F_X$ and $F_Y$ respectively denote the probability distribution functions for $X$ and $Y.$ Then we have $$F_Y(y) = P(Y \leq y) = P \left (X \leq \frac {y - 3} {4} \right ) = F_X \left (\frac {y - 3} {4} \right ).$$ Now take the derivative in both sides with respect to $y$ to find the probability density function of $Y.$ It will be the following $:$ $$f_Y (y) = \begin{cases} \frac {1} {8} (y - 3) & 3 \lt y \lt 7 \\ 0 & \text{otherwise} \end{cases}$$

Answer 2

Using your original approach:

If $Y = g(X)$ then,

$f_Y(y) = f_X(g^{-1}(y)) \cdot \frac{d}{dy} (g^{-1}(y))$

In this case, $Y = g(X) = 4X + 3 \implies X = g^{-1}(Y) = \frac{Y-3}{4}$

So, $f_Y(y) = f_X{\left(\frac{y-3}{4}\right)} \cdot \frac{d}{dy} (((y-3)/4)) ​= \frac{y-3}{8}$
Support of probability distribution is $3 \lt Y \lt 7$ as you mentioned.

Answer 3

This is a location-scale transformation. Let’s take it piece by piece:

$X \to 4X$ is equivalent to stretching the original pdf $f$ by a factor of 4 (I.e., $f(x) \to f(x/4)$). However, we haven’t adjusted the height of $f$ and so it will not integrate to 1, but to 4.

You can see this via simple substitution in calculus:

$$ \int f(x/4) dx, \; u=x/4 \implies du =dx/4 \implies dx = 4du $$

$$\implies 4\int f(u)du = 4$$

Therefore we need to scale by $1/4$ to ensure it’s a valid density.

$$4X \sim 1/4f(x/4)$$

Now the easier part. Adding 3 simply shifts the variable values to the right by 3 so we can implement this as

$$X+3 \sim f(x-3)$$

No scaling needed.

Putting together we get

$$Y =4X +3 \sim 1/4f\left(\frac{y-3}{4}\right)$$

Subbing into your $f_X$ we get:

$$f_Y = (y-3)/8$$

Not sure how you got 3/16