Find Probability Density Function of $Y = X^3$, where $X$ is an exponential distribution of parameter $4$.

Question

Random variable $X$ is an exponential distribution of parameter $4$.

I have another random variable, $Y = X^3$. I need to find probability density function of Y, $f_y(y)$.

Here's what I did: $$F_y(y) = P(Y < y) = P(X^3 < y) = F_y(x^3)$$

now, I took the derivative [I've used the chain rule], $$3\cdot x^2f_y(x^3)$$

does it make sense?

Answer 1

Here's what I did: $$F_y(y) = P(Y < y) = P(X^3 < y) = F_y(x^3)$$

The last step is clearly invalid. From where did the free variable $x$ come ? Just substituting $y=x^3$ is of no help.

The subscripts may indicate your confusion. The letters $Y$ and $y$ do not mean the same thing. $Y$ is a random variable, while $y$ is a value; likewise $X$ and $x$. The subscripts should be capitals, showing that they are the CDF for the Random Variables, like so: $F_Y(\cdot)$ and $F_X(\cdot)$ .

You do not want an expression in $x$ involving the function $F_Y(\cdot)$. Rather, you want an expression in $y$ involving the function $F_X(\cdot)$ .

So you should have:

$$\begin{align} F_{\small Y}(y) &= P(Y \leqslant y) \\&= P(X^3 \leqslant y)\\&= P(X\leqslant y^{1/3}) \\&= F_{\small X}(y^{1/3})\end{align}$$

And since $X\sim\mathcal{Exp}(4)$ then $F_{\small Y}(y) = (1-\mathrm e^{-4 y^{1/3}})\,\mathbf 1_{0\leqslant y}$

now, I took the derivative [I've used the chain rule],

That is the correct move, just from the wrong place. Apply it to:

$$f_{\small Y}(y)= \dfrac {\mathrm d~~}{\mathrm d y}F_{\small X}(y^{1/3})$$