# Transformation of random variable $Y=X^4$ when $X$ is continuous with uniform distribution

Given $X$ is a random variable with uniform distribution over $(-2, 1)$ and $Y=X^4$, where $Y$ is also a random variable. Find distribution function $F_Y$ and pdf $f_Y$.

I am using the monotonic (inverse) transformation method with $x= \pm\sqrt[4]y$. So$$F_Y(y)= F_X(\sqrt[4]y) - F_X(-\sqrt[4]y)\implies F_Y(y)= \dfrac23\sqrt{y}.$$

Since the value of $x$ is in $(-2,1)$ and that of $y$ in $(0,16)$, my final answer was$$F_Y(y) = \begin{cases} \dfrac23\sqrt{y}; & 0< y <16\\ \dfrac16\sqrt[4]{y^{-3}}; &0 < y < 16 \end{cases}$$

The answer was only given for $f_Y(y)$ and it was$$f_Y(y) = \begin{cases} \dfrac16\sqrt[4]{y^{-3}}; & 0 < y < 1\\ \dfrac1{12}\sqrt[4]{y^{-3}}; &1 < y < 16 \end{cases}$$

Any idea where I went wrong?

• $<$ is a special character in the markup (which is why it was cutting off your question at "0<y<1" before you edited it) . More generally, you should learn to use MathJax for readability. – spaceisdarkgreen Jun 4 '18 at 23:37

You can't use the inverse transform method because the mapping $X^4$ is not monotone. You can instead apply the monotone transform theorem to the sets where it is monotone, namely $(-2,0)$ and $(0,1)$.

Specifically, $X = g^{-1}(Y)$ is a well defined inverse on $(-2,1)$ and $(0,1)$ respectively. The Jacobian of this transformation is $\left|\frac{d}{dy}g^{-1}(y)\right| = \frac{1}{4}y^{-3/4}$. So, the pdf of $Y$ is given by,

$$f_Y(y) = \frac{1}{12}y^{-3/4}\chi_{(0,16)}(y) + \frac{1}{12}y^{-3/4}\chi_{(0,1)}(y) = \frac{1}{12}y^{-3/4}\chi_{(1,16)}(y)+\frac{1}{6}y^{-3/4}\chi_{(0,1)}(y)$$

• Thanks. Is my set up correct then that FY(y)= FX(y^1/4) - FX(-y^(1/4) then solve over both intervals (-2,0) and (0,1)? Or is it just FX(y^1/4) for (0,1) values of y since they are all positive, and FX(-y^1/4) for (-2,0) since they are all negative? – Altered Beast Jun 4 '18 at 23:50
• In this case, I would try to find the pdf directly. – Flowsnake Jun 5 '18 at 0:06

$$F_X(x) = \dfrac{2+x}{3}\mathbf 1_{x\in (-2;1)}+\mathbf 1_{x\in[1;\infty)}$$

$X$ is supported over $(-2;1)$ which is $(-2;-1)\cup[-1;0)\cup[0;1)$

Notice that applying $g:X\mapsto X^4$ to these subintervals gives: $\lower{3ex}{\begin{split}g(-2;1)&=(1;16)\\g[-1;0)&=(0;1]\\g[~~~0;1)&=[0;1)\end{split}}$

So remember to account for this fold of two subintervals into the same subinterval. $$\begin{split}F_Y(y) &= \begin{cases} 0&:& y\in(-\infty;0)\\ F_X(\sqrt[4]y)-F_X(-\sqrt[4]y) & :& y\in [0;1]\\ F_X(1)-F_X(-\sqrt[4]y) &:& y\in(1;16)\\ 1&:& y\in [16;\infty)\end{cases}\\ &=\dfrac{1}{3} \begin{cases} 0&:& y<0)\\ 2\sqrt[4]y & :& 0\leqslant y\leqslant 1\\ 1+\sqrt[4]y &:& 1<y<16\\ 1&:& 16\leqslant y\end{cases}\\[2ex] f_Y(y) & = \dfrac{\mathsf d~F_Y(y)}{\mathsf d ~y\qquad}\end{split}$$

And since $\dfrac{\mathsf d~\sqrt[4]y}{\mathsf d~y\quad}= \dfrac 1{4\sqrt[4]y^3}$, the rest is straightforward.