How to calculate the density of $Y=X^4$

Question

Let $X$ be a uniformly distributed variable on $[0,1]$. What is the density of $Y=X^4$? How do you calculate it? Thank you

score 4 · Answer 1 · answered Jun 22 '13 at 18:22

4

In general, if $g$ is monotonic, and $Y = g(X)$, then $Y$ has density $$f_Y(y) = \left| \frac{d}{dy} (g^{-1}(y)) \right| \cdot f_X(g^{-1}(y)),$$ where $f_X$ is the density of $X$, and $g^{-1}$ is the inverse of $g$. See e.g. Wikipedia.

answered Jun 22 '13 at 18:22

fuglede

6,716

score 4 · Accepted Answer · answered Jun 22 '13 at 18:22

4

We take the slow way, by first computing the cumulative distribution function $F_Y(y)$ of $Y$. So we want $\Pr(Y\le y)$.

First do the really easy parts. If $y\le 0$, then $F_Y(y)=0$. If $y\ge 1$, then $f_Y(y)=1$.

Now we deal with the interesting part, where $0\lt y\lt 1$. For such $y$, we have $$F_Y(y)=\Pr(Y\le y)=\Pr(X^4\le y)=\Pr(X\le y^{1/4})=y^{1/4}.$$

For the density function $f_Y(y)$ of $Y$, find the derivative of $F_Y(y)$.

answered Jun 22 '13 at 18:22

André Nicolas

507,029

1

You're too fast. I was still dredging up my memory of the vague outlines of how to approach this problem when you submitted your answer. – dfeuer Jun 22 '13 at 18:24
It helps to have taught basic probability theory for quite a while. – André Nicolas Jun 22 '13 at 18:29
so the density function is f(x)=1/4*x^(-3/4) right?? Thank you Andre – Ryan Jun 22 '13 at 21:23
You are welcome. The conventional name for the variable is $y$. And on should add that it is that on $(0,1)$, and $0$ elsewhere. – André Nicolas Jun 22 '13 at 21:35

How to calculate the density of $Y=X^4$

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