Suppose $X\sim\operatorname{Uniform}(0,1)$ and $Y = X^3$. What is the pdf for $Y$ ?
Answer is given as: $$\frac{1}{3y^2}$$
But I get: $$\frac{1}{3\sqrt[3]{y}^2}$$
Asked
Active
Viewed 458 times
1
-
It's often a lot easier to transform commutative distributions, as there are fewer subtleties to remember and take into account. Have you tried that? – Arthur Apr 18 '20 at 18:46
-
Source for this question is: https://ocw.mit.edu/courses/mathematics/18-05-introduction-to-probability-and-statistics-spring-2014/readings/reading-questions-5d/ – georgedeath Apr 07 '21 at 08:23
2 Answers
1
Actually, yours is the correct answer. Note in particular that $\int_0^1\frac{1}{3y^2}dy=\infty\ne1$. The source that claimed $\frac{1}{3\color{red}{y}^2}$ should have said $\frac{1}{3\color{blue}{x}^2}$, which of course agrees with your answer.
J.G.
- 115,835
0
Since $F_Y(y) = \textsf{P}(Y \leq y) = \textsf{P}(X \leq y^{1/3}) = F_X(y^{1/3})$, by the chain rule,
\begin{align} f_Y(y) = (F_Y)'(y) &= \frac13 y^{-2/3} (F_X)'(y^{1/3}) \\ &= \frac{1}{3 y^{2/3}}f_X(y^{1/3}) \end{align}
Now, recall that $f_X(x) = 1$ if $x \in (0,1)$ and $f_X(x) = 0$ otherwise.
azif00
- 20,792