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I have already consulted V.K. Rohatgi, and it has an example where it takes $Y=X^a$ where $a>0$ but the domain of $X$ is positive real values.
Even the theorem for transformation of continuous random values restricts the derivative of $Y$ w.r.t. $X$ to be positive or negative for the entire domain of $X$ where as in the case when $Y = X^3$ derivative of $Y$ w.r.t. $X$ is $0$ at $X=0$.
I am unable to proceed further.

Math Lover
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AshishSinha5
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  • Could you tell us what you want exactly? Do you want to find the density or the CDF? If you are interested in one of those have you tried using the usual expressions to determine an answer? Where did you get stuck? – Jan Sep 05 '18 at 15:20
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    $P(Y=0)=P(X=0)=0$ so you do not need to worry about the derivative being zero only at that single point – Henry Sep 05 '18 at 15:25

2 Answers2

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I 've been doing some questions from Rohtagi also. It's a very good book. And to answer your question.we are give $X\sim N(0,1),f(x)=\dfrac{1}{\sqrt{2\pi}}e^\frac{-x^{2}}{2} ;-\infty<x<\infty$

$Y=X^3$ is a monotone function and for the monotone function, you can directly apply the transformation formula given to us which is

$f(y)=f(x)\bigg|\dfrac{dx}{dy}\bigg|$

$Y=X^3 \implies X=Y^{\frac{1}{3}}$

$\dfrac{dx}{dy}=\dfrac{1}{3}y^{-\frac{2}{3}}$

$f(x)=f(y^{\frac{1}{3}})$

combining it we have $f(y)=f(y^{\frac{1}{3}})\dfrac{1}{3}y^{-\frac{2}{3}}=\dfrac{1}{\sqrt{2\pi}}e^\frac{{-y^ \frac{2}{3}}}{2}\cdot\dfrac{1}{3}y^{-\frac{2}{3}} ;-\infty<y<\infty$

Daman
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We have

\begin{align*} P(Y \leq y) &= P(X^3 \leq y)\\ &= P(X \leq y^\frac{1}{3}) \end{align*}

Now we take the derivative in this case with $F$ being the CDF of X and f being the pdf of X.

$$F(y^\frac{1}{3})' = \frac{1}{3}f(y^\frac{1}{3})y^{-\frac{2}{3}}$$

Thus you just subsitute in the PDF of X.

Hongyu Wang
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