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Given that $X$ follows a Gaussian distribution $e^{-x^2/2\sigma^2}$, what distribution is followed by $X^{1/3}$?

How does one start to solve this problem? I guess it isn't $e^{-x^{2/3}/2\sigma^{2/3}}$ but I'm not really certain how to start.

user1936752
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    check this http://math.stackexchange.com/questions/683359/distribution-of-higher-powers-than-2-of-a-gaussian-distribution – D.Castro Apr 10 '15 at 09:59
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    Review your question based on the fact that math is case-sensitive. If you don't take the effort of can't tell $X$ from $x$ you're hardly going to understand anything in probabilities. – Oskar Limka Apr 10 '15 at 10:11
  • Thanks for the link, I think that answers my question. Apologies for the notation, I didn't know how to write it but I do now. – user1936752 Apr 10 '15 at 10:30

2 Answers2

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You can also write, given any measurable function $g$,

$$E(g(X^{1/3}))=\int_\Bbb R g(x^{1/3})f_X(x)\mathrm{d}x$$

Now, with a change of variable $x=y^3$, and writing $Y=X^{1/3}$,

$$E(g(Y))=\int_\Bbb R g(y)3y^2f_X(y^3)\mathrm{d}y$$

Hence the density of $Y$ is $f_Y(y)=3y^2f_X(y^3)$.

Jean-Claude Arbaut
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  • is that right that $Y = |X|^{q}$ has Gamma distribution iff $q=2$ for standard normal Gaussian variable $X$? – tortue Jun 24 '19 at 16:47
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You use the distribution function: write $Y=X^{1/3}$ $$ P(Y<y)=P(X^{1/3}<y)=P(X<y^3), $$ since the transformation and the distribution function are bijective. You can then differentiate the cumulative function to get the density function.

Jean-Claude Arbaut
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Chappers
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  • I didn't realise you could do that. Does this always hold? $p (X^2 <y)=p (-sqrt {y}<X <sqrt {y}) for example? Thanks – Karl Apr 10 '15 at 10:48
  • @Karl You might think about it like this: $x^{1/3} < y \Leftrightarrow x < y^3$. So $P(X^{1/3} < y) = P(X < y^3)$. – Myridium Apr 10 '15 at 10:50
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    @Karl Yes, of course: $X^2 \leq y$ and $-\sqrt{y}\leq X\leq \sqrt{y}$ describe the same set, so they have the same probability. – Jean-Claude Arbaut Apr 10 '15 at 10:51
  • Thanks I honestly don't know how I've not seen it before. – Karl Apr 10 '15 at 10:57