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Let $X$ be a random variable with mean $0$ and variance $\sigma ^2$, i.e. $X \sim \mathcal{N}(0, \sigma ^2)$.What is the distribution of $Y= X^n$, $n \in \mathbb {N}.$ ?

I know what distributribution has $Y=aX+b$, but I have not found anything for this case.

Revised version: In the end, only the expectation of $Y$ is needed, not its distribution.

Did
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Edin_91
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    can you calculate the distributional function of $X^n$ ? – Chival Jun 28 '15 at 17:05
  • I don´t understand what you are trying to say, @Chival – Edin_91 Jun 28 '15 at 17:10
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    How do you compute the distribution of G(X), say when G is monotonous and one knows the distribution of X? – Did Jun 28 '15 at 17:54
  • If $Y=g(x)$, then $$f_Y (y)=f_X (g^{-1}(y)) \left| \frac{\partial (g^{-1}(y))}{\partial y} \right| $$ . But I don´t obtain any satisfactory result in this case.. – Edin_91 Jun 28 '15 at 18:10
  • Yet this yields a full answer for every odd exponent n. And now for the even exponents... – Did Jun 28 '15 at 20:31
  • What do you mean, @Did ? – Edin_91 Jun 28 '15 at 20:59
  • I mean that "this yields a full answer for every odd exponent n. And now for the even exponents..." – Did Jun 28 '15 at 21:02
  • But, what do you need to distinguish when $n \in \mathbb{N}$ is odd or even? – Edin_91 Jun 28 '15 at 21:08
  • @Edin_91 Even functions are not monotonic in X –  Jun 28 '15 at 22:02
  • Thanks, Bey:) I have found the next http://mathworld.wolfram.com/NormalProductDistribution.html but does not help me a lot... – Edin_91 Jun 28 '15 at 22:45
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    If $n$ is odd then $X^n \leq y$ is equivalent to $X \leq y^{1/n}$. If $n$ is even, then $X^n \leq y$ for $y \geq 0$ is equivalent to $X \in [-y^{1/n},y^{1/n}]$, and is impossible for $y < 0$. Work from there to get a CDF, and then differentiate to get a PDF. – Ian Jun 28 '15 at 23:03
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    When you google "powers gaussian", the first thing that comes up is this MSE post by Did. – BruceZ Jun 28 '15 at 23:06
  • I don´t finish to understand why I have to sepate the case that n is odd or even. – Edin_91 Jun 28 '15 at 23:11
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    @Edin_91 Because of differences in monotonicity. When $n$ is odd, $f(x)=x^n$ is an increasing function. When $n$ is even, it is not. This means that you have to handle events of the form $X^n \leq y$ differently in the two cases. Think about the graphs of $x^2$ and $x^3$; for $x^2$, the $y$ values are below $64$ on $(-8,8)$ (a bounded interval). For $x^3$, the $y$ values are below $64$ on $(-\infty,4)$ (an unbounded interval). – Ian Jun 28 '15 at 23:26
  • Oh, okay, I have undestood! So, with the formula $$f_Y (y)=f_X (g^{-1}(y)) \left| \frac{\partial (g^{-1}(y))}{\partial y} \right|$$ I can obtain the case when n is only? And the other case? – Edin_91 Jun 28 '15 at 23:28
  • I have realised that for solving my problem I don´t need the distribution of $X^n$, do I? It would be enought to compute $${\mathbb E} [ e^{Y}] = {\mathbb E} [e^{X^n}]=\int_{-\infty}^{+\infty} e^{x^n} \frac{1}{\sqrt{2\pi \sigma ^2}} e^{{-1 \over {2\sigma^2}}(x-\mu)^2}dx$$ and building a perfect square. – Edin_91 Jun 28 '15 at 23:38
  • uhm.. why not just use the moment generating function? –  Jun 29 '15 at 00:37

1 Answers1

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"In the end, only the expectation of Y is needed, not its distribution." This a typical famous university exam question.

As Did already mentioned $E(X^n)=0$ for n being odd. Now for $n$ even ($2m$ say and WLOG assume $\sigma^2=1$), $$E(X^{2m})=2\int_0^{\infty} \frac{x^{2m}\exp(-x^{2}/2)}{\sqrt{2\pi}}dx$$ Substitute, $y=x^2/2$ to make the above as a $Gamma$ integral. The final answer is $\frac{2^m \times\Gamma(m+0.5)}{\sqrt\pi}$ which is $1\times3\times5...\times (2m-1)$

Saty
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