📂Probability Distribution

Expectation of the Power of Normally Distributed Random Variables with Mean Zero

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Random Variable $X$ follows a Normal Distribution $N \left( 0 , \sigma^{2} \right)$, then the expectation of its powers $X^{n}$ can be recursively expressed as follows¹. $$ E \left( X^{n} \right) = (n - 1) \sigma^{2} E \left( X^{n-2} \right) $$ $E \left( X^{n} \right)$ is $0$ when $n$ is odd, and for even it is given by the following². $$ E \left( X^{2n} \right) = \left( 2n - 1 \right)!! \sigma^{2n} $$ Here, the symbol with two exclamation marks $k!! = k \cdot \left( k - 2 \right) \cdots$ represents the Double Factorial.

Explanation

As a well-known corollary, $E \left( X^{4} \right) = 3 \sigma^{4}$ holds, and it can be used in the derivation of the Ito table, among others.

Derivation

There are two methods to obtain the result. The method through Gauss integration serves as a shortcut via a generalized formula, making the derivation easy and fast. The method through partial integration is somewhat tricky, but through the process, one can also obtain recursive formulas. Both methods start with the integration of $E \left( X^{n} \right)$ as follows. $$ E \left( X^{n} \right) = \int_{-\infty}^{\infty} {\frac{ 1 }{ \sqrt{2 \pi} \sigma }} x^{n} e^{-x^{2} / 2 \sigma^{2}} dx $$

Method Using Gauss Integration

Generalization of Gauss Integration: Let $n$ be a natural number. $$ \begin{align*} \int_{-\infty}^{\infty} x^{2n} e^{-\alpha x^{2}}dx =& \dfrac{(2n)!}{n! 2^{2n}}\sqrt{\dfrac{\pi}{\alpha^{2n+1}}} \\ \int_{-\infty}^{\infty} x^{2n+1} e^{-\alpha x^{2}}dx =& 0 \end{align*} $$

Product of Consecutive Odd Numbers: For an integer $n \ge 0$, the following holds. $$ (2n-1) \cdot (2n-3) \cdots 5 \cdot 3 \cdot 1 = \dfrac{(2n)!}{2^{n} (n!)} = (2n-1)!! $$

If $n$ is odd, it becomes $E \left( X^{n} \right) = 0$ by the formula without further ado. If $n$ is even, substitute $\alpha = 1/2 \sigma^{2}$ to obtain the following. $$ \begin{align*} E \left( X^{2n} \right) =& {\frac{ 1 }{ \sqrt{2 \pi} \sigma }} \int_{-\infty}^{\infty} x^{2n} e^{-x^{2} / 2 \sigma^{2}} dx \\ =& {\frac{ 1 }{ \sqrt{2 \pi} \sigma }} \frac{(2n)!}{n! 2^{2n}}\sqrt{\pi 2^{2n+1} \sigma^{4n+2}} \\ =& \frac{(2n)!}{n! 2^{n} 2^{n}}\sqrt{2^{2n} \sigma^{4n}} \\ =& \frac{(2n)!}{n! 2^{n} 2^{n}} 2^{n} \sigma^{2n} \\ =& \frac{(2n)!}{n! 2^{n}} \sigma^{2n} \\ =& (2n-1) !! \sigma^{2n} \end{align*} $$

Method Using Partial Integration

$$ I_{n} := \int_{-\infty}^{\infty} t^{n} e^{-t^{2} / 2} dt $$ Let $I_{n}$ as shown above and use Integration by Parts. This part is somewhat tricky. $$ \begin{align*} - I_{n} =& - \int_{-\infty}^{\infty} t^{n} e^{-t^{2} / 2} dt \\ =& \int_{-\infty}^{\infty} t^{n-1} \cdot {\frac{ - 2 t }{ 2 }} e^{-t^{2} / 2} dt \\ =& \left[ t^{n-1} \cdot e^{-t^{2} / 2} \right]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} (n-1) t^{n-2} e^{-t^{2} / 2} dt \\ =& 0 - (n-1) I_{n-2} \end{align*} $$ To summarize, it is $I_{n} = (n-1) I_{n-2}$, and applying it to the process of calculating $E \left( X^{n} \right)$. Let $t = x / \sigma$, then it becomes $dx = \sigma dt$, thus $$ \begin{align*} E \left( X^{n} \right) =& \int_{-\infty}^{\infty} {\frac{ 1 }{ \sqrt{2 \pi} \sigma }} x^{n} e^{-x^{2} / 2 \sigma^{2}} dx \\ =& \int_{-\infty}^{\infty} {\frac{ 1 }{ \sqrt{2 \pi} \sigma }} \left( \sigma t \right)^{n} e^{-t^{2} / 2 } \cdot \sigma dt \\ =& {\frac{ \sigma^{n} }{ \sqrt{2 \pi} }} \int_{-\infty}^{\infty} t^{n} e^{-t^{2} / 2 } dt \\ =& {\frac{ \sigma^{n} }{ \sqrt{2 \pi} }} I_{n} \\ =& {\frac{ \sigma^{n} }{ \sqrt{2 \pi} }} (n-1) I_{n-2} \\ =& (n-1) {\frac{ \sigma^{2} \cdot \sigma^{n-2} }{ \sqrt{2 \pi} }} \int_{-\infty}^{\infty} t^{n-2} e^{-t^{2} / 2} dt \\ =& (n-1) \sigma^{2} \int_{-\infty}^{\infty} {\frac{ \sigma^{n-2} }{ \sqrt{2 \pi} }} t^{n-2} e^{-t^{2} / 2} dt \\ =& (n-1) \sigma^{2} E \left( X^{n-2} \right) \end{align*} $$ is obtained. Assuming $X$ follows a normal distribution with mean $0$, we have $E \left( X^{1} \right) = 0$, and for all odd $n$, it is $E \left( X^{n} \right) = 0$. For even values, by expanding the recursive formula, the following result is obtained. $$ \begin{align*} E \left( X^{2n} \right) =& (2n-1) \sigma^{2} E \left( X^{2n-2} \right) \\ =& (2n-1) \sigma^{2} \cdot (2n-3) \sigma^{2} E \left( X^{2n-4} \right) \\ \vdots& \\ =& [ (2n-1) (2n-3) \cdots 1 ] \sigma^{2(n-1)} E \left( X^{2} \right) \\ =& (2n-1)!! \sigma^{2n} \end{align*} $$

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