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Given a random variable $X\sim\mathcal N(0,\sigma^2)$, how can we prove that $E[X^4]=3\sigma^4$? I am having trouble even starting with the proof.

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  • $\begingroup$ Hint: $\int x^4p(x)dx$. Integrate by parts. $\endgroup$
    – user65203
    Sep 7, 2016 at 8:38
  • $\begingroup$ ... and then do partial integration a couple times :) $\endgroup$
    – b00n heT
    Sep 7, 2016 at 8:39
  • $\begingroup$ Or compute a bunch of derivatives of the moment generating function of $X$. $\endgroup$
    – Surb
    Sep 7, 2016 at 8:44
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    $\begingroup$ Hint: let $X=\sigma U$ where $U$ has standard normal distribution. Then $\mathbb EX^4=\mathbb E\sigma^4U^4=\sigma^4\mathbb EU^4$. It remains to prove that $\mathbb EU^4=3$. For this see the other hints. If in calculations parameters can be avoided then do so. $\endgroup$
    – drhab
    Sep 7, 2016 at 8:44
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    $\begingroup$ 1) Using the following web keywords "fourth order moment normal distribution proof", I have obtained at once for example this 2) The technical name for the fourth order moment is "kurtosis". $\endgroup$
    – Jean Marie
    Sep 7, 2016 at 9:07

5 Answers 5

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First with $\sigma=1$, omitting the range $(-\infty,\infty)$ for convenience and integrating twice by parts

$$E[X^4]=\frac{\displaystyle\int x^4e^{-x^2/2}dx}{\displaystyle\int e^{-x^2/2}dx}=\frac{-x^3e^{-x^2/2}+3\displaystyle\int x^2e^{-x^2/2}dx}{\displaystyle\int e^{-x^2/2}dx}=\frac{0-3xe^{-x^2/2}+3\displaystyle\int e^{-x^2/2}dx}{\displaystyle\int e^{-x^2/2}dx}=3.$$

Then by rescaling the variable,

$$3\sigma^4.$$


By observing the pattern, you easily generalize to

$$E[X^{2n}]=(2n-1)!!\sigma^{2n}.$$

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I list some hints below.

The probability density function of a normally distributed random variable with mean $0$ and variance $\sigma^2$ is

\begin{equation} f(x) = \frac{1}{\sqrt{2 \pi \sigma^2}} \mathrm{e}^{-\frac{x^2}{2\sigma^2}}. \end{equation}

In general, you compute an expectation of a continuous random variable as

\begin{equation} \mathbb{E}[g(X)] = \int_{-\infty}^\infty g(x) f(x) \, \mathrm{d}x. \end{equation}

For your particular question we have that $g(x) = x^4$ and therefore

\begin{equation} \mathbb{E}[X^4] = \int_{-\infty}^\infty x^4 f(x) \, \mathrm{d}x = \frac{1}{\sqrt{2 \pi \sigma^2}} \int_{-\infty}^\infty x^4 \mathrm{e}^{-\frac{x^2}{2\sigma^2}} \, \mathrm{d}x. \end{equation}

You can solve this integral by using partial integration a number of times.

An alternative approach is to determine the moment generating function and differentiate. The moment generating function of a continuous random variable $X$ is defined as

\begin{equation} M_X(t) := \mathbb{E}[\mathrm{e}^{tX}] = \int_{-\infty}^\infty \mathrm{e}^{t x} f(x) \, \mathrm{d}x, \quad t \in \mathbb{R}. \end{equation}

For your random variable $X$ we have

\begin{equation} M_X(t) = \frac{1}{\sqrt{2 \pi \sigma^2}} \int_{-\infty}^\infty \mathrm{e}^{t x} \mathrm{e}^{-\frac{x^2}{2\sigma^2}} \, \mathrm{d}x. \end{equation}

Conveniently

\begin{equation} \mathbb{E}[X^n] = \frac{\mathrm{d}^n}{\mathrm{d}t^n} M_X(t) \bigg\vert_{t = 0}. \end{equation}

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    $\begingroup$ When using the PDF with $g(x) = x^4$, at what point does the $\frac{1}{\sqrt{2 \pi \sigma^2}}$ term disappear? $\endgroup$
    – M Smith
    Apr 18, 2018 at 13:33
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@Yves Daoust already has a great answer, but we can omit a round of integration by parts, by just working it out in the most straight-forward way.

Applying integration by parts once, $$ \mathbb{E} (x^4) = \int x^4 \varphi(x) dx = 0 + 3\sigma^2 \int x^2 \varphi(x) dx $$

where $\varphi(x)$ is normal PDF and $\varphi(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{x^2}{2\sigma^2}}$

We recognise $\int x^2 \varphi(x) dx = \sigma^2$. So,

$$ \mathbb{E} (x^4) = 3\sigma^2 \sigma^2 = 3 \sigma^4 $$

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Here's a solution with no integrals.

Let $Y, Z \sim \mathcal{N}(0, \sigma^2 / 2)$ be independent Gaussian random variables. Then $Y + Z \sim \mathcal{N}(0, \sigma^2)$. We wish to determine the value of $R = \mathbb{E} [(Y + Z)^4]$.

To do so, note that $$ \begin{align} R &= \mathbb{E}[(Y + Z)^4] \\ &= \mathbb{E}[Y^4 + 3Y^3Z + 6Y^2Z^2 + 3YZ^3 + Z^4] \\ &= \mathbb{E}[Y^4] + 3\mathbb{E}[Y^3]\mathbb{E}[Z] + 6\mathbb{E}[Y^2]\mathbb{E}[Z^2] + 3\mathbb{E}[Y] \mathbb{E}[Z^3] + \mathbb{E}[Z^4] \\ &= R / 4 + 0 + 6 \sigma^4 / 4 + 0 + R / 4. \end{align} $$ Thus $R = R / 2 + 6\sigma^4 / 4$, so $R = 3\sigma^4$.

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  • $\begingroup$ never seen a proof like this! $\endgroup$
    – Car Loz
    May 21, 2023 at 4:10
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The moment generating function of the standard normal $$M_{Z}(t)= e^{t^2/2}$$ has fouth derivative $$M_{Z}''''(t)=3M_{Z}''(t)+tM_{Z}'''(t)$$ Setting $t=0$ results in $E(Z^4) = 3$. Now, $$ E(X^4)=\sigma^4E(Z^4)=3\sigma^4 $$

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  • $\begingroup$ Where does $\sigma^4$ come from? $\endgroup$ Oct 28, 2022 at 18:30
  • $\begingroup$ @RylanSchaeffer It is assumed that $X\sim N(0,\sigma^2)$ and $Z = X/\sigma \sim N(0,1)$. $\endgroup$
    – Car Loz
    Oct 31, 2022 at 20:36

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