[Math] How to derive the expected value of even powers of a standard normal random variable

expectationnormal distributionprobability

I am trying to prove that, for a standard normal random variable $Z \sim N(0,1)$, ${\mathbb E}[z^n]=n!!$ for even values of $n$. What I'm doing is integrating the p.d.f. of $Z$ which is $\frac{1}{\sqrt{2\pi}}exp(-x^2/2)$, i.e.,

$${\mathbb E}[z^n] = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}x^n exp(-x^2/2)$$

While trying to use integration by parts, I'm unable to get either function into a form where the above reduces to some sort of a recurrence relation. Can you help?

Best Answer

\begin{align*} {\mathbb E}[z^n] &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}x^n exp(-x^2/2)\\ &=\frac{1}{\sqrt{2\pi}}\left[(n-1)x^{n-2}xe^{-x^2/2}\right]^{\infty}_{-\infty}-\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}-(n-1)x^{n-2}e^{-x^2/2}dx\\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(n-1)x^{n-2}e^{-x^2/2}dx\\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(n-1)!!e^{-x^2/2}dx\\ &= (n-1)!! \end{align*}

Related Solutions

Probability Distributions – Characteristic Function of a Standard Normal Random Variable

I will give two answers:

Do it without complex numbers, notice that

$$ \begin{eqnarray} \mathcal{F}(\omega) = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x &=& \int_{0}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x + \int_{-\infty}^0 \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x \\ &=& \int_{0}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x + \int_{0}^{\infty} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{-j \omega x} \mathrm{d} x \\ &=& 2 \int_{0}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \cos(\omega x) \mathrm{d} x \end{eqnarray} $$ Now, compute $\mathcal{F}^\prime(\omega)$, and integrate by parts: $$\begin{eqnarray} \mathcal{F}^\prime(\omega) &=& -\frac{2}{\sqrt{2\pi}} \int_0^\infty \mathrm{e}^{-\frac{x^2}{2}} x \sin(\omega x) \mathrm{d} x = \frac{2}{\sqrt{2\pi}} \int_0^\infty \sin(\omega x) \mathrm{d} \left( \mathrm{e}^{-\frac{x^2}{2}} \right) \\ &=& \frac{2}{\sqrt{2\pi}} \left. \mathrm{e}^{-\frac{x^2}{2}} \sin(\omega x) \right|_0^\infty - \frac{2}{\sqrt{2\pi}} \int_0^\infty \mathrm{e}^{-\frac{x^2}{2}} \omega \cos(\omega x) \mathrm{d} x \\ &=& - \omega \mathcal{F}(\omega) \end{eqnarray} $$ The solution to so obtained ODE, $\mathcal{F}^\prime(\omega) = - \omega \mathcal{F}(\omega)$ is $\mathcal{F}(\omega) = c \exp\left( - \frac{\omega^2}{2} \right)$, and the integration constant is seen to be one from normalization requirement $\mathcal{F}(0)=1$ of the Gaussian probability density.

Complex integration: As you have started, complete the square: $$ \left( -\frac{x^2}{2} + j \omega x \right) = \left( -\frac{x^2}{2} + j \omega x + \frac{\omega^2}{2} \right) - \frac{\omega^2}{2} = -\frac{1}{2} \left( x - j \omega \right)^2 - \frac{\omega^2}{2} $$ We then have: $$ \mathcal{F}(\omega) = \mathrm{e}^{-\frac{\omega^2}{2}} \cdot \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{1}{2} \left( x - j \omega \right)^2 } \mathrm{d} x $$ The integral $I = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{1}{2} \left( x - j \omega \right)^2 } \mathrm{d} x$ is indeed $1$. To see this, consider $$ \begin{eqnarray} I_L &=& \int_{-L}^L \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{1}{2} \left( x - j \omega \right)^2 } \mathrm{d} x = \int_{-L-j \omega}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z \\ &=& \left(\int_{-L-j \omega}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z\right) + \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z \\ &=& \left(\int_{-L-j \omega}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z\right) + \mathcal{I}_L \end{eqnarray} $$ Here we denoted $\mathcal{I}_L = \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z$. Notice that $\lim\limits_{L \to \infty} \mathcal{I}_L = 1$.

Consider a complex contour $\mathcal{C}$, $ -L \to L \to L - j \omega \to -L - j \omega \to -L$: $$ \begin{eqnarray} I_L - \mathcal{I}_L &=&\left(\int_{-L-j \omega}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z\right) \\ &=& -\int_\mathcal{C} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{L}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L-j \omega}^{-L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z \end{eqnarray} $$ The integral over $\mathcal{C}$ is zero, since the integrand is holomorphic. Therefore: $$ I-1 = \lim_{L \to \infty} (I_L-\mathcal{I}_L) = \lim_{L \to \infty} \left( - \int_{L}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L-j \omega}^{-L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z \right) $$ And the limit above is easily seen to vanish. Indeed: $$ \lim_{L\to\infty} \left| \mathrm{e}^{-\frac{-(-L - j \omega t)^2}{2}} \right| = \lim_{L\to\infty} \left| \mathrm{e}^{-\frac{-(L^2 + \omega^2 t^2)}{2}} \right| =0. $$

Probability – Expected Value of Normal Distribution Given Positive Distribution

Let $f(x)$ be the density of $X$; let $F(x)$ be its CDF.

Then the density of $X$, conditional on it being positive, is $f(x)/P(X \ge 0)$ if $x \ge 0$, and $0$ otherwise.

Of course $P(X \ge 0) = 1/2$ by symmetry, so the density of $X$ conditional on $X \ge 0$ is $2f(x)$ (on $x \ge 0$).

So you need to do the integral $$ \int_0^\infty 2xf(x) \: dx $$ which is twice the integral you've written.

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Probability Distributions – Characteristic Function of a Standard Normal Random Variable

Probability – Expected Value of Normal Distribution Given Positive Distribution

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