Some Normal Distribution Trivia

Problem 1. Let $f$ be a continuously differentiable function. Evaluate $\displaystyle \int x f'(x^2)\, \mathrm dx$ . Hence, evaluate

$\displaystyle \int x e^{-x^2}\, \mathrm dx,\quad \int \frac{x}{1+x^4}\, \mathrm dx.$

(Click for Solution)

Solution. Make the substitution $u = x^2 \Rightarrow \mathrm du = 2x\, \mathrm dx$ so that

$\begin{aligned} \int x f'(x^2)\, \mathrm dx &= \int x f'(u) \cdot \frac 1{2x}\, \mathrm du \\ &= \frac 12 \int f'(u)\, \mathrm du \\ &= \frac 12 f(u) + C \\ &= \frac 12 f(x^2) + C. \end{aligned}$

Particularise the result to $f(x) = -e^{-x}$ and $f(x) = \tan^{-1}(x)$ to obtain

$\displaystyle \int x e^{-x^2}\, \mathrm dx = -\frac 12 e^{-x^2} + C,\quad \int \frac{x}{1+x^4}\, \mathrm dx = \frac 12 \tan^{-1}(x) + C.$

Assume the following integral identity

$\displaystyle \left( \int_{-\infty}^{\infty} e^{-x^2}\, \mathrm dx \right)^2 = \int_0^{\infty} 2 \pi x e^{-x^2}\, \mathrm dx.$

Problem 2. Evaluate $\displaystyle \int_{-\infty}^{\infty} x^n e^{-x^2}\, \mathrm dx$ for $n = 0, 1, 2$ .

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Solution. For $n = 0$ , we use the given identity

$\begin{aligned} \left( \int_{-\infty}^{\infty} e^{-x^2}\, \mathrm dx \right)^2 &= \int_0^{\infty} 2 \pi x e^{-x^2}\, \mathrm dx \\ &= 2\pi \int_0^{\infty} xe^{-x^2}\, \mathrm dx \\ &= 2\pi \left[-\frac 12 e^{-x^2}\right]_0^\infty \\ &= 2\pi \left(-\frac 12 \cdot 0 + \frac 12 \cdot 1\right) = \pi. \end{aligned}$

Hence, $\displaystyle \int_{-\infty}^{\infty} e^{-x^2}\, \mathrm dx = \sqrt{\pi}$ . For $n = 1$ ,

$\begin{aligned}\int_0^{\infty} x e^{-x^2}\, \mathrm dx &= \left[-\frac 12 e^{-x^2}\right]_{-\infty}^\infty = -\frac 12 \cdot 0 + \frac 12 \cdot 0 = 0. \end{aligned}$

For $n = 2$ , we first integrate by parts to get

$\begin{aligned} \int x^2 e^{-x^2}\, \mathrm dx &= \int x \cdot x e^{-x^2}\, \mathrm dx \\ &= \underbrace{ -\frac 12 e^{-x^2} }_{\mathrm I} \underbrace{ x }_{\mathrm S} - \int \underbrace{ -\frac 12 e^{-x^2} }_{\mathrm I} \underbrace{ 1 }_{\mathrm D}\, \mathrm dx \\ &= -\frac 12 xe^{-x^2} + \frac 12 \int e^{-x^2}\, \mathrm dx. \end{aligned}$

Plugging in the limits,

$\displaystyle \begin{aligned} \int_{-\infty}^{\infty} x^2 e^{-x^2}\, \mathrm dx &= \left[ -\frac 12 xe^{-x^2} \right]_{-\infty}^{\infty} + \frac 12 \int_{-\infty}^{\infty} e^{-x^2}\, \mathrm dx \\ &= (0 - 0) + \frac 12 \sqrt{\pi} = \frac 12 \sqrt{\pi}.\end{aligned}$

Problem 3. Prove that for any $C > 0$ and $n \in \mathbb N$ ,

$\displaystyle \int_{-\infty}^{\infty} Cxe^{-x^2/n}\, \mathrm dx = 0.$

Compute $C, n$ so that

$\begin{aligned} \int_{-\infty}^{\infty} Ce^{-x^2/n}\, \mathrm dx &= \int_{-\infty}^{\infty} Cx^2 e^{-x^2/n}\, \mathrm dx = 1. \end{aligned}$

(Click for Solution)

Solution. Make the substitution $x^2/n = t^2 \Rightarrow \mathrm dx = \sqrt n\, \mathrm dt$ so that

$\begin{aligned} \int_{-\infty}^{\infty} Cxe^{-x^2/n}\, \mathrm dx &= Cn \int_{-\infty}^{\infty} t e^{-t^2}\, \mathrm dt = Cn \cdot 0 = 0 \end{aligned}$

by Problem 2. Similarly,

$\begin{aligned} 1 = \int_{-\infty}^{\infty} Ce^{-x^2/n}\, \mathrm dx &= C \cdot \int_{-\infty}^{\infty} e^{-x^2/n}\, \mathrm dx \\ &= C\sqrt{n} \cdot \int_{-\infty}^{\infty} e^{-t^2}\, \mathrm dt \\ &= C\sqrt{n} \cdot \sqrt{\pi}\end{aligned}$

so that $C = 1/\sqrt{n \pi}$ . Furthermore,

$\begin{aligned} 1 = \int_{-\infty}^{\infty} Cx^2 e^{-x^2/n}\, \mathrm dx &= Cn\sqrt n \int_{-\infty}^{\infty} t^2 e^{-t^2}\, \mathrm dt \\ &= Cn \sqrt n \cdot \frac 12 \sqrt{\pi}. \end{aligned}$

Comparing results yields $n = 2$ , implying $C = 1/\sqrt{2\pi}$ .

Define $\phi(t) := Ce^{-t^2/n}$ , where $C,n$ are determined from Problem 3. It is obvious that $\phi(-t) = \phi(t)$ for any $t \in \mathbb R$ . Define

$\displaystyle \Phi(z) := \int_{-\infty}^z \phi(t)\, \mathrm dt.$

Here, $\phi, \Phi$ are the probability density function and cumulative distribution function respectively of the standard normal distribution, commonly denoted $\mathcal N(0, 1)$ .

Problem 4. Prove the following properties:

For $z \geq 0$ , $\Phi(-z) = 1 - \Phi(z)$ .
$\Phi(0) = 1/2$ .
For any $a < b$ , $\displaystyle \int_a^b \phi(t)\, \mathrm dt = \Phi(b) - \Phi(a)$ .
$\displaystyle \lim_{T \to \pm \infty} \Phi(T) = 1/2 \pm 1/2$ .

(Click for Solution)

Solution. We observe that

$\begin{aligned} \Phi(-z) = \int_{-\infty}^{-z} \phi(t)\, \mathrm dt &= \int_{-\infty}^{-z} \phi(-t)\, \mathrm dt \\ &= \int_{z}^{\infty} \phi(t)\, \mathrm dt \\ &= 1 - \int_{-\infty}^z \phi(t)\, \mathrm dt = 1 - \Phi(z). \end{aligned}$

Setting $z = 0$ , $\Phi(0) = 1 - \Phi(0)$ yields $\Phi(0) = 1/2$ . For the third result,

$\begin{aligned} 1 = \int_{-\infty}^{\infty} \phi(t)\, \mathrm dt &= \int_{-\infty}^a \phi(t)\, \mathrm dt + \int_a^b \phi(t)\, \mathrm dt + \int_b^\infty \phi(t)\, \mathrm dt \\ &= \Phi(a) + \int_a^b \phi(t)\, \mathrm dt + (1- \Phi(b)) \\ &= 1 - (\Phi(b) - \Phi(a)) + \int_a^b \phi(t)\, \mathrm dt. \end{aligned}$

Algebra yields the desired result. For the final result, the case $T \to \infty$ is immediate since

$\displaystyle \lim_{T \to \infty} \Phi(T) = \lim_{T \to \infty} \int_{-\infty}^{T} \phi(t)\, \mathrm dt = \int_{-\infty}^{\infty} \phi(t)\, \mathrm dt = 1.$

Furthermore,

$\displaystyle \lim_{T \to -\infty} \Phi(T) = \lim_{T \to \infty} \Phi(-T) = 1 - \lim_{T \to \infty} \Phi(T) = 1 - 1 = 0.$

Problem 5. For any $\mu, \sigma \in \mathbb R$ with $\sigma > 0$ , define $\phi_{\mu, \sigma}$ by

$\displaystyle \phi_{\mu, \sigma}(x) := K\phi\left( \frac{x-\mu}{\sigma} \right),$

where $K > 0$ is a normalising constant i.e. $\displaystyle \int_{-\infty}^{\infty} \phi_{\mu, \sigma}(x)\, \mathrm dx = 1$ .

Evaluate $K$ . Furthermore, evaluate

$\displaystyle \int_{-\infty}^{\infty} Kx\phi_{\mu, \sigma}(x)\, \mathrm dx , \quad \int_{-\infty}^{\infty} Kx^2\phi_{\mu, \sigma}(x)\, \mathrm dx .$

(Click for Solution)

Solution. Make the substitution $t = (x-\mu)/\sigma \Rightarrow \mathrm dx = \sigma\, \mathrm dt$ so that $\phi_{\mu,\sigma}(x) = K\phi(t)$ , and

$\begin{aligned} 1 = \int_{-\infty}^{\infty} \phi_{\mu, \sigma}(x)\, \mathrm dx &= K\sigma \int_{-\infty}^{\infty} \phi(t)\, \mathrm dt = K \sigma \cdot 1 = K \sigma. \end{aligned}$

Hence, $K = 1/\sigma$ . Similarly,

$\begin{aligned}\int_{-\infty}^{\infty} x\phi_{\mu, \sigma}(x)\, \mathrm dx &= \int_{-\infty}^{\infty} K\cdot (\mu + \sigma t)\cdot \phi(t) \cdot \sigma \, \mathrm dt \\ &= \mu \cdot K\sigma \int_{-\infty}^{\infty} \phi(t)\, \mathrm dt + K\sigma^2 \int_{-\infty}^{\infty} t \phi(t)\, \mathrm dt \\ &= \mu \cdot 1 \cdot 1 + K \sigma^2 \cdot 0 = \mu. \end{aligned}$

Likewise,

$\begin{aligned} \int_{-\infty}^{\infty} x^2 \phi_{\mu, \sigma}(x)\, \mathrm dx &= \int_{-\infty}^{\infty} K \cdot (\mu + \sigma t)^2 \cdot \phi(t) \cdot \sigma\, \mathrm dt \\ &= \int_{-\infty}^{\infty} (\mu^2 + 2 \mu \sigma t + \sigma^2 t^2) \cdot \phi(t) \, \mathrm dt \\ &= \mu^2 \int_{-\infty}^{\infty} \phi(t)\, \mathrm dt + 2 \mu \sigma \mu^2 \int_{-\infty}^{\infty} t\phi(t)\, \mathrm dt + \sigma^2 \mu^2 \int_{-\infty}^{\infty} t^2 \phi(t)\, \mathrm dt \\ &= \mu^2 \cdot 1 + 2 \mu \sigma \cdot 0 + \sigma^2 \cdot 1 \\ &= \mu^2 + \sigma^2. \end{aligned}$

Problem 6. Prove that $\displaystyle \phi_{\mu, \sigma}(x) = \frac 1{\sigma \sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}$ .

(Click for Solution)

Solution. By the previous problems,

$\displaystyle \begin{aligned} \phi_{\mu, \sigma}(x) &= K \phi\left( \frac{x - \mu}{\sigma} \right) \\ &= KC e^{-\frac{(x-\mu)^2/\sigma^2}{n}} \\ &= \frac 1{\sigma} \cdot \frac{1}{\sqrt{2\pi}} e^{-\frac{(x-\mu)^2/\sigma^2}{2}} \\ &= \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}. \end{aligned}$

Remark 1. Problem 6 gives us the standard formula for the probability distribution function of a normal distribution $\mathcal N(\mu, \sigma^2)$ with mean $\mu$ and variance $\sigma^2$ :