Solving PDF Integrals of Normal Distribution

Sep 12, 2024
normal distribution statistics

We know that a random variable distribution must satisfy the fundamental probability density function (PDF) requirements for a continuous random variable:

$f(x) \geq 0$ , for all $x \in \mathbb{R}$
$\int_{-\infty}^{\infty} f(x)\,dx = 1$
$P(a < X < b) = \int_a^b f(x)\,dx$

These conditions ensure that the function is a valid probability model over the continuous sample space.

To prove that the total area under the normal distribution curve is equal to 1, we evaluate:

P(-\infty < X < \infty) = \int_{-\infty}^{\infty} f(x)\,dx

Substituting the normal distribution PDF:

= \int_{-\infty}^{\infty} \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx

Standardization Transformation

Let:

z = \frac{x-\mu}{\sigma}

Then:

x = \sigma z + \mu

dx = \sigma dz

Substituting:

= \int_{-\infty}^{\infty} \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(\sigma z + \mu - \mu)^2}{2\sigma^2}} \sigma dz

Simplifying:

= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}} dz

= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-z^2/2} dz

Relating to the Gaussian Integral

Recall:

\int_{-\infty}^{\infty} e^{-y^2}\,dy = \sqrt{\pi}

To match this form, let:

y = \frac{z}{\sqrt{2}}

Then:

z = \sqrt{2}y

dz = \sqrt{2}\,dy

Substituting:

= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-(\sqrt{2}y)^2/2} \sqrt{2}\,dy

Simplifying:

= \frac{1}{\sqrt{\pi}} \int_{-\infty}^{\infty} e^{-y^2} dy

Using the Gaussian integral result:

= \frac{1}{\sqrt{\pi}} \cdot \sqrt{\pi}

= 1

Final Result

P(-\infty < X < \infty) = \int_{-\infty}^{\infty} f(x)\,dx = 1

Thus, it is proven that the total area under the normal distribution curve is exactly 1, satisfying the fundamental requirement of a valid probability density function.