The standard Normal distribution¶

The standard normal distribution is perhaps the most recognizable continuous distribution. It has that characteristic bell shape. We write:

\[ Z \sim N(0,1), \]

and we read it as:

\(Z\) follows the standard Normal.

Now the 0 in \(N(0,1)\) is the expected value and the 1 is the variance. We will see that later.

The PDF of the standard Normal¶

We commonly use the function \(\phi(z)\) to represent the PDF of the standard Normal. It is:

\[ \phi(z) = \frac{1}{\sqrt{2\pi}}\exp\left\{-\frac{z^2}{2}\right\}. \]

So, this is an exponential that has \(z^2\) inside it. The term \(\frac{1}{\sqrt{2\pi}}\) is there so that the PDF is normalized, i.e.:

\[ \int_{-\infty}^{+\infty}\phi(z)dz = 1. \]

Here is how you can make a standard Normal in scipy.stats:

import matplotlib.pyplot as plt
%matplotlib inline
import seaborn as sns
sns.set(rc={"figure.dpi":100, 'savefig.dpi':300})
sns.set_context('notebook')
sns.set_style("ticks")
from IPython.display import set_matplotlib_formats
set_matplotlib_formats('retina', 'svg')
import numpy as np
import scipy.stats as st

Z = st.norm()

And here are some samples from it:

Z.rvs(size=10)

array([ 0.07650491, -0.55433992,  0.35401783,  0.26572118, -0.5496704 ,
        0.47381155, -0.53989222,  1.13163748,  1.23093609,  0.42745344])

And here is the PDF of the standard normal:

fig, ax = plt.subplots()
zs = np.linspace(-6.0, 6.0, 100)
ax.plot(zs, Z.pdf(zs))
ax.set_xlabel('$z$')
ax.set_ylabel('$\phi(z)$');

Here are some important properties of the PDF of the standard Normal:

First, \(\phi(z)\) is positive for all \(z\).
Second, as \(z\) goes to \(-\infty\) or \(+\infty\), \(\phi(z)\) goes to zero.
Third, \(\phi(z)\) has a unique mode (maximum) at \(z=0\). In other words, \(z=0\) is the most probable point under this distribution.
Fourth, \(\phi(z)\) is symmetric about \(z=0\). Mathematically:

\[ \phi(-z) = \phi(z). \]

Let’s test some of them using scipy.stats.

Z.pdf(np.inf)

0.0

Z.pdf(-np.inf)

0.0

Z.pdf(-5) == Z.pdf(5)

True

Expectation of the standard normal¶

The expectation of \(Z\) is:

\[ \mathbf{E}[Z] = \int_{-\infty}^{+\infty}z\phi(z)dz = 0. \]

You can prove this quite easily by invoking the fact that \(\phi(-z) = \phi(z)\).

Here it is in scipy.stats:

Z.expect()

0.0

Variance of the standard normal¶

The variance of \(Z\) is:

\[ \mathbf{V}[Z] = \int_{-\infty}^{+\infty}z^2\phi(z)dz = 1. \]

You need integration by parts to prove this. Let’s do it.

\[\begin{split} \begin{split} \mathbf{V}[Z] &= \int_{-\infty}^{+\infty}z^2\phi(z)dz\\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} z^2\exp\left\{-\frac{z^2}{2}\right\}dz\\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} (-z)\cdot\left[-z\exp\left\{-\frac{z^2}{2}\right\}\right]dz\\ &=-\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} z\cdot\frac{d}{dz}\left[\exp\left\{-\frac{z^2}{2}\right\}\right]dz\\ &= -\frac{1}{\sqrt{2\pi}}\left\{\left[z\exp\left\{-\frac{z^2}{2}\right\}\right]_{-\infty}^{+\infty} -\int_{-\infty}^{+\infty} \exp\left\{-\frac{z^2}{2}\right\}dz\right\}\\ &= -\frac{1}{\sqrt{2\pi}}\left(0 -\sqrt{2\pi}\right)\\ &= 1. \end{split} \end{split}\]

Note that the standard deviation of \(Z\) is also 1 (\(Z\) does not have units).

Again, here it is in scipy.stats:

Z.var()

1.0

The CDF of the standard Normal¶

The CDF of \(Z\) gives you the probability that \(Z\) is smaller than a number \(z\). It is common to use \(\Phi(z)\) to denote the CDF. There is no closed form. But you can write this:

\[ \Phi(z) = \int_{-\infty}^z \phi(z') dz', \]

where \(\phi(z)\) is the PDF of \(Z\). Let’s plot it.

fig, ax = plt.subplots()
ax.plot(zs, Z.cdf(zs))
ax.set_xlabel('$z$')
ax.set_ylabel('$\Phi(z)$');

Some properties of the CDF of the standard normal¶

First, note that:

\[ \Phi(0) = 0.5. \]

This follows very easily from the symmetry of the PDF \(\phi(z)\) about zero. Remember that \(\Phi(0)\) is the probability that \(Z\) is smaller than zero and because of symmetry that probability is exactly 50%. A point \(z\) with such a property (the probability that the random variable is smaller than it is 0.5) is called the median of the random variable.

Now here is a non-trivial property. Take any number \(z\). Then, we have:

\[ \Phi(-z) = 1 - \Phi(z). \]

Before attempting to prove this property, let’s demonstrate it visually. Take \(z\) to be some positive number. Then \(\Phi(-z)\) is the probability that \(Z\) is smaller than \(-z\), or this area below the PDF:

\[ \Phi(-z) = \int_{-\infty}^{-z}\phi(z')dz'. \]

On other hand, \(1 - \Phi(z)\) is the following area:

\[\begin{split} \begin{split} 1 - \Phi(z) &= \int_{-\infty}^{+\infty}\phi(z')dz' - \int_{-\infty}^z\phi(z')dz'\\ &= \int_{-\infty}^{z}\phi(z')dz' + \int_{z}^{+\infty}\phi(z')dz' - \int_{-\infty}^z\phi(z')dz'\\ &= \int_{z}^{+\infty}\phi(z')dz'. \end{split} \end{split}\]

Notice that we used the fact that \(\phi(z)\) is normalized and a standard property of the integral.

Alright, so visually, the expression \(\Phi(-z) = 1 - \Phi(z)\) means that the red and the blue areas in the following plot are the same for any \(z\):

fig, ax = plt.subplots()
ax.plot(zs, Z.pdf(zs))
z = 1
zsb = np.linspace(-6.0, -z, 100)
zsa = np.linspace(z, 6.0, 100)
ax.fill_between(zsb, 0.0, Z.pdf(zsb), color='b', alpha=0.5)
ax.fill_between(zsa, 0.0, Z.pdf(zsa), color='r', alpha=0.5)
ax.set_xlabel('$z$')
ax.set_ylabel('$\phi(z)$');

And this makes a lot of sense since \(\phi(z)\) is symmetric. So, in words the property says:

The probability that \(Z\) is smaller than \(-z\) is the same as the probability that \(Z\) is greater than \(z\).

The formal proof is actually trivial. It goes like this:

\[\begin{split} \begin{split} \Phi(-z) &= \int_{-\infty}^{-z}\phi(z')dz'\\ &= \int_{+\infty}^{z}\phi(\tilde{z})(-1)d\tilde{z}, \end{split} \end{split}\]

after applying the transformation \(\tilde{z} = -z'\). And finally:

\[ \int_{+\infty}^{z}\phi(\tilde{z})(-1)d\tilde{z} = -\int_{+\infty}^{z}\phi(\tilde{z})d\tilde{z} = \int_{z}^{+\infty}\phi(\tilde{z})d\tilde{z}, \]

which as we saw above is the same as \(1 - \Phi(z)\).

Let’s demonstrate this in scipy.stats:

print('p(Z <= -1) = {0:1.3f}'.format(Z.cdf(-1)))

p(Z <= -1) = 0.16

And this should be the same as the probability that \(Z\) is greater than 1, which is:

print('p(Z >= 1) = {0:1.3f}'.format(1 - Z.cdf(1)))

p(Z >= 1) = 0.16

What is the probability that \(Z\) is between \(-1\) and \(1\)? It is:

\[ p(-1 < Z < 1) = 1 - p(Z \le -1) - p(Z \ge 1) = 1 - 2 \Phi(-1). \]

In scipy.stats:

print('p(-1 < Z < 1) = {0:1.3f}'.format(1 - 2.0 * Z.cdf(-1)))

p(-1 < Z < 1) = 0.683

Questions¶

Modify the code above to find the probability that \(Z\) is between -2 and 2.
Repeat for \(Z\) between -3 and 3.

Introduction to Data Science for Mechanical Engineers (Lecture Book)

The standard Normal distribution

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