# 16.5 - The Standard Normal and The Chi-Square

16.5 - The Standard Normal and The Chi-Square## Theorem

We have one more theoretical topic to address before getting back to some practical applications on the next page, and that is the relationship between the normal distribution and the chi-square distribution. The following theorem clarifies the relationship.

If \(X\) is normally distributed with mean \(\mu\) and variance \(\sigma^2>0\), then:

\(V=\left(\dfrac{X-\mu}{\sigma}\right)^2=Z^2\)

is distributed as a chi-square random variable with 1 degree of freedom.

### Proof

To prove this theorem, we need to show that the p.d.f. of the random variable \(V\) is the same as the p.d.f. of a chi-square random variable with 1 degree of freedom. That is, we need to show that:

\(g(v)=\dfrac{1}{\Gamma(1/2)2^{1/2}}v^{\frac{1}{2}-1} e^{-v/2}\)

The strategy we'll take is to find \(G(v)\), the cumulative distribution function of \(V\), and then differentiate it to get \(g(v)\), the probability density function of \(V\). That said, we start with the definition of the cumulative distribution function of \(V\):

\(G(v)=P(V\leq v)=P(Z^2 \leq v)\)

That second equality comes, of course, from the fact that \(V=Z^2\). Now, taking note of the behavior of a parabolic function:

we can simplify \(G(v)\) to get:

\(G(v)=P(-\sqrt{v} < Z <\sqrt{v})\)

Now, to find the desired probability we need to integrate, over the given interval, the probability density function of a standard normal random variable \(Z\). That is:

\(G(v)= \int^{\sqrt{v}}_{-\sqrt{v}}\dfrac{1}{ \sqrt{2\pi}}\text{exp} \left(-\dfrac{z^2}{2}\right) dz\)

By the symmetry of the normal distribution, we can integrate over just the positive portion of the integral, and then multiply by two:

\(G(v)= 2\int^{\sqrt{v}}_0 \dfrac{1}{ \sqrt{2\pi}}\text{exp} \left(-\dfrac{z^2}{2}\right) dz\)

Okay, now let's do the following change of variables:

\(\text{Let}\ z=\sqrt{y}=y^{1/2}\)

\(\text{So, }\ dz=\frac{1}{2}y^{-1/2}\ dy=\frac{1}{z\sqrt{y}}dy\)

\(z^2=y,\ \\ \begin{aligned} z&=0 \Rightarrow y=0 \\ z&=\sqrt{v} \Rightarrow y=v\end{aligned}\)

Doing so, we get:

\(G(v)= 2\int^v_0 \dfrac{1}{ \sqrt{2\pi}}\text{exp} \left(-\dfrac{y}{2}\right) \left(\dfrac{1}{2\sqrt{y}}\right) dy\)

\(G(v)= \int^v_0 \dfrac{1}{ \sqrt{\pi}\sqrt{2}} y^{\frac{1}{2}-1} \text{exp} \left(-\dfrac{y}{2}\right) dy\)

for \(v>0\). Now, by one form of the Fundamental Theorem of Calculus:

\(\text{Given, }\)

\(\int^x_a f(t)dt=F(x)-F(a)\)

\(\frac{d}{dx}\int^x_a f(t)dt=F'(x)=f(x)\)

we can take the derivative of \(G(v)\) to get the probability density function \(g(v)\):

\(g(v)=G'(v)= \dfrac{1}{ \sqrt{\pi}\sqrt{2}} v^{\frac{1}{2}-1} e^{-v/2}\)

for \(0<v<\infty\). If you compare this \(g(v)\) to the first \(g(v)\) that we said we needed to find way back at the beginning of this proof, you should see that we are done if the following is true:

\(\Gamma \left(\dfrac{1}{2}\right)=\sqrt{\pi}\)

It is indeed true, as the following argument illustrates. Because \(g(v)\) is a p.d.f., the integral of the p.d.f. over the support must equal 1:

\(\int_0^\infty \dfrac{1}{ \sqrt{\pi}\sqrt{2}} v^{\frac{1}{2}-1} e^{-v/2} dv=1\)

Now, change the variables by letting \(v=2x\), so that \(dv=2dx\). Making the change, we get:

\(\dfrac{1}{ \sqrt{\pi}} \int_0^\infty \dfrac{1}{ \sqrt{2}} (2x)^{\frac{1}{2}-1} e^{-x}2dx=1\)

Rewriting things just a bit, we get:

\(\dfrac{1}{ \sqrt{\pi}} \int_0^\infty \dfrac{1}{ \sqrt{2}}\dfrac{1}{ \sqrt{2}} x^{\frac{1}{2}-1} e^{-x}2dx=1\)

And simplifying, we get:

\(\dfrac{1}{ \sqrt{\pi}} \int_0^\infty x^{\frac{1}{2}-1} e^{-x} dx=1\)

Now, it's just a matter of recognizing that the integral is the gamma function of \(\frac{1}{2}\):

\(\dfrac{1}{ \sqrt{\pi}} \Gamma \left(\dfrac{1}{2}\right)=1\)

Our proof is complete.

So, now that we've taken care of the theoretical argument. Let's take a look at an example to see that the theorem is, in fact, believable in a practical sense.

## Example 16-4

Find the probability that the standard normal random variable \(Z\) falls between −1.96 and 1.96 in two ways:

- using the standard normal distribution
- using the chi-square distribution

#### Solution

The standard normal table (Table V in the textbook) yields:

\(P(-1.96<Z<1.96)=P(Z<1.96)-P(Z>1.96)=0.975-0.025=0.95\)

The chi-square table (Table IV in the textbook) yields the same answer:

\(P(-1.96<Z<1.96)=P(|Z|<1.96)=P(Z^2<1.96^2)=P(\chi^2_{(1)})<3.8416)=0.95\)