# 16.4 - Normal Properties

16.4 - Normal PropertiesSo far, all of our attention has been focused on learning how to use the normal distribution to answer some practical problems. We'll turn our attention for a bit to some of the theoretical properties of the normal distribution. We'll start by verifying that the normal p.d.f. is indeed a valid probability distribution. Then, we'll derive the moment-generating function \(M(t)\) of a normal random variable \(X\). We'll conclude by using the moment generating function to prove that the mean and standard deviation of a normal random variable \(X\) are indeed, respectively, \(\mu\) and \(\sigma\), something that we thus far have assumed without proof.

## The Normal P.D.F. is Valid

Recall that the probability density function of a normal random variable is:

\(f(x)=\dfrac{1}{\sigma \sqrt{2\pi}} \text{exp}\left\{-\dfrac{1}{2} \left(\dfrac{x-\mu}{\sigma}\right)^2\right\}\)

for \(-\infty<x<\infty\), \(-\infty<\mu<\infty\), and \(0<\sigma<\infty\). Also recall that in order to show that the normal p.d.f. is a valid p.d.f, we need to show that, firstly \(f(x)\) is always positive, and, secondly, if we integrate \(f(x)\) over the entire support, we get 1.

Let's start with the easy part first, namely, showing that \(f(x)\) is always positive. The standard deviation \(\sigma\) is defined to be positive. The square root of \(2\pi\) is positive. And, the natural exponential function is positive. When you multiply positive terms together, you, of course, get a positive number. Check... the first part is done.

Now, for the second part. Showing that \(f(x)\) integrates to 1 is a bit messy, so bear with me here. Let's define \(I\) to be the integral that we are trying to find. That is:

\(I=\int_{-\infty}^\infty \dfrac{1}{\sigma \sqrt{2\pi}} \text{exp}\left\{-\dfrac{1}{2\sigma^2} (x-\mu)^2\right\}dx\)

Our goal is to show that \(I=1\). Now, if we change variables with:

\(w=\dfrac{x-\mu}{\sigma}\)

our integral \(I\) becomes:

\(I=\int_{-\infty}^\infty \dfrac{1}{\sqrt{2\pi}} \text{exp}\left\{-\dfrac{1}{2} w^2\right\}dw\)

Now, squaring both sides, we get:

\(I^2=\left(\int_{-\infty}^\infty \dfrac{1}{\sqrt{2\pi}} \text{exp}\left\{-\dfrac{x^2}{2} \right\}dx\right) \left(\int_{-\infty}^\infty \dfrac{1}{\sqrt{2\pi}} \text{exp}\left\{-\dfrac{y^2}{2} \right\}dy\right)\)

And, pulling the integrals together, we get:

\(I^2=\dfrac{1}{2\pi}\int_{-\infty}^\infty \int_{-\infty}^\infty \text{exp}\left\{-\dfrac{x^2}{2} \right\} \text{exp}\left\{-\dfrac{y^2}{2} \right\}dxdy\)

Now, combining the exponents, we get:

\(I^2=\dfrac{1}{2\pi}\int_{-\infty}^\infty \int_{-\infty}^\infty \text{exp}\left\{-\dfrac{1}{2}(x^2+y^2) \right\} dxdy\)

Converting to polar coordinates with:

\(x=r\cos\theta\) and \(y=r\sin\theta\)

we get:

\(I^2=\dfrac{1}{2\pi}\int_0^{2\pi}\left(\int_0^\infty \text{exp}\left\{-\dfrac{r^2}{2} \right\} rdr\right)d\theta \)

Now, if we do yet another change of variables with:

\(u=\dfrac{r^2}{2}\) and \(du=rdr\)

our integral \(I\) becomes:

\(I^2=\dfrac{1}{2\pi}\int_0^{2\pi}\left(\int_0^\infty e^{-u}du\right)d\theta \)

Evaluating the inside integral, we get:

\(I^2=\dfrac{1}{2\pi}\int_0^{2\pi}\left\{-\lim\limits_{b\to \infty} [e^{-u}]^{u=b}_{u=0}\right\}d\theta \)

And, finally, completing the integration, we get:

\(I^2=\dfrac{1}{2\pi} \int_0^{2\pi} -(0-1) d \theta= \dfrac{1}{2\pi}\int_0^{2\pi} d \theta =\dfrac{1}{2\pi} (2\pi)=1\)

Okay, so we've shown that \(I^2=1\). Therefore, that means that \(I=+1\) or \(I=-1\). But, we know that \(I\) must be positive, since \(f(x)>0\). Therefore, \(I\) must equal 1. Our proof is complete. Finally.

## The Moment Generating Function

### Theorem

The moment generating function of a normal random variable \(X\) is:

\(M(t)=\text{exp}\left\{\mu t+\dfrac{\sigma^2 t^2}{2}\right\}\)

### Proof

Well, I better start this proof out by saying this one is a bit messy, too. Jumping right into it, using the definition of a moment-generating function, we get:

\(M(t)=E(e^{tX})=\int_{-\infty}^\infty e^{tx}f(x)dx=\int_{-\infty}^\infty e^{tx}\left[\dfrac{1}{\sigma \sqrt{2\pi}} \text{exp}\left\{-\dfrac{1}{2\sigma^2} (x-\mu)^2\right\} \right]dx\)

Simply expanding the term in the second exponent, we get:

\(M(t)=\int_{-\infty}^\infty \dfrac{1}{\sigma \sqrt{2\pi}}\text{exp}\{tx\} \text{exp}\left\{-\dfrac{1}{2\sigma^2} (x^2-2x\mu+\mu^2)\right\} dx\)

And, combining the two exponents, we get:

\(M(t)=\int_{-\infty}^\infty \dfrac{1}{\sigma \sqrt{2\pi}} \text{exp}\left\{-\dfrac{1}{2\sigma^2} (x^2-2x\mu+\mu^2)+tx \right\} dx\)

Pulling the \(tx\) term into the parentheses in the exponent, we get:

\(M(t)=\int_{-\infty}^\infty \dfrac{1}{\sigma \sqrt{2\pi}} \text{exp}\left\{-\dfrac{1}{2\sigma^2} (x^2-2x\mu-2\sigma^2tx+\mu^2) \right\} dx\)

And, simplifying just a bit more in the exponent, we get:

\(M(t)=\int_{-\infty}^\infty \dfrac{1}{\sigma \sqrt{2\pi}} \text{exp}\left\{-\dfrac{1}{2\sigma^2} (x^2-2x(\mu+\sigma^2 t)+\mu^2) \right\} dx\)

And, simplifying just a bit more in the exponent, we get:

Now, let's take a little bit of an aside by focusing our attention on just this part of the exponent:

\((x^2-2(\mu+\sigma^2t)x+\mu^2)\)

If we let:

\(a=\mu+\sigma^2t\) and \(b=\mu^2\)

then that part of our exponent becomes:

\(x^2-2(\mu+\sigma^2t)x+\mu^2=x^2-2ax+b\)

Now, complete the square by effectively adding 0:

\(x^2-2(\mu+\sigma^2t)x+\mu^2=x^2-2ax+a^2-a^2+b\)

And, simplifying, we get:

\(x^2-2(\mu+\sigma^2t)x+\mu^2=(x-a)^2-a^2+b\)

Now, inserting in the values we defined for \(a\) and \(b\), we get:

\(x^2-2(\mu+\sigma^2t)x+\mu^2=(x-(\mu+\sigma^2t))^2-(\mu+\sigma^2t)^2+\mu^2\)

Okay, now stick our modified exponent back into where we left off in our calculation of the moment-generating function:

\(M(t)=\int_{-\infty}^\infty \dfrac{1}{\sigma \sqrt{2\pi}}\text{exp}\left\{-\dfrac{1}{2\sigma^2}\left[(x-(\mu+\sigma^2t))^2-(\mu+\sigma^2t)^2+\mu^2\right]\right\}dx\)

We can now pull the part of the exponent that doesn't depend on \(x\) through the integral getting:

\(M(t)=\text{exp}\left\{-\dfrac{1}{2\sigma^2}\left[-(\mu+\sigma^2t)^2+\mu^2\right]\right\} \int_{-\infty}^\infty \dfrac{1}{\sigma \sqrt{2\pi}}\text{exp}\left\{-\dfrac{1}{2\sigma^2}\left[(x-(\mu+\sigma^2t))^2 \right]\right\}dx\)

Now, we should recognize that the integral integrates to 1 because it is the integral over the entire support of the p.d.f. of a normal random variable \(X\) with:

mean \(\mu+\sigma^2t\) and variance \(\sigma^2\)

That is, because the integral is 1:

\(\int_{-\infty}^\infty \dfrac{1}{\sigma \sqrt{2\pi}}\text{exp}\left\{-\dfrac{1}{2\sigma^2}\left[(x-(\mu+\sigma^2t))^2 \right]\right\}dx = 1\)

\(\text{Since, }\ N(\mu+\sigma^2t, \sigma^2)\)

our moment-generating function reduces to this:

\(M(t)=\text{exp}\left\{-\dfrac{1}{2\sigma^2}\left[-\mu^2-2\mu\sigma^2t-\sigma^4t^2+\mu^2\right]\right\}\)

Now, it's just a matter of simplifying:

\(M(t)=\text{exp}\left\{\dfrac{2\mu\sigma^2t+\sigma^4t^2}{2\sigma^2}\right\}\)

and simplifying a bit more:

\(M(t)=\text{exp}\left\{\mu t +\dfrac{\sigma^2t^2}{2}\right\}\)

Our second messy proof is complete!

## The Mean and Variance

### Theorem

The mean and variance of a normal random variable \(X\) are, respectively, \(\mu\) and \(\sigma^2\).

### Proof

We'll use the moment generating function:

\(M(t)=\text{exp}\left\{\mu t +\dfrac{\sigma^2t^2}{2}\right\}\)

to find the mean and variance. Recall that finding the mean involves evaluating the derivative of the moment-generating function with respect to \(t\) at \(t=0\):

So, we just found that the first derivative of the moment-generating function with respect to \(t\) is:

\(M'(t)=\text{exp}\left(\mu t +\dfrac{\sigma^2t^2}{2}\right)\times (\mu+\sigma^2t)\)

We'll use it to help us find the variance: