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# Moment Generating Function

schedule Mar 5, 2023
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Before reading this guide, please make sure you're familiar with the concept of moments. If not, please check out the first part of our guide (TODO) here first.

# What is moment generating function?

We can describe a random variable $X$ using numerical measures such as its mean and standard deviation. However, these numerical measures do not uniquely capture the distribution of $X$ since many different distributions can still possess the same mean and standard deviation. One way of uniquely characterizing a distribution is by referring to its moment-generating function.

This makes moment-generating functions extremely useful when proving certain theorems such as the reproductive property of the normal distribution.

Definition.

# Moment-generating function

The moment-generating function $M_X(t)$ of a random variable $X$ is defined as:

$$M_X(t)=\mathbb{E}(e^{tx})$$
Example.

## Finding the moment-generating function of a discrete random variable

Find the moment-generating function of a discrete random variable $X$ with the following probability mass function:

$$p_X(x)= \begin{cases} 0.2\text{ if } x=3\\ 0.8\text{ if } x=4 \end{cases}$$

Solution. Using the definition of moment-generating function:

\begin{align*} M_X(t)&=\mathbb{E}(e^{tx})\\ &=\sum^2_{i=1}e^{tx_i}\cdot{p_X(x_i)}\\ &=(e^{3t})(0.2)+(e^{4t})(0.8)\\ &=0.2e^{3t}+0.8e^{4t} \end{align*}
Example.

## Finding the moment-generating function of the standard normal distribution

Suppose a random variable $X$ follows the standard normal distribution:

$$f_X(x)=\frac{1}{\sqrt{2π}} \mathrm{exp}\left(\frac{-x^2}{2}\right)$$

Prove that the moment-generating function of $X$ is:

$$M_X(t)= \exp\Big(\frac{t^2}{2}\Big)$$

Solution. Using the definition of moment-generating function:

\begin{align*} M_X(t)&=\mathbb{E}(e^{tx})\\ &=\int^\infty_{-\infty} \frac{1}{\sqrt{2\pi}}\exp(tx)\cdot\exp\left(\frac{-x^2}{2}\right)\;dx\\ &=\int^\infty_{-\infty} \frac{1}{\sqrt{2\pi}}\exp\left(\frac{-x^2}{2}+tx\right)\;dx\\ &=\int^\infty_{-\infty} \frac{1}{\sqrt{2\pi}}\exp\left(\frac{-x^2+2tx}{2}\right)\;dx\\ &=\int^\infty_{-\infty} \frac{1}{\sqrt{2\pi}}\exp\left(\frac{-(x^2-2tx+t^2)+t^2}{2}\right)\;dx\\ &=\int^\infty_{-\infty} \frac{1}{\sqrt{2\pi}}\exp\left(\frac{-(x-t)^2}{2}+\frac{t^2}{2}\right)\;dx\\ &=\exp\Big(\frac{t^2}{2}\Big) \int^\infty_{-\infty} \frac{1}{\sqrt{2\pi}} \exp\left(\frac{-(x-t)^2}{2}\right)\;dx\\ \end{align*}

Now, here's the 🤯 step - the function inside the integral is actually a normal distribution with mean $t$ and variance $1$. Since the area of all probability density functions must equal one, the integral above will equal one:

$$M_X(t)= \exp\Big(\frac{t^2}{2}\Big)$$

This completes the proof.

Theorem.

# Derivative of moment-generating function

Let $X$ be any random variable. The $k$-th derivative of the moment-generating function of $X$ evaluated at $t=0$ gives the $k$-th moment of $X$ about the origin:

$$M_X^{(k)}(0)=\mathbb{E}(x^k)=\mu'_k$$

If you don't know what the $k$-th moment of $X$ about the origin means, please consult the the first section of our guide on moments. TODO

Proof. From Taylor's series, we know that $e^{tx}$ can be represented as follows:

$$e^{tx}= \frac{(tx)^0}{0!} +\frac{(tx)^1}{1!} +\frac{(tx)^2}{2!} +\frac{(tx)^3}{3!} +\cdots +\frac{(tx)^n}{n!}$$

Let's now take expected value of both sides:

\label{eq:E8GLN2o41NCrdpqjVLd} \begin{aligned}[b] M_X(t)&= \mathbb{E}(e^{tx})\\&= \mathbb{E}\Big(\frac{(tx)^0}{0!}+\frac{(tx)^1}{1!}+\frac{(tx)^2}{2!} +\cdots +\frac{(tx)^n}{n!}\Big)\\ &= \mathbb{E}\Big(\frac{(tx)^0}{0!}\Big) +\mathbb{E}\Big(\frac{(tx)^1}{1!}\Big) +\mathbb{E}\Big(\frac{(tx)^2}{2!}\Big) +\cdots +\mathbb{E}\Big(\frac{(tx)^n}{n!}\Big)\\ &=\frac{t^0}{0!}\mathbb{E}(x^0) +\frac{t^1}{1!}\mathbb{E}(x^1) +\frac{t^2}{2!}\mathbb{E}(x^2) +\cdots +\frac{t^n}{n!}\mathbb{E}(x^n)\\ \end{aligned}

Using the definition of expected value:

\label{eq:ruURQ7S1eURQjBz05Jy} \begin{aligned}[b] M_X(t)&= \frac{t^0}{0!}\mathbb{E}(x^0) +\frac{t^1}{1!}\mathbb{E}(x^1) +\frac{t^2}{2!}\mathbb{E}(x^2) +\cdots +\frac{t^n}{n!}\mathbb{E}(x^n)\\ &=\frac{t^0}{0!}\sum_x(x^0p(x)) +\frac{t^1}{1!}\sum_x(x^1p(x)) +\frac{t^2}{2!}\sum_x(x^2p(x)) +\cdots +\frac{t^n}{n!}\sum_x(x^np(x))\\ \end{aligned}

Now, let's define $\mu'_i$ like so:

$$$$\label{eq:v5cgn94QCe5jSpdeTpN} \mu'_i= \mathbb{E}(x^i) =\sum_x{x^ip(x)}$$$$

Using \eqref{eq:v5cgn94QCe5jSpdeTpN}, we can write \eqref{eq:E8GLN2o41NCrdpqjVLd} as follows:

\begin{align*} M_X(t)&= \frac{t^0}{0!}\mu'_0 +\frac{t^1}{1!}\mu'_1 +\frac{t^2}{2!}\mu'_2 +\cdots +\frac{t^n}{n!}\mu'_n\\ &= \sum_{i=0}^n\frac{t^i}{i!}\mu'_i \end{align*}

The magic ✨ happens when we start taking the derivatives of $M_X(t)$. Let' use the expression of $M_X(t)$ in \eqref{eq:E8GLN2o41NCrdpqjVLd}. The first derivative of $M_X(t)$ is:

\begin{align*} M_X^{(1)}(t) &=\frac{d}{dt}\Big( \frac{t^0}{0!}\mathbb{E}(x^0) +\frac{t^1}{1!}\mathbb{E}(x^1) +\frac{t^2}{2!}\mathbb{E}(x^2) +\cdots +\frac{t^n}{n!}\mathbb{E}(x^n)\Big)\\ &= \frac{t^0}{0!}\mathbb{E}(x^1) +\frac{t^1}{1!}\mathbb{E}(x^2) +\frac{t^2}{2!}\mathbb{E}(x^3) +\cdots +\frac{t^{n-1}}{n-1!}\mathbb{E}(x^n) \end{align*}

If we set $t=0$, then we have:

$$M^{(1)}_X(0)=\mathbb{E}(x^1)=\mu'_1$$

We get the first moment of a random variable about the origin!

Now, lets take the second derivative:

\begin{align*} M_X^{(2)}(t)&= \frac{d}{dt}M_X^{(1)}(t)\\ &=\frac{d}{dt}\Big(\frac{t^0}{0!}\mathbb{E}(x^1) +\frac{t^1}{1!}\mathbb{E}(x^2) +\frac{t^2}{2!}\mathbb{E}(x^3) +\cdots +\frac{t^{n-1}}{n-1!}\mathbb{E}(x^n)\Big)\\ &=\frac{t^0}{0!}\mathbb{E}(x^2) +\frac{t^1}{1!}\mathbb{E}(x^3) +\cdots +\frac{t^{n-2}}{n-2!}\mathbb{E}(x^n) \end{align*}

If we set $t=0$:

$$M_X^{(2)}(0)=\mathbb{E}(x^2)=\mu'_2$$

We now get the second moment of a random variable about the origin!

Here, we can easily see where the moment-generating function gets its’ name from. The moment-generating functions can generate the $k$-th moment of a random variable about the origin! For the general case, we observe the following:

$$M_X^{(k)}(0)=\mathbb{E}(x^k)=\mu'_k$$

This completes the proof.

Example.

## Deriving mean and variance of standard normal random variables

Let $X$ be a random variable drawn from a standard normal distribution. Using the derivative of moment-generating function, compute the mean $\mathbb{E}(X)$ and variance $\mathbb{V}(X)$.

Solution. In our previous example question, we have derived the moment-generating function of a standard normal random variable $X$ to be:

$$M_X(t)= \exp\Big(\frac{t^2}{2}\Big)$$

Let's take the first derivative of the moment generating function:

\begin{align*} M_X^{(1)}(t)&= \frac{d}{dt}\exp\Big(\frac{t^2}{2}\Big)\\ &=t\cdot\exp\Big(\frac{t^2}{2}\Big) \end{align*}

From theorem, we know that setting $t=0$ will give us the first moment of $X$ about the origin:

\begin{align*} \mathbb{E}(X)&=M_X^{(1)}(0)\\ &=0 \end{align*}

Let's now take the second de

\begin{align*} M_X^{(2)}(t)&= \frac{d}{dt}(M_X^{(1)}(t))\\ &=\frac{d}{dt}\Big(t\cdot\exp\Big(\frac{t^2}{2}\Big)\Big)\\ &=\exp\Big(\frac{t^2}{2}\Big)+t^2\exp\Big(\frac{t^2}{2}\Big)\\ \end{align*}

Once again, from theorem, we know that setting $t=0$ for $M_X^{(2)}(t)$ will give us the second moment of $X$ about the origin:

\begin{align*} \mathbb{E}(X^2)&=M_X^{(2)}(0)\\ &= \exp\Big(\frac{0^2}{2}\Big)+(0)^2\exp\Big(\frac{0^2}{2}\Big)\\ &=1 \\ \end{align*}

Now, we know from the theorem that the variance can be computed as:

\begin{align*} \mathbb{V}(X)&=\mathbb{E}(X^2)-(\mathbb{E}(X))^2\\ &=1-(0)^2\\ &=1 \end{align*}

Therefore, as we would expect, the mean and variance of a standard normal variable $X$ is:

\begin{align*} \mathbb{E}(X)=0\\ \mathbb{V}(X)=1\\ \end{align*}

# Properties of moment-generating functions

Theorem.

## Moment-generating function of random variable X+a

Let $X$ be any random variable and $a$ be some scalar. The moment-generating function of $X+a$ can be expressed as:

$$M_{X+a}(t)=e^{at}M_X(t)$$

Proof. We can use the definition of moment generating functions to prove this:

\begin{align*} M_{X+a}(t)&=\mathbb{E}(e^{t(x+a)})\\ &=\mathbb{E}(e^{tx+ta})\\ &=\mathbb{E}(e^{tx}e^{ta})\\ &=e^{ta}\mathbb{E}(e^{tx})\\ &=e^{ta}M_X(t) \end{align*}

This completes the proof.

Theorem.

## Moment-generating function of random variable aX

Let $X$ be any random variable and $a$ be some scalar. The moment-generating function of $aX$ can be expressed as:

$$M_{aX}(t)=M_X(at)$$

Proof. We can use the definition of moment generating functions to prove this:

\begin{align*} M_{aX}(t)&=\mathbb{E}(e^{t(ax)})\\ &=\mathbb{E}(e^{x(at)})\\ &=M_X(at) \end{align*}

This completes the proof.

Theorem.

If X_1$$X_2, \cdots, X_n are independent random variables with moment-generating function M_{X_1}(t) M_{X_2}(t), \cdots, and M_{X_n}(t) respectively and Y=X_1+X_2+⋯+X_n, then we have that:$$\begin{align*} M_Y(t)=M_{X_1}(t)M_{X_2}(t)\cdots{M_n(t)} \end{align*}$$Proof. Once again, we use the definition of moment-generating functions:$$\begin{align*} M_Y(t)&=M_{X_1+X_2+\cdots+X_n}(t)\\ &=\mathbb{E}(e^{t(x_1+x_2+\cdots+x_n)})\\ &=\mathbb{E}(e^{tx_1+tx_2+\cdots+tx_n})\\ &=\mathbb{E}(e^{tx_1}e^{tx_2}\cdots{e^{tx_n}}) \end{align*}$$Since X_1$$X_2, \cdots$$X_n are independent:$$\begin{align*} M_Y(t)&= \mathbb{E}(e^{tx_1}) \mathbb{E}(e^{tx_2}) \cdots \mathbb{E}(e^{tx_n})\\ &=M_{X_t}(t)M_{X_2}(t)\cdots{M_{X_n}(t)} \end{align*}$This completes the proof. Theorem. ## Uniqueness Theorem Let$X$and$Y$be two random variables with moment-generating functions$M_X(t)$and$M_Y(t)$respectively. If$M_X(t)=M_Y(t)$for all values of$t$, then$X$and$Y\$ have the same probability distribution. In other words, if the moment-generating function of two random variables are the same, then they must have the same probability distribution.

The proof of the uniqueness theorem is quite complex and requires graduate-level mathematics, so we will omit the proof here 😞. The uniqueness theorem is one of most important properties of the moment-generating function and is used to prove theorems such as the reproductive property of the normal distribution (TODO)!

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