CH 43 - Moment Generating Functions

Definition for a Generation Function

If $a_1,a_2,\dots ,a_n$ is an interesting sequence of numbers from which we want to build a generating function $g(t)$, we write:

$$g(t)=\sum _{j=0}^{\infty }\frac{a_j}{j!}{t}^j=\frac{a_0}{0!}{t}^0+\frac{a_1}{1!}{t}^1+\frac{a_2}{2!}{t}^2+\frac{a_3}{3!}{t}^3+\cdots +\frac{a_n}{n!}{t}^n$$

We get something like:

$a_j=g^{(j)}(0)$

an example:

$$a_0=g^{(0)}(0)={a_0+a_{1}t+\frac{a_2}{2}{t}^2+\dots |}_{t=0}$$

which is quite boring and just becomes:

$$=a_0$$

same would yield for $a_1$.

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Moment Generating Functions

Definition for a Moment Generating Funciton

The moment generating function $M_X(t)$ of a random variable $X$ is equal to the expected value of $e^{tX}$:

$$M_X(t)=\sum _{j=0}^{\infty }\frac{E(X^j)}{j!}{t}^j=E\left(\sum _{j=0}^{\infty }\frac{(tX{)}^j}{j!}\right)=E(e^{tX})$$

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Here we define $a_j=E(X^j)$ as the$j^{th}$ moment about the value of a Random variable $X$

$$\text{mean}\mu x=E(x^1)$$

you can also get $E(X^2)$:

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New Exercise

Example

Starting at 6:00AM, birds perch on a power line with time between consecutive birds is exponential with an average of 7.5 minutes.

Hint

From this we can note that we will use:

$$X\approx Exp(.133)\text{ for }\lambda$$

And we can use the exponential equation:

$$E(x)=7.5=\frac{1}{\lambda }\Rightarrow \lambda =0.133=\frac{1}{7.5}$$

1. Given that there has not been a bird perched in the previous 5 minutes, what is the probability that a bird will not perch in the next 10 minutes.

$$P(X>15\mid X>5)=\frac{P(X>15\cap X>5)}{P(X>5)}$$

We will use the Memoryless Property For Exponentials

$$\begin{array}{rl}P(X>10+5)\mid X>5)& =P(X>10)\\ & =1-F(10)\\ & =1-(1-e^{-0.133(10)})\\ & =e^{-.133(10)}=\boxed{0.2636}\end{array}$$

2. How many birds are expected to perch in the next hour?

We expect that 1 bird will perch every 7.5 minutes, and thus we will switch to poisson.

$$E(X)=\frac{60\text{ minutes}}{7.5\text{ minutes}}=\text{8 birds in an hour }={\lambda }_P$$

Now that we have the average number of birds per hour, we can use the Poisson distribution to find the probability of a certain number of birds perching in the next hour.

$$\begin{array}{rl}P(X=k)& =\frac{e^{-\lambda }{\lambda }^k}{k!}\end{array}$$

For example, the probability of exactly 8 birds perching in the next hour is:

$$\begin{array}{rl}P(X=8)& =\frac{e^{-8}{8}^8}{8!}\\ & \approx 0.1396\end{array}$$

And the probability of at least 10 birds perching in the next hour is:

$$\begin{array}{rl}P(X\ge 10)& =1-P(X\le 9)\\ & =1-\sum \_{k=0}^9\frac{e^{-8}{8}^k}{k!}\\ & \approx 0.2798\end{array}$$