Moment-generating function

[azwethinkweiz]

The pH of a certain reactant is measured by the random variable X whose density function is given by:

\(\displaystyle f(x)=\left\{\begin{array}{rcl}a(x-3.8), \;\ if \;\ 3.8\leq x\leq 4 \\ a(4.2-x), \;\ if \;\ 4<x<4.2\\ 0 \;\ otherwise\end{array}\right\)

Find \(\displaystyle M_{x}(t)\) , the moment generating function of X and show that \(\displaystyle M'_{x}(0) = E(x)\)

I'm really confused about the moment generating function. First of all, how do I know if this is discrete or continuous? After that, I'm not sure how to continue

 

[mmm4444bot]

I'm guessing that your pronoun "this" refers to function f.

Function f is certainly not discrete because its domain is all Real numbers.

f(x) is continuous over its entire domain, too, but it is not differentiable at x = 3.8, x = 4, or x = 4.2 because the graph has cusps at those locations.

If you're familiar with linear functions, piecewise functions, and their graphs, you can verify that the graph of f(x) travels along the x-axis (from left to right) until it reaches x = 3.8.

Assuming that parameter a is positive, the graph then becomes a line segment with positive slope a until x = 4.

The next piece is a line segment with negative slope a that goes back to the x-axis at x = 4.2. The graph then continues along the x-axis forever.

         /\
        /  \
_______/    \_______ 
    3.8   4  4.2     x

If the value of a is negative, then the two line segments intersect below the x-axis, and the graph is flipped.

Also, changing the slope (a or -a) does not alter the x-intercepts of those lines. I mean, you can change the value of a, and the two segments will still intersect at x = 4, and the lines will still "hit" the axis at x = 3.8 and x = 4.2.

I hope this information helps you, somewhat.

I do not know what function E is.

 

[azwethinkweiz]

By E, I mean the Expectance.

As for the graphing information you posted, is it possible to find the cumulative frequency distribution and then graph it too? Would I not need to find out what constant a is, at first?

 

[mmm4444bot]

Yes, if you want to generate a graph, you'll need to have a specific value for a.

I looked up "random variable density function", and I see that f(x) must be non-negative; therefore, the parameter a must represent a positive number.

I know next to nothing about moment-generating functions and expectance, but the cummulative frequency distribution of x is probably the area between the graph of f(x) and the x-axis, and therefore, is the area from x = 3.8 through some point up to or including x = 4.2.

You could write an integral of f(x), working symbolically with a.

I can't help much with this exercise; I tried to provide information about f(x).

 

[galactus]

Here is a little info about the MGF. What I remember from stats back in the day. It's been a while since I pondered this stuff.

When X is continuous:

\(\displaystyle M_{x}(t)=E(e^{tx})=\int e^{tx}\cdot f(x)dx\)

Example:

Say we want to find \(\displaystyle M_{x}(t)\) for

\(\displaystyle f(x)=\left\{\begin{array}{rcl}e^{-x}, \;\ for \;\ x>0\\ 0, \;\ \text{elsewhere}\end{array}\)

\(\displaystyle {\mu}'_{1}\) is the mean of the distribution of X.

\(\displaystyle {\mu}'_{1}=E(X)=\int x\cdot f(x)dx\)

By defintion:

\(\displaystyle M_{x}(t)=E(e^{tx})=\int_{0}^{\infty}e^{tx}\cdot e^{-x}dx=\frac{1}{1-t}, \;\ for \;\ t<1\)

When |t|<1, the MacLaurin series for this MGF is:

\(\displaystyle M_{x}(t)=1+t+t^{2}+t^{3}+....+t^{r}+....\)

\(\displaystyle =1+t+2!\cdot \frac{t^{2}}{2!}+3!\cdot \frac{t^{3}}{3!}+.....+r!\cdot \frac{t^{r}}{r!}+....\)

Thus, hence, and therefore, \(\displaystyle {\mu}'_{r}=r!\), for r=0,1,2,....

\(\displaystyle \frac{d^{r}M_{x}(t)}{dt^{r}}|_{t=0}={\mu}'_{r}\)

If a function is expanded as a power series in t, the coefficient of \(\displaystyle \frac{t^{r}}{r!}\) is the rth derivative of the function

with respect to t at t=0.