How can I find moment generating function?

[magic448]

Question: What is moment generating function good for?
How can I find the MGF of the following probability funtion?


f(y) = 19/27 for y = 0
8/27 for y = 1

thank you

[Jo_M.]

Question: What is moment generating function good for?
How can I find the MGF of the following probability funtion?


f(y) = 19/27 for y = 0
8/27 for y = 1

thank you

The moment generating function is defined as E[e^ty]. Therefore, in your case, My(t)= e^t(0) * 19/27 + e^t(1) * 8/27 = 19/27 + e^t * 8/27

An important aspect of the moment generating function is that differentiating it with t=0 will yield the different moments of the distribution. That is, d^n/dy^n My(0) = E[Y^n]. Thus, the first derivative of the moment generating function will give you the expected value of X, the second E[X^2], etc.

So in your example, the E[Y]=8/27*e^0 = 8/27

(You also could have noticed that your distribution is simply a bernouilli distribution where p=8/27 and q=19/27, in which case it is useful to know by heart that the mgf of a bernouilli is q+pe^t. Similarly, the mgf of a binomial distribution, which is merely the sum of n bernouilli rv's, is (q+pe^t)^n.... etc. You should definitely learn the mgfs of all the most common discrete and continuous distributions)

[Andy Medina]

The definition of the moment generating function of X is M(t) = E(e^tx).

The main useful thing about a moment generating function is that the nth-derivative evaluated at t=0 is the nth-moment of X. In other words:

d^n/dt^n (M(t)) | t=0 = E(X^n)

Another useful property is that the second derivative of the natural log of a moment generating function evaluated at t = 0 is the variance of X.

It would be best to show you with an example. With yours, we can find E(e^tx).

E(e^ty)=sum (e^ty*P(Y=y))=e^(t*0) * P(Y=0) + e^(t*1) * P(Y=1) = 1*19/27 + 8e^t/27

So, the moment generating function of Y is M(t) = 19/27 + 8e^t/27

We know that the E(Y) = sum (y*P(Y=y)) = 0*19/27 + 1*8/27 = 8/27.

We can find E(Y) an alternate way by taking the derivative of the moment generating function of Y and setting t = 0.

M'(t) = 0 + 8e^t/27
M'(0) = 8e^0/27 = 8/27

If we want E(Y^2), we can do it the usual way like before or we can take the second derivative of the moment generating function of Y and set t = 0.

This may seem like a long and hard way to do it. For the Bernoulli, it's best to find any moment you want probably by just using the definition of E(Y) (and not the moment generating function). However, there are times when this is useful. It is much easier to find say E(X^5) if X~exponential using the moment generating function than it is to derive it (well unless you use the integration trick). There are also times when you, say want to prove that the sum of exponentials is Gamma or the sum of poissons is poisson. It's easiest to do this using the moment generating function. It is also easy to use the MGF to find expected values.

So, there are uses of it but at first it seems like a long and hard way to do the same thing you already know how to do.

[magic448]

oh, yeah i just realized it is a bernouilli distribution. I can use the shortcut.

But one more question
M(t) = E(e^tx) Is x an expoment?
How come after substitution

e^t(1) * 8/27 = e^t * 8/27

8/27 became a leading coefficient? do we have to take derivative somewhere?

[djerry81]

This is a Bernoulli MGF....From my test in May07 there were alot of MGF questions that are easy if you memorized them

This is one of the questions from the SOA123

something you might also want to know

lets say
x=0 .25
x=1 .1
x=2 .65

then the MGF would be .25e^0t+.1e^1t+.65e^2t

[Jo_M.]

oh, yeah i just realized it is a bernouilli distribution. I can use the shortcut.

But one more question
M(t) = E(e^tx) Is x an expoment?
How come after substitution

e^t(1) * 8/27 = e^t * 8/27

8/27 became a leading coefficient? do we have to take derivative somewhere?

Yes, xt is the exponent. I should have written: Mx(t)=E[e^(tx)].

As for your second question, 8/27 was always a leading coefficient since you need to do the following multiplication to find E[e^(xt)]: (e^(xt))*P(X=x) --- then sum for all possible values of x...

The derivative part only comes into play once you want to find the first moment from the mgf. In this case, M'(0) = 0 + 8/27*(e^0)*1=8/27


Did I understand your question right?

[magic448]

This time i got it :laugh:

[actuaZane]

Moment Generating Functions

Hello everyone.

I'm new to this forum and have joined, because of the very intriguing name, "actuary.com". I'm currently a first year Actuarial Science student.

Today our lecturer gave us a moment generating function and we are to determine the probability distribution that it represents. After some time trying to notice a pattern, I just cannot. Perhaps somebody can help me.

This is the function:
m(t) = 1/6 + (1/6)e^t + (2/6)e^2t + (2/6)e^4t.

I do note that there is no "+ ..." added to the end of the function. This is how it was given.

Could anybody possibly help me with this problem?

Kind Regards
Zane