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I am not from a Math field, but use statistics. I am having a tough time understanding the concept of a moment generating function.

I use PDF's mostly to characterize my data. What are the possible advantages of working with MGFs?

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I just want to add some advantages of MGF that @BCLC does not mention.

  1. If two MGFs of two random variables coincide (maybe with some little more conditions), then the distributions of the two random variables are identical. This helps you to determine the distribution of some random variables with what you already know. It is also possible to recover the cumulative distribution by MGF.

  2. If $X_n$'s MGF $M_n$ converges to a MGF $M $( maybe with some little more condition), then we know the limit distribution of $X_n$ is the distribution corresponding to $M$. This really helps in statistics as the limit distribution is usually hard to compute but MGF is easy.

  3. MGF may not usually exists but they are easier to manipulate than the characteristic functions. Usually, people compare MGF, Cumulant and characteristic function. MGF is more practical than the other two since they only involve real numbers and already tells you a lot of the information of the random variables(all the moments and distribution to some extend).

Brian Ding
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PDFs as in Probability Distribution Functions? They don't always exist. They are taught to beginners to characterize data but really advanced studies use Cumulative Distribution Functions.

MGFs are close to characteristic functions which always exist. Not really sure.

I guess another advantage that MGFs give you moments from just differentiating and then plugging in zero while you need to do integration from PDFs to get moments. Take a look at these, and then tell me that you would rather integrate $x^n f(x)$ rather than just differentiate the MGF. :P

Not an expert. Wait for a better answer.

BCLC
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Can be very useful when you want to know the distribution of the sum of random variables, or even a random sum of random variables. (i.e. $\sum_1^N{x}$ where N is random as well as x.)

user121049
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This is late, but I happened to stumble upon this question. I am not an expert, but wanted to add another insight:

To obtain a marginal mgf from a joint mgf (say of X and Y), you simply have to calculate $M_{X,Y}(t,0) = M_X(t)$, which may be easier to work with in the case that the marginal pdf ($f_X(x)=\int_{-\infty}^{\infty}f_{X,Y}(x,y)dy$) is difficult to calculate because of a complicated integral.

After calculating the mgf of X and Y, you could check for independence by testing whether $M_X(t)*M_Y(u)=M_{X,Y}(t,u)$, rather than checking whether $f_X(x)*f_Y(y)=f_{X,Y}(x,y)$.

bob
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