Statistics Definitions > Moment Generating Function (MGF)
If you aren’t familiar with moments, you may want to read this article first: what is a moment?
What is a Moment Generating Function?
Moment generating functions are a way to find moments like the mean(μ) and the variance(σ2). They are an alternative way to represent a probability distribution with a simple one-variable function. Each probability distribution has a unique MGF, which means they are especially useful for solving problems like finding the distribution for sums of random variables. They can also be used as a proof of the Central Limit Theorem.
There isn’t an intuitive definition for exactly what an MGF is; it’s merely a computational tool.
How to Find an MGF
Finding an MGF for a discrete random variable involves summation; for continuous random variables, calculus is used. It’s actually very simple to create moment generating functions if you are comfortable with summation and/or differentiation and integration:
For the above formulas, f(x) is the probability density function of X and the integration range (listed as -∞ to ∞) will change depending on what range your function is defined for.
Example: Find the MGF for e-x.
Step 2: Integrate. The MGF is 1 / (1-t).
The moment generating function only works when the integral converges on a particular number. The above integral diverges (spreads out) for t values of 1 or more, so the MGF only exists for values of t less than 1. You’ll find that most continuous distributions aren’t defined for larger values (say, above 1). This is usually not an issue: in order to find expected values and variances, the mgf only needs to be found for small t values close to zero.
Using the MGF
Once you’ve found the moment generating function, you can use it to find expected value, variance, and other moments.
M(0) = 1,
M'(0) = E(X),
M”(0) = E(X2),
M”'(0) = E(X3)
and so on;
Var(X) = M”(0) − M'(0)2.
Example: Find E(X3) using the MGF (1-2t)-10.
Step 1: Find the third derivative of the function (the list above defines M”'(0) as being equal to E(X3); before you can evaluate the derivative at 0, you first need to find it):
M”'(t) = (−2)3(−10)(−11)(−12)(1 − 2t)-13
Step 2: Evaluate the derivative at 0:
M”'(0) = (−2)3(−10)(−11)(−12)(1 − 2t)-13
Solution: E(X3) = 10,560.------------------------------------------------------------------------------
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