Basics of moment generating functions

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The discussion focuses on understanding moment generating functions (MGFs), specifically for the function f_X(x) = e^{-x} for x > 0. The moment generating function is defined as m_X(u) = E(e^{uX}), and the user is attempting to compute it by integrating e^{ux}e^{-x} from 0 to infinity. There is some confusion regarding the application of expectation and the role of the variable u in the calculations. A suggestion is made to refer to the Wikipedia page on moment-generating functions for further clarification. The user expresses relief that deriving the MGF will simplify finding expectations later.
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I missed my class that introduced us to moment generating functions, and my notes are missing some pretty essential parts to helping me understand them, so here it is:


Homework Statement



Find the moment generating function of f_X(x)=e^{-x}, x>0



Homework Equations



E(X^R)=\int_{-\infty}^{\infty}X^Rf_X(x)dx

R must be a natural number I believe.

m_X(u) = E(e^{uX})



The Attempt at a Solution



I'm unsure if the expectation is applied in the same way, in other words, would this be correct?

m_X(u) = E(e^{ux})

= \int_{-\infty}^{\infty}e^{uX}f_X(x)dx

= \int_{0}^{\infty}e^{ux}e^{-x}dx

= \int_{0}^{\infty}e^{x(u-1)}dx

And I'm not quite sure what the u in this case is either.
 
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Oh awesome, thanks vela.

I can also see that the expectation can be found quite easily after deriving the moment generating function, which is good because I'll be needing it :smile:
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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