MHB Moment-Generating Function question

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The moment-generating function \( M_x(t) = \frac{t}{t-1} \) cannot correspond to any random variable because \( M_x(0) \) does not equal 1, which is a requirement for valid moment-generating functions. Differentiating the function yields a mean of 1 and a variance of 1, which initially appears acceptable. However, the failure of \( M_x(0) \) to equal 1 indicates that it does not satisfy the properties of a moment-generating function. This inconsistency confirms that no random variable can be represented by this moment-generating function. Understanding these properties is crucial for identifying valid moment-generating functions in probability theory.
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Q:Explain why there can be no random variable for which $M_x(t)=\frac{t}{t-1}$
($M_x(t)$ is the moment-generating function of the random variable $x$.)

A: I tried differentiating it twice and got mean of $x=1$ and variance of $x=1$ which seems fine. Maybe its because $M_x(0)$ is not equal to $1$?Thanks in advance for any help with this problem.
 
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JGalway said:
Maybe its because $M_x(0)$ is not equal to $1$?

Yes, that's exactly the reason.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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