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I am working on the following problem, and I am having trouble understanding why (a) is not a moment-generating other than it doesn't satisfy the general form of the MGF (i.e. $E[e^{tX}]$), and that (b) the sum is a moment-generating function. Are there some properties that an MGF must satisfy for it to be considered an MGF?

Problem: Let $p_r > 0$ and $a_r \in R$ with $1 \le r \le n$. Which of the following is a moment-generating function, and for what random variable?

(a) $M(t)=1+\sum_{r=1}^{n}p_{r}\, t^{r}$

(b) $M(t)=\sum_{r=1}^{n}p_r\,e^{a_r t}$

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    $\begingroup$ For (b) I suspect you need $\sum\limits_1^n p_r=1$ to ensure $M(0)=1$. You then have a mixture distribution. $\endgroup$
    Henry
    –  Henry
    2024-10-29 07:46:12 +00:00
    Commented Oct 29, 2024 at 7:46

1 Answer 1

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In (a), $M(t)$ is a polynomial, so there is some large $k$ such that the $2k$th derivative of $M$ is zero. But then $E[X^{2k}] = M^{(2k)}(0) = 0$. Since $X^{2k} \ge 0$, the only way it can have expectation equal to zero is if $P(X^{2k} = 0)=1$. But then $P(X=0)=1$.

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  • $\begingroup$ Thanks, I noticed it was a polynomial, but I didn't consider differentiating. $\endgroup$
    Harry Lofi
    –  Harry Lofi
    2024-10-29 13:21:58 +00:00
    Commented Oct 29, 2024 at 13:21

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