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I have trouble understanding the term of second cumulant generating function. By the definition of cumulant generation function, it is defined by the logarithm of moment generating function $M_X(t)=E(e^{tX})$. How can I know the second cumulant is variance? Thanks.

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    $\begingroup$ Have you actually tried to figure out $\left.\frac{d^2}{dx^2}\right|_{x=0}(\log M_X(x))$ by direct computation? $\endgroup$ – Sangchul Lee Jun 15 '18 at 22:20
  • $\begingroup$ So the problem is to show that if $$ M = 1 + at + b\frac{t^2} 2 + \cdots $$ then $$\log M = at + (b-a^2)\frac {t^2} 2 +\cdots.$$ $\endgroup$ – Michael Hardy Jun 15 '18 at 22:30
  • $\begingroup$ After direct computation, I figured out my problem. Thanks! :) $\endgroup$ – Chen Jun 15 '18 at 22:32

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