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Cumulant Generating Function

  1. May 6, 2010 #1
    Cumulative generating function is
    [tex]K(t)=K_1(t)t+K_2(t)\frac{t^2}{2!}+K_3(t)\frac{t^3}{3!}+...[/tex]
    where
    [tex]K_{n}(t)=K^{(n)}(t)[/tex]

    Now
    [tex]K(t)=ln M(t)=ln E(e^{ty})=ln E(f(0)+f'(0)\frac {t}{1!}+f''(0)\frac{t^2}{2!}+...)=ln E(1+\frac{t}{1!}y+\frac{t^2}{2!} y^2+...)=ln [1+\frac{t}{1!} E(Y)+\frac{t^2}{2!} E(Y^2)+...]=ln [1+\frac{t}{1!}\mu'_1+\frac{t^2}{2!}\mu'_2+...][/tex]
    where [tex]\mu'_n=E(Y^n)[/tex]
    [tex]=>K(0)=ln1=0[/tex]

    Also
    [tex]K'(t)=\frac{1}{M(t)}M'(t)[/tex]
    where
    [tex]M(0)=1; M'(t)=\mu'_1+\frac{t}{1}\mu'_2+\frac{t^2}{2!}\mu'_3+...[/tex]
    [tex]=>M'(0)=\mu'_1[/tex]

    In fact
    [tex]M^{(n)}(0)=\mu'_n[/tex]

    So
    [tex]K'(0)=\frac{\mu'_1}{1}=\mu'_1[/tex]

    Furthermore
    [tex]K''(t)=\frac{M''(t)M(t)-[M'(t)]^2}{[M(t)]^2}[/tex]
    [tex]=>K''(0)=\frac{\mu'_2*1-(\mu'_1)^2}{1^2}=\mu'_2-(\mu'_1)^2=E(Y^2)-[E(Y)]^2=\sigma^2[/tex]

    Is this correct?
     
    Last edited: May 6, 2010
  2. jcsd
  3. May 11, 2010 #2
    Correction:
    K(t) is:
    [tex]K(t)=K_1t+K_2\frac{t^2}{2!}+...[/tex]
    where
    [tex]K_n=K^{(n)}(0)[/tex]
     
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