donutmax
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Cumulative generating function is
K(t)=K_1(t)t+K_2(t)\frac{t^2}{2!}+K_3(t)\frac{t^3}{3!}+...
where
K_{n}(t)=K^{(n)}(t)
Now
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+...]
where \mu'_n=E(Y^n)
=>K(0)=ln1=0
Also
K'(t)=\frac{1}{M(t)}M'(t)
where
M(0)=1; M'(t)=\mu'_1+\frac{t}{1}\mu'_2+\frac{t^2}{2!}\mu'_3+...
=>M'(0)=\mu'_1
In fact
M^{(n)}(0)=\mu'_n
So
K'(0)=\frac{\mu'_1}{1}=\mu'_1
Furthermore
K''(t)=\frac{M''(t)M(t)-[M'(t)]^2}{[M(t)]^2}
=>K''(0)=\frac{\mu'_2*1-(\mu'_1)^2}{1^2}=\mu'_2-(\mu'_1)^2=E(Y^2)-[E(Y)]^2=\sigma^2
Is this correct?
K(t)=K_1(t)t+K_2(t)\frac{t^2}{2!}+K_3(t)\frac{t^3}{3!}+...
where
K_{n}(t)=K^{(n)}(t)
Now
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+...]
where \mu'_n=E(Y^n)
=>K(0)=ln1=0
Also
K'(t)=\frac{1}{M(t)}M'(t)
where
M(0)=1; M'(t)=\mu'_1+\frac{t}{1}\mu'_2+\frac{t^2}{2!}\mu'_3+...
=>M'(0)=\mu'_1
In fact
M^{(n)}(0)=\mu'_n
So
K'(0)=\frac{\mu'_1}{1}=\mu'_1
Furthermore
K''(t)=\frac{M''(t)M(t)-[M'(t)]^2}{[M(t)]^2}
=>K''(0)=\frac{\mu'_2*1-(\mu'_1)^2}{1^2}=\mu'_2-(\mu'_1)^2=E(Y^2)-[E(Y)]^2=\sigma^2
Is this correct?
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