- #1
donutmax
- 7
- 0
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?
[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: