If X has generating function G, then the expectation(X) = G'(1) and variance(X) = G''(1) + G'(1) - G'(1)^2.
Use this to prove that if X and Y are independent, then variance(X+Y) = variance(X) + variance(Y).
If X has generating function G, then the expectation(X) = G'(1) and variance(X) = G''(1) + G'(1) - G'(1)^2.
Use this to prove that if X and Y are independent, then variance(X+Y) = variance(X) + variance(Y).
Let $\displaystyle X$ have generating function $\displaystyle G$, and $\displaystyle Y$ have generating function $\displaystyle H$.
Then as $\displaystyle X$ and $\displaystyle Y$ are independent the generating function of $\displaystyle X+Y$ is:
$\displaystyle F(z)=H(z)G(z).$
and as these are probability generating functions $\displaystyle H(1)=G(1)=1$.
You should now be able to complete this yourself, if not let us know and
we will finnish this off for you.
RonL
variance(X+Y) = F''(1) + F'(1) - F'(1)^2
Now use the product rule repeatedly on F(z)=H(z)G(z) to get F'(z) and F''(z)
in terms of H''(z), H'(z), H(z), G''(z), G'(z) an G(z). Now put z=1 and
substitute into the equation for the variance of X+Y and the result drops out.
RonL
okay so
F'(z) is H(z)G'(z) + G(z)H'(z).
F''(z) is 2*(H'(z)G'(z)) + H(z)G"(z) + G(z)H"(z)
so variance(X+Y) = F"(1)+ F'(1) - F'(1)^2
so it is the same as:
2*(H'(1)G'(1)) + H(1)G"(1) + G(1)H"(1) +
H(1)G'(1) + G(1)H'(1) +
(H(1)G'(1) + G(1)H'(1))^2
however, how do i reduce this?
First replace all the G(1)'s and H(1)'s by 1, so:
2*(H'(1)G'(1)) + H(1)G"(1) + G(1)H"(1) + H(1)G'(1) + G(1)H'(1) - (H(1)G'(1) + G(1)H'(1))^2
.... = 2*(H'(1)G'(1)) + G"(1) + H"(1) + G'(1) + H'(1) - (G'(1) + H'(1))^2
.... = 2*(H'(1)G'(1)) + G"(1) + H"(1) + G'(1) + H'(1) - (G'(1))^2 - (H'(1))^2 -2*(H'(1)G'(1))
.... = G"(1) + H"(1) + G'(1) + H'(1) - (G'(1))^2 - (H'(1))^2
.... = G"(1) + G'(1) - (G'(1))^2 + H"(1) + H'(1) - (H'(1))^2
.... = var(X) + var(Y)
RonL