Combining independent distributions

Question

Suppose that $X_{1} \sim B(n_{1},p_{1})$ and $X_{2} \sim B(n_{2},p_{2})$ are independent. Write down the moment generating function for the random variable $Y=X_{1}+X_{2}$.

I know the moment generating function $M_{Y}(t)$ is defined at $E[e^{tY}]$.

$$E[e^{tY}] = \sum^{\infty}_{y=0}e^{ty}f_{Y}(y)$$ $$= \sum^{n_{1}+n_{2}}_{x_{1}+x_{2}=0}e^{t(x_{1}+x_{2})}f_{X_{1}+X_{2}}(x_{1}+x_{2})$$

But now I'm confused in what to do next.

score 3 · Accepted Answer · answered Apr 24 '14 at 15:30

3

When you add independent RV's $X$ and $Y$, your moment generating functions multiply: $E[e^{t (X+Y)}] = E[e^{tX}e^{tY}] = E[e^{tX}]E[e^{tY}]$ where the last step follows by independence of $X$ and $Y$ implies functions of them are independent (so $e^{tX}$ and $e^{tY}$ are independent).

answered Apr 24 '14 at 15:30

Batman

19,390

To check then $$M_{Y}(t) = (p_{1} e^{t}+1-p_{1})^{n_{1}}(p_{2}e^{t}+1-p_{2})^{n_{2}}$$ Is this correct? – user2850514 Apr 24 '14 at 16:01
can you check the above? – user2850514 Apr 24 '14 at 16:28
It is correct.. – Batman Apr 24 '14 at 16:46

score 3 · Answer 2 · answered Apr 24 '14 at 15:30

3

Hint: If $X_1$ and $X_2$ are independent: $$ \mathbb E(e^{tX_1}e^{tX_2})=\mathbb E(e^{tX_1})\mathbb E(e^{tX_2}). $$

answered Apr 24 '14 at 15:30

Arash

11,131

Combining independent distributions

2 Answers2