1. ## moment-generating function

$\displaystyle Y$ have pdf $\displaystyle P(Y=k)=(k+1)p^2 q^k$, with $\displaystyle 0<p<1$and $\displaystyle q=1-p$ and $\displaystyle k=0,1,2,...$
Is easy prove that $\displaystyle M_{Y}(t)=p^2 (1-qe^t)^2$. Let $\displaystyle X=Y_1+\cdots Y_{24}$, with $\displaystyle Y_{i}$ with same distribution of $\displaystyle Y$
Calculate $\displaystyle M_{X}(t)$

Note: $\displaystyle M_Y (t)=E(e^{tk})$

2. Originally Posted by jorgeston
$\displaystyle Y$ have pdf $\displaystyle P(Y=k)=(k+1)p^2 p^k$, with $\displaystyle 0<p<1$and $\displaystyle q=1-p$ and $\displaystyle k=0,1,2,...$
Is easy prove that $\displaystyle M_{Y}(t)=p^2 (1-qe^t)^2$. Let $\displaystyle X=Y_1+\cdots Y_{24}$, with $\displaystyle Y_{i}$ with same distribution of $\displaystyle Y$
Calculate $\displaystyle M_{X}(t)$

Note: $\displaystyle M_Y (t)=E(e^{tk})$
I have no time now. I'll reply later if no-one else does.

3. $\displaystyle X = 24Y$. Note that $\displaystyle M_{aY+b} = e^{bt} M_{Y}(at)$. In this case, $\displaystyle b = 0$ and $\displaystyle a = 24$. So $\displaystyle M_{24Y}(t) = M_{X}(t) = p^2(1-qe^{24t})^{2}$.

4. Originally Posted by tukeywilliams
$\displaystyle X = 24Y$. Note that $\displaystyle M_{aY+b} = e^{bt} M_{Y}(at)$. In this case, $\displaystyle b = 0$ and $\displaystyle a = 24$. So $\displaystyle M_{24Y}(t) = M_{X}(t) = p^2(1-qe^{24t})^{2}$.
If you have reason, i don't understand why $\displaystyle Y_1=Y_2=\cdots = Y_{24}$

If this is true, implies $\displaystyle X=24Y$, but i am confused

5. Originally Posted by jorgeston
If you have reason, i don't understand why $\displaystyle Y_1=Y_2=\cdots = Y_{24}$

If this is true, implies $\displaystyle X=24Y$, but i am confused
*Ahem* Didn't you say in your question:

"with with same distribution of " ....?