1. [SOLVED] Contradiction between expectation and MGF

Sorry, it was a stupid question. You can ignore it. But it was a good LATEX practice for me though.

In the text, for
$\displaystyle$
$\displaystyle P\left (X=x \right )=\binom{x-1}{r-1}p^r\left (1-p \right )^{x-r}; Supp\left (X \right )=\left \{r,r+1,r+2,... \right \}$
We have
$\displaystyle E(X)=\frac{r}{p}$ and $\displaystyle Var(X)=\frac{r(1-P)}{p^2}$ ; therefore, $\displaystyle E(X^2)=\frac{r^2+r-rp}{p^2}$
I then did that transformation [tex]Y=X-r{/math] to obtain
$\displaystyle P\left (Y=y \right )=\binom{y+r-1}{r-1}p^r\left (1-p \right )^y; Supp\left (Y \right )=\left \{0,1,2,3,... \right \}$
I then found the MGF to be:
$\displaystyle M_Y(t)=E(e^{tY})=\sum_{y=0}^{\infty }e^{ty}\binom{y+r-1}{r-1}p^r\left (1-p \right )^y=$$\displaystyle \frac{p^r}{(1-(1-p)e^t)^r};t\neq -\ln (1-p)$
I then tried us use the MGF to find $\displaystyle E(X)$ and $\displaystyle E(X^2)$

$\displaystyle E(Y)=\frac{d }{d x}M_Y(0)=\frac{r(1-p)}{p}$
and
$\displaystyle E(Y^2)=\frac{d^2 }{d x^2}M_Y(0)=\frac{r(1-p)[(r(1-p)+1]}{p^2}$
therefore
$\displaystyle Var(Y)=\frac{r(1-P)}{p^2}$

So here's my question:
Why does
$\displaystyle E(X)\neq E(Y)$
and
$\displaystyle E(X^2) \neq E(Y^2)$
but
Var(Y)=Var(X)?