Show that $E((q/p)^Y)=1$ if $P(Y=1)=p$ and $P(Y=-1)=q=1-p$

Question

I'm having a little trouble with this expectations question.

Problem: Let $P(Y=1)=p$ and $P(Y=-1)=q=1-p$. Show that E($\frac{q}{p})^Y = 1.$

My solution: We know that $E[Y]= 1( p) + -1(1-p) = p -1+p = 2p-1.$

Therefore, $p = \frac{E[Y]+1}{2}$

Now, E($\frac{q}{p})^Y = E(\frac{1-p}{p})^Y = E[\frac{1}{p}-1]^Y = E[\frac{2}{E[y]+1} -1] $

Not sure where to go beyond this point, or if my method so far is even correct. Help is much appreciated.

Answer 1

$Y$ only takes $2$ values.

$$\mathbb{E}\left( \left( \frac{q}p\right)^Y\right) = p \cdot \frac{q}p + q \left( \frac{q}p\right)^{-1} =q+p=1$$