In this statement "If a random variable Y is symmetrically distributed around its mean $\mu_Y$, that is, $Y=\mu_Y+U$, where U is symmetrically distributed aronud zero- then the odd moments of Y are powers of $\mu_Y$, i.e. $E[(Y)^j]= (\mu_Y)^j $ for any odd integer j, assuming the moment exists".
I think this statement is true for both even and odd moments. What is the significance of the odd moment? And is it true when it comes to the even moment?
Hint: All you need to know is that $U$ is symmetrically distributed with zero mean. This implies that all odd moments of $U$ will be 0.
So, we know that
$$E[(Y-\mu_Y)^3]=E[Y^3 -3 Y^2\mu_Y+3 Y\mu_Y^2-\mu_Y^3]=0 \implies$$ $$ E[Y^3] = E[3 Y^2\mu_Y-3 Y\mu_Y^2+\mu_Y^3]=3 \mu_Y E[Y^2]-3 \mu_Y^2 E[Y]+\mu_Y^3 \implies $$
$$E[Y^3]= 3 \mu_Y E[Y^2]-2\mu_Y^3 $$
This value depends on the value for $E[Y^2]$, which we can derive as follows:
$$Var[Y] = E[Y^2]-E[Y]^2 \implies E[Y^2] = Var[Y] + \mu_Y^2$$
However, $Var[Y]=Var[U]$ so
$$E[Y^2] = Var[U] + \mu_Y^2$$
So, assuming $Var[U]>0$, what can you conclude?