So, suppose I have a binomial random variable $X$ with parameters $n=6$ and $p=(1/5)$. Now, if I have a function of $X$ like $Y=(X-2)^2$, how do I go about finding it's properties - and in particular, $P(Y=y), P(Y<y)$ and $E[Y]$?
In the case of a continuous random variable, I would use the distribution function or change-of-variable techniques to find the CDF or PDF of $Y$, but in this case $X$ is a (discrete) binomial random variable. Furthermore, $Y$ is two-to-one, which further complicates things.
All I've figured out is that $Y$ should share the same parameter p as $X$ (I think?) and that the space of $Y$ in this case should be:
$Y \in (1,4,9,16)$ since $X \in (0,1,2,3,4,5,6)$ and $Y=(X-2)^2$.
Is the mean of $Y$ (or $E[Y]$) still equal to $np$? How do I find it's PMF? How would I go about evaluating $P(Y<y)$?
Just some guidance would be appreciated. Thanks
If you get $P(Y=y)$ you can get $P(Y<y)$ and $E[Y]$, so i'll give you a hint for getting $P(Y=y)$.
$P(X=0) = (4/5)^6$, $P(X=4) = 15(1/5)^4(4/5)^2$.
Now,
$Y=4$ iff $X=0$ or $X=4$, so $P(Y=4) = P(X=0) + P(X=4)$
Also $Y$ can take the value $0$ if $X=2$