Probability density function help If a random variable is given by $Y=aX+b$ where $X$ is a random variable does that mean $aX+b$ is $Y$'s pdf?? and if i wanted to find $E[Y^2]$ would this just be the same as finding $E[(aX+b)^2]$ and if so why is it not the case that if $X=f(x)$ $E[X]=E[f(x)^2]$ but $= \int(x^2f(x))dx$ ?? please help  
 A: There are three things in your question: random variables, probability density functions and the law of the unconscious statistician.
If $X$ is a random variable, then $aX+b$ is also a random variable and you can calculate the expected value of $Y$ using the properties of expected value:
$$
\operatorname E(aX+b)^2=\operatorname E[a^2X^2+2abX+b^2]=a^2\operatorname EX^2+2ab\operatorname EX+b^2.
$$
But $aX+b$ is not $Y$'s probability density function. In fact, $Y$ might not even have a probability density function at all. Only continuous random variables have probability density functions and we do not know whether $Y$ is a continous random variable.
The law of the unconscious statistician helps us find an expected value of a random variable $Y$ given by $g(X)$ with a measurable function $g$ without explicitly knowing the distribution of $Y$. We only need to know the distribution of $X$.
So when we calculate
$$
\operatorname EX^2=\int_{-\infty}^\infty x^2f(x)\mathrm dx
$$
for a continuous random variable $X$, we are actually using the law of the unconscious statistician.
