Difference between $\operatorname{Var}(Y)$ and $\operatorname{Var}(Y\mid X)$? What is the difference between $\mathrm{var}(Y)$ and $\mathrm{var}(Y\mid X)$? If $Y = c + \beta X$ and $\operatorname{var}(X)=\sigma^2$, won't both come out to be the same, i.e., $\beta^2\sigma^2$?
 A: Note that we always have $E[f(X)|X] = f(X)$. Loosely speaking, given $X$ there is no randomness in $f(X)$, so we expect the conditional variance to be zero.
Let $f(x) = c+ \beta x$.
\begin{eqnarray}
\operatorname{var} (f(X)|X) &=& E [ (f(X)-E[f(X)|X])^2 | X] \\
&=& E [ (f(X)-f(X))^2 | X] \\
&=& E[0 | X] \\
&=& 0
\end{eqnarray}
On the other hand (note that in this case we have $E[f(X)] = f(EX)$).
\begin{eqnarray}
\operatorname{var} (f(X)) &=& E [ (f(X)-E[f(X)])^2 ] \\
&=& E [ (f(X)-f(EX))^2] \\
&=& E[(\beta(X-EX))^2] \\
&=& \beta ^2 E[(X-EX)^2] \\
&=& \beta^2 \operatorname{var} X \\
&=& \beta^2 \sigma^2
\end{eqnarray}
A: $\operatorname{Var}(Y)=\mathbb{E}(Y-\mathbb{E}(Y))^2$ is a number.  Var(Y|X) is the conditional variance of $Y$ given $X$:
$$
\operatorname{Var}(Y\mid X):=\mathbb{E}[(Y-\mathbb{E}[Y\mid X])^2\mid X]
$$
is a function of the random variable $X$.
A: Note that given the value of $X$, $Y$ ceases to be random (presuming $\beta$ and $c$ are constants). Therefore, $$\mathrm{var}(Y|X) = 0.$$ 
On the other hand, if $X$ is not known then $Y$ can take on different values. Therefore, $\mathrm{var}(Y) \neq 0$. In fact, $\mathrm{var}(Y) = \beta^2 \mathrm{var}(X)=\beta^2 \sigma^2$.
