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What is the difference between $mathrmvar(Y)$ and $mathrmvar(Ymid X)$? If $Y = c + beta X$ and $operatornamevar(X)=sigma^2$, won't both come out to be the same, i.e., $beta^2sigma^2$?

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 $f(x) = c+ beta x$.
begineqnarray
operatornamevar (f(X)|X) &=& E [ (f(X)-E[f(X)|X])^2 | X] \
&=& E [ (f(X)-f(X))^2 | X] \
&=& E[0 | X] \
&=& 0
endeqnarray
On the other hand (note that in this case we have $E[f(X)] = f(EX)$).
begineqnarray
operatornamevar (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 operatornamevar X \
&=& beta^2 sigma^2
endeqnarray

$operatornameVar(Y)=mathbbE(Y-mathbbE(Y))^2$ is a number. Var(Y|X) is the conditional variance of $Y$ given $X$:
$$
operatornameVar(Ymid X):=mathbbE[(Y-mathbbE[Ymid X])^2mid X]
$$
is a function of the random variable $X$.

Note that given the value of $X$, $Y$ ceases to be random (presuming $beta$ and $c$ are constants). Therefore, $$mathrmvar(Y|X) = 0.$$
On the other hand, if $X$ is not known then $Y$ can take on different values. Therefore, $mathrmvar(Y) neq 0$. In fact, $mathrmvar(Y) = beta^2 mathrmvar(X)=beta^2 sigma^2$.