Difference between $\operatorname{Var}(Y)$ and $\operatorname{Var}(Y\mid X)$?

Question

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., $\beta^2\sigma^2$?

Answer 1

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}

Answer 2

$\operatorname{Var}(Y)=\mathbb{E}(Y-\mathbb{E}(Y))^2$ 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 $X$.

Answer 3

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, $\mathrm{var}(Y) = \beta^2 \mathrm{var}(X)=\beta^2 \sigma^2$.