This seems counter-intuitive to me since variance is a difference of expectations and afaik, unconditional expectation is a real number.
Apparently, $X_t$ where $dX_t = Y_t dW_t$, where $Y_t$ is an independent Brownian motion to $W_t$, has random variance.
Solving the SDE gives $X_t = X_0 + \int_0^t Y_s dW_s$
Computing the first and second moments give:
$E[X_t] = E[X_0] + E[\int_0^t Y_s dW_s]$
$E[X_t^2] = E[X_0^2] + 2E[X_0 \int_0^t Y_s dW_s] + E[\int_0^t Y_s^2 ds]$ by Itō isometry
$ = E[X_0^2] + 2E[X_0 \int_0^t Y_s dW_s] + \int_0^t E[Y_s^2] ds$ by Tonelli's theorem
I'm guessing that $E[\int_0^t Y_s dW_s]$ or $E[X_0 \int_0^t Y_s dW_s]$ is random? Why? It seems that $\int_0^t Y_s dW_s$ is a random variable with a unique mean given by $E[\int_0^t Y_s dW_s]$.
What am I not getting here?
I'm not too familiar with Brownian motions, but I think the problem is basically the same as in the case of individual random variables as in muaddib's answer: "Take the process $X$ where $X_0$, $X_1$ are iid positive random variables and $X_2$ is a normal random variable with variance $X_0+X_1$ and mean zero." Here $X_2$ could be said to have "random variance", namely in the sense that the conditional variance $\operatorname{Var}[X_2\mid X_0+X_1]=X_0+X_1$ is a random variable, which could be expressed as "the variance of $X_2$ depends on the value of $X_0+X_1$". Nevertheless, you are of course right that $X_2$ is a standard random variable with a standard non-random variance, which according to the law of total variance is given by
\begin{align} \operatorname{Var}[X_2]&=E_{X_0+X_1}(\operatorname{Var}[X_2\mid X_0+X_1])+\operatorname{Var}_{X_0+X_1}[E(X_2\mid X_0+X_1)]\\ &=E_{X_0+X_1}(X_0+X_1)+\operatorname{Var}_{X_0+X_1}\cdot\,0\\ &=E(X_0+X_1)\;. \end{align}