Why is my variance negative? Let $Y_1 \sim Gamma(2,1)$, and $Y_2|Y_1 \sim f_{Y_2}(Y_2)$, where $y_1\geq y_2\geq 0$.
Where
$$f_{Y_2}(y_2) = \frac{1}{y_1}$$
$$f_{Y_1}(y_1) = y_1 \exp( -y_1)$$
Goal: Find standard deviation of $Y_2 - Y_1$
Start with
$$Var(Y_1) = 1, E(Y_1) = 1$$
$$E(Y_2|Y_1) = \int^{y_1}_0 \frac{y_2}{y_1}dy_2 = \frac{y_1}{2}$$
$$E(Y_2^2|Y_1)= \int^{y_1}_0 \frac{y_2^2}{y_1}dy_2 = \frac{y_1^2}{3} $$
$$Var(Y_2|Y_1) = E(Y^2_2|Y_1) - (E(Y_2|Y_1))^2 = \frac{y_1^2}{3} - \frac{y_1^2}{4}  = \frac{y_1^2}{12}$$
\begin{align*}
Var(Y_2) = &  E(Var(Y_2|Y_1))+ Var(E(Y_2|Y_1))\\
         = &  E\left (\frac{y_1^2}{12} \right ) + Var\left (\frac{y_1}{2}\right )\\
         = & E\left (\frac{y_1^2}{12}\right ) + Var\left (\frac{y_1}{2}\right )\\
         = & \frac{2}{12} + \frac{1}{4}\\
         = & \frac{5}{12}
\end{align*}
The formula I am using is $Var(Y_1 - Y_2) = Var(Y_1) + Var(Y_2) - 2Cov(Y_1, Y_2)$
So now I need
\begin{align*}
Cov(Y_1, Y_2) = & E(Y_1 Y_2) - E(Y_1)E(Y_2)\\
              = & E(Y_1 Y_2) - E(Y_1)E(E(Y_2|Y_1))\\
              = & E(Y_1 Y_2) - (1)\frac{1}{2}
\end{align*}
Now find
\begin{align*}
E(Y_1Y_2) = &\int^{\infty}_0 \int^{y_1}_0 y_1 y_2 \exp(-y_1) dy_2 dy_1\\
          = & 3
\end{align*}
Thus $Cov(Y_1, Y_2) = 3 - \frac{1}{2} = \frac{5}{2}$
Hence
\begin{align*}
Var(Y_1-Y_2) = & Var(Y_1) + Var(Y_2) - 2Cov(Y_1, Y_2) \\
             = & 1 + \frac{5}{12} - 5
\end{align*}
Which clearly can't be true.  I don't where I went wrong.
I tried another approach by transforming the random variables and finding the distribution of $Y_1 - Y_2$ which did produce a realistic answer.  However, I would like to know what was wrong with this approach.
 A: I found my mistake.
$$E(Y_1) = 2$$
$$Var(Y_1) = 2$$
Therefore
\begin{align*}
Var(Y_2) = & E\left (Var(Y_2|Y_1)\right) + Var\left ( E(Y_2|Y_1)\right ) \\
         = & E \left ( \frac{y^2_1}{12}\right ) + Var \left( \frac{y_1}{2} \right ) \\
         = & \frac{6}{12} + \frac{2}{4}\\
         = &1
\end{align*}
In addition, the $Cov(Y_1, Y_2) = 3 - (2)\frac{2}{2} = 1$
\begin{align*}
Var(Y_1 - Y_2) = &Var(Y_1) + Var(Y_2) - 2 Cov(Y_1, Y_2)\\
               = & 2 + 1 - 2(1)\\
               = & 1
\end{align*}
A: $$Y_1\sim\mathcal{Gamma}(2,1) \iff [~\mathsf E(Y_1)=2, \mathsf{Var}(Y_1)=2~]$$

$$Y_2\mid Y_1\sim\mathcal U(0,Y_1) \iff \left[\mathsf E(Y_2\mid Y_1)=\tfrac 12Y_1,\mathsf{Var}(Y_2\mid Y_1)=\tfrac 1{12}Y_1^2\right]$$

$$\begin{align}\mathsf{Var}(Y_2)&=\mathsf E(\mathsf{Var}(Y_2\mid Y_1))+\mathsf{Var}(\mathsf{E}(Y_2\mid Y_1))\\[1ex]&=\tfrac1{12}\mathsf E(Y_1^2)+\tfrac 14\mathsf{Var}(Y_1)\\[1ex]&=\tfrac 1{12}(\mathsf{Var}(Y_1)+\mathsf{E}(Y_1)^2)+\tfrac 14\mathsf{Var}(Y_1)\\[1ex]&=\tfrac 1{12}(2+2^2)+\tfrac 14\cdot 2\\[1ex]&=1\\[2ex]\mathsf{Cov}(Y_1,Y_2)&=\mathsf E(\mathsf{Cov}(Y_1,Y_2\mid Y_1))+\mathsf{Cov}(\mathsf E(Y_1\mid Y_1),\mathsf E(Y_2\mid Y_1))\\&=\mathsf E(0)+\mathsf{Cov}(Y_1,\tfrac 12Y_1)\\&=\tfrac 12\mathsf{Var}(Y_1)\\&=1\\[2ex]\mathsf {Var}(Y_2-Y_1) &= \mathsf{Var}(Y_2)+\mathsf{Var}(Y_1)-2\mathsf{Cov}(Y_1,Y_2)\\[1ex]&=1+2-2\\[1ex]&=1\end{align}$$
