Conditional Variance of PDF A random vector $(X,Y)$ has a continuous distribution with a density function $$f(x,y)=\begin{cases}c⋅x & \text{when }0 ≤ x ≤ 2, \max\{0,1−x\} ≤ y ≤2−x\\ 0& \text{otherwise}\end{cases}$$ where $c > 0$ is a constant. Find variance of a $Y$ conditioned on $X = 1.5$, $Var(Y |X = 1.5)$.
I found $c = \frac 67$ with the given integral. Now I want to ask how can I find variance?
I found variance as -87/3136. Is it possible ?
Here is my attempt

Thank you

You have the region being the area trapped between the triangles $B$ and $A$ . Let it be denoted by $S$ .

You have $$\iint_{S}f(x,y)\,dxdy=1$$
$$\iint_{S}f(x,y)\,dxdy=\iint_{B}f(x,y)\,dxdy-\iint_{A}f(x,y)\,dxdy=\frac{8c}{6}-\frac{c}{6}=\frac{7c}{6}$$.
Hence $c=\frac{6}{7}$ as you correctly calculated.
Now the marginal density for $X$ is found by integrating the joint pdf over $y$ .
Hence $f_{X}(x)=\begin{cases}\int_{1-x}^{2-x}cx\,dy\,,0\leq x< 1\\\int_{0}^{2-x}cx\,dy\,,1\leq x\leq 2\end{cases}$
Hence $f_{X}(x)=\begin{cases}cx\,,0\leq x<1\\cx(2-x)\,,1\leq x\leq 2\end{cases}$
Now $$f_{Y|X=x}(y)=\frac{f(x,y)}{f_{X}(x)}$$
Hence $f_{Y|X=x}(y)=\begin{cases}\frac{cx}{cx}\,,0\leq x <1,\,,\max\{0,1-x\}\leq y\leq 2-x\\ \frac{cx}{cx(2-x)}\,,1\leq x<2 ,\,\max\{0,1-x\}\leq y\leq 2-x\end{cases}$
That is : $$f_{Y|X}=\begin{cases}1\,,0\leq x<1 \,,\max\{0,1-x\}\leq y\leq 2-x \\ \frac{1}{2-x}\,,1\leq x \leq 2 ,\max\{0,1-x\}\leq y\leq 2-x\end{cases}$$
Hence as $x=\frac{3}{2}$ we have
$f_{Y|X=x}(y)=2\,,0\leq y\leq 2-\frac{3}{2}$
That is $$f_{Y|X=\frac{3}{2}}(y)=2\cdot\mathbf{1}_{\{0\leq y\leq \frac{1}{2}\}}$$ .
Then $$\Bbb{E}(Y|X=\frac{3}{2})=\int_{0}^{\frac{1}{2}}2y\,dy=\frac{1}{4}$$
And $$\Bbb{E}(Y^{2}|X=\frac{3}{2})=\int_{0}^{\frac{1}{2}}2y^{2}\,dy=\frac{1}{12}$$
Hence $$\text{Var}(Y|X=\frac{3}{2})=\frac{1}{12}-\frac{1}{4^{2}}=\frac{1}{12}-\frac{1}{16}=\frac{1}{48}$$ .
