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 $X = 1.5$, $Var(Y |X = 1.5)$.
Here is my attempt
I found $c = \frac 67$ with the given integral. Is it correct ? Also, I want to ask how can I find variance?
Thank you
$c=6/7$ is correct.
Similarly, :
$$\mathsf E(g(X,Y)) = \frac 67\left(\int_0^1\int_{1-x}^{2-x} xg(x,y)\,\mathrm dy\,\mathrm dx + \int_1^2 \int_{0}^{2-x} x g(x,y)\,\mathrm dy\,\mathrm dx\right)$$
And for the conditional variance
$$\mathsf{Var}(Y\mid X=1.5)=\dfrac {\mathsf E(Y^2\,\mathbf 1_{X=1.5})}{\mathsf E(\mathbf 1_{X=1.5})} -\left(\dfrac{\mathsf E(Y\,\mathbf 1_{X=1.5})}{\mathsf E(\mathbf 1_{X=1.5})}\right)^2$$
Where: $$\mathsf E(Y^j\,\mathbf 1_{X=k}) = \frac 67\left(\mathbf 1_{k\in[0,1)}~k\int_{1-k}^{2-k} y^j\,\mathrm dy + \mathbf 1_{k\in[1,2]}~k \int_{0}^{2-k} y^j\,\mathrm dy\right)$$