Given joint pdf, how can I find $\operatorname{Var}(X)$ without using the marginal distribution of $X$?

Question

$(X,Y)$ has joint pdf $\frac{1}{y}$ for $0<x<y<1$.

I would usually use the marginal pdf to get the expected value. But the question doesn't let me use the marginal distribution of $X$. I think this means that I am not allowed to use the marginal pdf... Is there any other way to do this?

Answer 1

The variance is $\int\int x^{2} f(x,y)dxdy -(\int\int x f(x,y)dxdy)^{2}$. This can be written as $(\int_0^{1} \int_o^{y}x^{2} \frac 1 y dxdy-\int_0^{1} \int_o^{y}x \frac 1 y dxdy)^{2}$. I will leave the rest to you.

Answer 2

$$\mathbb{E}f\left(X\right)=\int_{0}^{1}\int_{0}^{y}y^{-1}f\left(x\right)dxdy=\int_{0}^{1}y^{-1}\int_{0}^{y}f\left(x\right)dxdy$$

Apply this on the functions $x\mapsto x$ and $x\mapsto x^2$ in order to find $\mathbb EX$ and $\mathbb EX^2$.

Then use: $$\mathsf{Var}(X)=\mathbb EX^2-(\mathbb EX)^2$$

Answer 3

Given $Y$, $X$ has a uniform distribution over the segment $(0,y)$, so $E(X\vert Y)=\tfrac{y}{2}$ and $V(X\vert Y)=\tfrac{y^2}{12}$. In addition, $f_Y(y)=\int_0^y \tfrac{1}{y}dx=1$, so the marginal distribution of $Y$ is $U(0,1)$.

From the law of total variance, $$V(X)=E(V(X\vert Y))+V(E(X\vert Y))=V(\tfrac{Y}{2})+E(\tfrac{Y^2}{12})=\tfrac{1}{2^2\cdot12}+\tfrac{1}{12}(V(Y)+E^2(Y))=\tfrac{1}{48}+\tfrac{1}{12}(\tfrac{1}{12}+\tfrac{1}{4})=\tfrac{7}{144}$$