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If $X,Y$ are jointly continuous random variables with the following joint PDF:

$$f_{X,Y}(x,y)=\begin{cases} \tfrac{21}{4}x^2y & : x^2< y<1 \\ 0 & : \textsf{otherwise} \end{cases}$$

How do I find the expected value of $Y$ and the variance of $Y$ I know the definition of the expected value (of $X$) is the integral of $x\cdot f(x)$ but as this a joint expectation I'm unsure how to create this integral.

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The real-valued support $x^2<y<1$ means that: $0<y<1$ and $-\sqrt y<x<\sqrt y$, so:

$$\begin{align}\mathsf E(Y^k) & = \iint_{\Bbb R^2} y^k\, f_{X,Y}(x,y)\operatorname d x\operatorname d y \\[1ex] & = \tfrac{21}{4} \int_0^1 y^{k+1} \int_{-\sqrt y}^{\sqrt y} x^2 \operatorname d x\operatorname d y \end{align}$$

Then find $\mathsf E(Y)=\mathsf E(Y^1)$ and $\mathsf {Var}(Y)=\mathsf E(Y^2)-\mathsf E(Y^1)^2$

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