What is the domain of my marginal PDF function

Question

Consider the joint pdf of $(X, Y)$ given by

$$f(x, y) = \begin{cases} 25x^{4}y^{4} & \text{ if } |x| \leq y, 0 < y < 1 \\ 0, & \text{ otherwise.} \end{cases} $$

Then,

$$f_{X}(x) = \int_{0}^{1} 25x^{4}y^{4 }\mathop{dy}$$

$$= 5x^{4}. $$

But what is the restriction to the domain? Is it just $|x| \leq y$? Or is it $x \in \mathbb{R}$? Why? $y$ is no longer in the function. Does it matter? Also to compute $\mathbb{E}[X]$, is the following correct:

$$\mathbb{E}[X] = \int_{-y}^{y} 5x^{4} \mathop{dy} = 2y^{5} $$

Answer 1

Your calculation of $f_X$ is wrong. The correct value is $\int_{|x|}^{1} 25x^{4}y^{4}dy=5x^{4}(1-|x|^{5})$ for $|x| \leq 1$. Also $EX=\int_{-1}^{1}5x^{5}(1-|x|^{5})dx=0$ because the integral of an odd function from $-1$ to $1$ is $0$.

Answer 2

Just a way of working that might prevent you from making mistakes in cases like this.

---

In general let $\left[\text{conditions on }x,y\right]$ be a function that takes value $1$ if the conditions are satisfied and takes value $0$ otherwise.

Then $f\left(x,y\right)=\left[\left|x\right|\leq y,0<y<1\right]25x^{4}y^{4}$ so that

$$f_{X}\left(x\right)=\int\left[\left|x\right|\leq y,0<y<1\right]25x^{4}y^{4}dy=\int_{0}^{1}\left[\left|x\right|\leq y\right]25x^{4}y^{4}dy$$

Now discern two cases:

• If $\left|x\right|\geq1$ then $f_{X}\left(x\right)=0$

• If $\left|x\right|<1$ then $f_{X}\left(x\right)=\int_{\left|x\right|}^{1}25x^{4}y^{4}dy=\left[5x^{4}y^{5}\right]_{\left|x\right|}^{1}=5x^{4}\left(1-\left|x\right|^{5}\right)$

---

Use this if you like it. If you think it is too cumbersome then just leave it aside.