Marginal density functions of $X$ and $Y$

Question

Let $R$ be a region bounded by $y=0$,$y=1$,$x=1$,$y=\sqrt{2x}$. Choose a point $(X, Y)$ uniformly at random from the bounded region. I know that $$f_{X,Y}(x,y) = \frac{1}{\text{area}(R)} = \frac65, \qquad (x,y) \in R$$ However, I can't find the correct upper bound and lower bound of the integral when computing the marginal density functions of $X$ and $Y$. For $X$, I guess the upper bound is $1$ and lower bound is $\sqrt{2x}$. For $Y$, I guess the upper bound is $1$ and lower bound is $\frac{1}{2}y^2$. However, I don't think it is correct, since it is not the bounded region after I plot it. What is the correct way to do it? Thank you!

Answer 1

The Marginals are as follows:-

$$f_{X}(x)=\int_{0}^{\sqrt{2x}}f(x,y)\,\mathbf{1}_{\{0\leq x\leq \frac{1}{2}\}}\,dy+\int_{0}^{1}f(x,y)\mathbf{1}_{\{\frac{1}{2}\leq x\leq 1\}}\,dy$$

Which is simplified as $f_{X}(x)=\begin{cases}\frac{6}{5}\sqrt{2x}\,\,,0\leq x\leq\frac{1}{2}\\ \frac{6}{5}\cdot\frac{1}{2}\,\,,\frac{1}{2}\leq x\leq 1\end{cases}$

And $$f_{Y}(y)=\bigg(\int_{\frac{y^{2}}{2}}^{\frac{1}{2}}\,f(x,y)\,dx+\int_{\frac{1}{2}}^{1}f(x,y)\,dx\bigg)\mathbf{1}_{\{0\leq y\leq 1\}}=\frac{6}{5}(1-\frac{y^{2}}{2})\mathbf{1}_{\{0\leq y\leq 1\}}$$

Which written simply is $\displaystyle f_{Y}(y)=\frac{6}{5}(1-\frac{y^{2}}{2}) \,,\,0\leq y\leq 1$

See the figure below where I have divided the region into a parabolic part and a rectangular part (given by P and R) to see how I chose the limits of integration .