# Joint distribution integral limits for marginal

Given a joint distribution as: $$f(x,y)=\frac{21}{4} \cdot x^2 y \,\,\text{ where } x^2\le y\le 1$$

I have problem finding the limits for $$x$$ to do the integral to find marginal of $$y.$$

• The range of $x$ is from $-\sqrt y$ to $\sqrt y$. – Shubham Johri Dec 6 '19 at 6:52
• How to get the range of x? @ShubhamJohri – Mel Dec 6 '19 at 6:58
• See my comment under my answer for further information @Mel. – xiA Dec 6 '19 at 7:52
• Note that the marginal support for $Y$ is $[0,1]$ – Henry Dec 6 '19 at 9:08

To find the marginal distribution here you must integrate over the joint distribution, in this case, with respect to $$x.$$
$$f_Y(y) = \int_{a}^{b} f(x,y) \ dx , \ x \in [a,b]$$
• @Mel If you have shaded the region of non-zero joint probability density in the $xy$ plane, you can see that for a particular value of $y\in[0,1],x$ ranges from $-\sqrt y$ to $\sqrt y$. This is a consequence of the inequality $x^2\le y$. – Shubham Johri Dec 6 '19 at 7:16
• @Mel Go a head and plot $x^{2}\le y\le1$ what you will find is a parabola type of shape but bounded at $y \le 1$. You can then go ahead and confirm by plotting $x^2$ for a sanity check which should overlay your existing plot. Once you have done that check, ask yourself, what is the inverse of $x^2$? Namely, solve $y=x^2 for x$ If you work that one out you will arrive at $\pm \sqrt{y}$ as Shubham Johri has stated. – xiA Dec 6 '19 at 7:45