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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 $y.$

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    $\begingroup$ The range of $x$ is from $-\sqrt y$ to $\sqrt y$. $\endgroup$ – Shubham Johri Dec 6 '19 at 6:52
  • $\begingroup$ How to get the range of x? @ShubhamJohri $\endgroup$ – Mel Dec 6 '19 at 6:58
  • $\begingroup$ See my comment under my answer for further information @Mel. $\endgroup$ – xiA Dec 6 '19 at 7:52
  • $\begingroup$ Note that the marginal support for $Y$ is $[0,1]$ $\endgroup$ – Henry Dec 6 '19 at 9:08
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To find the marginal distribution here you must integrate over the joint distribution, in this case, with respect to $x.$

$$ f_Y(y) = \int_{a}^{b} f(x,y) \ dx , \ x \in [a,b]$$

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  • $\begingroup$ Problem is how do you find the limits to that integral? thanks! @xiA $\endgroup$ – Mel Dec 6 '19 at 7:05
  • $\begingroup$ @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 $x^2\le y$. $\endgroup$ – Shubham Johri Dec 6 '19 at 7:16
  • $\begingroup$ @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. $\endgroup$ – xiA Dec 6 '19 at 7:45
  • $\begingroup$ thank you! @xiA $\endgroup$ – Mel Dec 6 '19 at 7:54
  • $\begingroup$ Please upvote if you found anything useful. Thanks. $\endgroup$ – xiA Dec 6 '19 at 8:04

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