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A joint pdf, $f_{X,Y}(x,y)=5$, is given with the following intervals:

$-1<x<1$

$x^2<y<x^2+{1\over{10}}$

I am trying to find marginal pdf of $f_Y(y)$ but I am stuck. Trying for hours.Help would be appreciated.

enter image description here

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    $\begingroup$ Have you plotted the area where the joint density is positive? $\endgroup$ – JimB Jan 20 at 17:01
  • $\begingroup$ I did.it is the area between two quadratic curves. y=x^2 and y=x^2+0.1 and -1<x<+1. Would it be y^1/2 and (y-0.1)^(1/2). This is my best answer so far, but I feel like I am not scanning all of the area with this. $\endgroup$ – Kamuran Karam Jan 20 at 17:12
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    $\begingroup$ From that figure you'll see that for some values of $y$ there are either 0, 1, or 2 sections of $x$ to integrate over. (Or you can simplify that a bit by noting the symmetry.) $\endgroup$ – JimB Jan 20 at 17:21
  • $\begingroup$ That is the problem. How many regions are there? Please! $\endgroup$ – Kamuran Karam Jan 20 at 17:27
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The joint density can be expressed as $5$ times indicator functions:

$$f_{X,Y}(x,y) = 5\times\mathbb{I}_{-1<x<1}\times\mathbb{I}_{x^2<y<x^2+1/10}$$

The marginal density of $Y$, $f_Y(y)$, is therefore the integral of the joint $f_{X,Y}(x,y)$ in $x$ over the conditional integration domain:

$$\{x; -1<x<1,\ y-1/10<x^2<y\}$$

which can also be expressed as

$$\{x; \max(-1,-\sqrt{y})<x<-\sqrt{\max(0,y-1/10)}\}\cup\{x; \min(1,\sqrt{y})>x>\sqrt{\max(0,y-1/10)}\}$$

with the marginal support of $Y$ obtained by the extreme values of its conditional domain:

$$y>\min_{-1<x<1}x^2=0\quad\text{ and }\quad y<\max_{-1<x<1}x^2+1/10=11/10$$

meaning that$$f_Y(y)=\int_{\{x;\ f_{X,Y}(x,y)>0\}}f_{X,Y}(x,y)\,\text{d}x$$is given by [using the fact that $f_{X,Y}(x,y)=f_{X,Y}(-x,y)$ is symmetric]

$$f_Y(y)=2\int_{\max(0,y-1/10)^{1/2}}^{\min(1,\sqrt{y})} 5\,\text{d}x$$

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Each particular value of $y$ corresponds to a horizontal line.

If $0.1 < y < 1,$ then as you go from left to right, that line passes through the red curve, then the blue curve, then the blue curve again, then the red curve again, so you get $$ f_Y(y) = \int_{-\sqrt y}^{-\sqrt{y-0.1}} 5 \, dx + \int_{\sqrt{y-0.1}}^{\sqrt y} 5 \, dx \quad \text{if } 0.1<y<1. $$

If $0\le y \le 0.1$ then the horizontal line passes through the red curve twice, so you get $$ f_Y(y) = \int_{-\sqrt y}^{+\sqrt y} 5\, dx \quad \text{if } 0\le y\le 0.1. $$

If $1\le y\le 1.1$ then the horizontal line passes through the blue curve twice, so you get $$ f_Y(y) = \int_{-1}^{-\sqrt{y-0.1}} 5\,dx + \int_{\sqrt{y - 0.1}}^1 5\, dx \quad \text{if } 1\le y\le1.1. $$

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  • $\begingroup$ Typo: the lower bound of the first integral on the last line is -1 and not 0. $\endgroup$ – Xi'an Jan 21 at 12:21

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