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 the marginal pdf of $f_Y(y)$ but I am stuck.
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3$\begingroup$ Have you plotted the area where the joint density is positive? $\endgroup$– JimBCommented Jan 20, 2019 at 17:01
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$\begingroup$ 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 KaramCommented Jan 20, 2019 at 17:12
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1$\begingroup$ 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.) $\endgroup$– JimBCommented Jan 20, 2019 at 17:21
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$\begingroup$ That is the problem. How many regions are there? Please! $\endgroup$– Kamuran KaramCommented Jan 20, 2019 at 17:27
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2 Answers
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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 $$ f_Y(y) = \int_{-1}^{-\sqrt{y-0.1}} 5\,dx + \int_{\sqrt{y - 0.1}}^1 5\, dx \quad \text{if } 1\le y\le1.1. $$
answered Jan 21, 2019 at 1:38
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$\begingroup$ Typo: the lower bound of the first integral on the last line is -1 and not 0. $\endgroup$– Xi'anCommented Jan 21, 2019 at 12:21