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I am trying to find the $f_Y(y)$ marginal pdf of a joint pdf that has some particular ranges on the variables:

$$ f(x,y) = \begin{cases} \frac{4x^2}{15}, & \text{1 $\leq$ $x$ $\leq$ 2 and 0 $\leq$ $y$ $\leq$ $x$} \\ 0, & \text{otherwise} \end{cases} $$

Specifically, I need to find the marginal $f_Y(y)$ for two different scenarios:

  1. 0 $\leq$ $y$ $\leq$ $1$
  2. 0 $<$ $y$ $\leq$ $2$

I normally would just integrate over $x$ to get to $f_Y(y)$, but I don´t understand what do I need to do different in splitting the marginal pdf in these two scenarios or how to find the marginal for each scenario.

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2 Answers 2

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The way you want to approach the problem is drawing the region of integration (it should be a trapezoid with the top being the line $y=x$). If you fix $0\leq y \leq 1$, you will notice that the range of integration in the trapezoid is the entire base (i.e. $x=1$ to $x=2$).

On the other hand, if $1<y\leq 2$, the $x$ integration range is from the line $x=y$ to the vertical line $x=2$.

Outside of these ranges, there are no $x$ in the range to integrate.

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  • $\begingroup$ Thank you so much! I could not get the second part before but now I get it :) $\endgroup$
    – paswort
    Mar 10, 2020 at 21:42
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To evaluate the marginal for $Y$, you need to express the support in terms of $y$ relative to contants, and $x$ relative to $y$.   The support is provided as an expression the other way around, so we need to do a little work.

We know that the maximum for $x$ is $2$ so since $y\leq x$ is a constraint, the maximum for $y$ must also be $2$.   So now look at values for $y$ from $0$ to $2$ and find how $x$ is constrained.   Well when $y<1$, the minumum value for $x$ is $1$, but otherwise it is $y$.

The support may be partitioned by whether $y\lt 1$ or not. $${\quad\{\langle x,y\rangle: 0\leq y\leq x ~\land~ 1\leq x\leq 2\}\\=\{\langle x,y\rangle: (0\leq y\lt 1~\land~ 1\leq x\leq 2)~\lor~(1\leq y\leq 2~\land~ y\leq x\leq 2)\}}$$

So therefore

$$f_Y(y)=\mathbf 1_{0\leq y\lt 1} \int_1^2 f_{X,Y}(x,y)\,\mathrm d x+\mathbf 1_{1\leq y\leq 2}\int_y^2f_{X,Y}(x,y)\,\mathrm d x$$

Or if you prefer:

$$f_Y(y)=\begin{cases}\displaystyle\int_1^2 f_{X,Y}(x,y)\,\mathrm d x&:&0\leq y\lt 1\\[1ex]\displaystyle\int_y^2f_{X,Y}(x,y)\,\mathrm d x&:&1\leq y\leq 2\\[1ex]0&:&\textsf{otherwise}\end{cases}$$

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