Let $R$ be the region in $\mathbb{R}^2$ bounded by the three lines $y=2.1$, $y=x$, $y=−x$ and contains the point $(0,\frac{1}{2})$. Let $(X,Y)$ be the coordinate of a point chosen uniformly at random on the region $R$. Let $f_X$ be the marginal density of the random variable $X$ that is a continuous function. Find $f_X(-0.84)$. Give your answer in four decimal places.
Progress: I have found the area of $R$ to be equal to $4.41$, and so $R$ has density $\frac{1}{4.41}$ at the support.
So does this mean $f_x(-0.84)=\int_{0.84}^{2.1}\frac{1}{4.41}dy\,$?
Best Answer
In short, yes that is correct.
As you obtained, $f_{X,Y}(x, y) = \frac{1}{4.41}$ where supported. The support of the distribution is $\{0 \lt y \lt 2.1, -y \lt x \lt y \}$ or, alternatively, $\{-2.1\lt x\lt 2.1, \lvert x\rvert \lt y\lt 2.1\}$ .
So if we find the marginal density of $X$ first,
$ \displaystyle f_X(x) = \int_{|x|}^{2.1} \frac{1}{4.41} ~dy = \frac{2.1 - |x|}{4.41}\quad[-2.1<x<2.1]$
In your case, $x = -0.84$
That is same as your answer.