Determining the CDF of an absolutely continuous random vector, given its density.

Problem. I'm given the following problem:

Let $V$ be the random vector with density:

\begin{cases} {1\over 900}(45-x-2y) & x\leq y\leq 3x\;\;and\;\;0\leq x\leq 6\\ 0 & else \end{cases}

Find the CDF of $V$.

Attempt.

While I'm familiar with the discrete case, neither the course nor the text cover the mechanics of solving such a problem in the continuous case in much depth. Here's how I've approached it.

I begin by sketching the support of $V$.

Recall that the CDF may be found by integrating over the density. I then identify $6$ regions of of $\mathbb{R}^2$ over which the integral takes on a unique form.

I then determine the CDF region-by-region.

In region $\mathbb{I}$, the CDF is clearly $0$.

In region $\mathbb{II}$, the CDF at $(p,q)$ is given by $\int_0^p\int_{p}^q {1\over 900}(45-x-2y)\;dydx$.

In region $\mathbb{III}$, the CDF at $(p,q)$ is given by $\int_0^6\int_p^q{1\over 900}(45-x-2y)\;dydx$.

In region $\mathbb{IIII}$, the CDF at $(p,q)$ is given by $\int_0^p\int_p^{3p} {1\over 900}(45-x-2y)\;dydx$.

In region $\mathbb{V}$, the CDF is clearly $1$.

Yielding density:

\begin{cases} {{p(-3p^2+p(1+90)+2(q-45)q)}\over 1800} & p \leq q \leq 3p\;\; and\;\; 0 \leq p \leq 6\\ {{(-p^2+42p+(q-42)q)}\over 150} & p > 6 \;\;and\;\; 0 \leq q \leq 18\\ {{p(p^2-p(q+30)-6(q-15)q)}\over 900} & 0\leq p\leq 6\;\;and\;\; q \geq 3p\\ ? & 0 < p < 6 \;\; and \;\; q < p\\ 1 & p > 6 \;\;and\;\; q > 18\\ 0 & x < 0 \;\; or \;\; y < 0\\ \end{cases}

Unfortunately, I'm a bit stuck on determining the form of the integral for region $\mathbb{VI}$. More importantly, however, I'm not sure if this is a valid approach, and, if so, whether I've properly written the integrals for each region.

Is this a valid solution? How may the integral for the remaining region be written?

Solution 1:

$\renewcommand{\dd}[1]{\,\mathrm{d}#1}$Since this is mainly about the region and integration limits, for convenience I'll use $f$ to denote the nonzero functional form of the density.

Region $\mathbf{\mathbb{II}}$, red triangle where the density $f(p,q)$ is non-zero.

Integrating "vertical stripes first", the dummy $y$ is confined by the dummy $x$ in the manner of $x<y<3x$: $$\begin{align} \sideset{_{\mathbb{II}}}{}{\Pr\{X < p\,,~Y<q\} } &= \int_{x=0}^p\int_{y=x}^{\min(3x,q)} f \dd{y}\dd{x}~, \qquad \text{split at}\quad 3x \leq q \Leftrightarrow x \leq q/3 \\ &= \int_{x=0}^{q/3}\int_{y=x}^{3x} f \dd{y}\dd{x} + \int_{x=q/3}^p\int_{y=x}^q f \dd{y}\dd{x} \end{align}$$ Note that by definition of being in this region, we have $q<3p~$, namely, the integration limit for $x$ can be double-checked: $~p>q/3$.

Equivalently, integrating "horizontal stripes first", the dummy $x$ is confined by the dummy $y$ in the manner of $\frac{y}3<x<y$: $$\begin{align} \sideset{_{\mathbb{II}}}{}{\Pr\{X < p\,,~Y<q\} } &= \int_{y=0}^q\int_{x=y/3}^{\min(y,p)} f \dd{x}\dd{y}~, \qquad \text{split at}\quad y \leq p \\ &= \int_{y=0}^p\int_{x=y/3}^y f \dd{x}\dd{y} + \int_{y=p}^q\int_{x=y/3}^p f \dd{y}\dd{x} \end{align}$$ Again, by definition of being in this region, we have $q>p~$ for the $y$ limit.

Also note that the boundary values $x=6$ or $y=18$ do not come into play at all.¹ Basically the boundary values only appear when the rectangle $\{X < p\,,~Y<q\}$ cuts across the boundaries, like the next case.

Region $\mathbf{\mathbb{III}}$, yellow, where $f(p,q)=0$ yet the rectangle $\{X < p\,,~Y<q\}$ includes part of the red triangle.

This should be split into two regions, and the boundary is a horizontal ray emanating from $(6,6)$ towards right. Denote the lower region that is adjacent to the purple as $\mathbb{III}_L$, and denote the upper one as $\mathbb{III}_U$ that touches the red and orange.

For region $\mathbb{III}_L$ where by definition $p>6$ and $q<6$, doing "horizontal stripes first" is more convenient (just a single integral). $$\sideset{_{\mathbb{III}_L}}{}{\Pr\{X < p\,,~Y<q\} } = \int_{y=0}^p\int_{x=y/3}^y f \dd{x}\dd{y} $$

For region $\mathbb{III}_U$ that is by definition $p>6$ and $6<q<18$, the integration region always contain the lower half of the red triangle. $$\sideset{_{\mathbb{III}_U}}{}{\Pr\{X < p\,,~Y<q\} } = \Pr\{X < 6\,,~Y<6\} + \int_{y=6}^q\int_{x=y/3}^6 f \dd{x}\dd{y} $$

Whenever the support is not an axis-aligned rectangle, it is a common feature of the CDF "outside the support" to be building on top of "lower-left" regions. Note that $\mathrm{CDF}(p, q) = \mathrm{CDF}(6, q)$ for any $p>6$, and $\mathrm{CDF}(6,6) = \Pr\{X < 6\,,~Y<6\}$.

It takes some time getting used to doing integration for regions where the density is zero (outside the support). Whenever in doubt, please go back to the basics and follow the definition. What is the CDF $\Pr\{X < p\,,~Y<q\}$ really encompassing? Sketch the rectangle $\{-\infty < X < p\,,~ -\infty<Y<q\}$ to see what regions should be included in the integration (and you'll see that part of the support is involved).

Footnote 1.
If so desired, one can check the continuity at $p=6^{-}$ coming from region $\mathbb{II}$ versus $p=6^{+}$ coming from region $\mathbb{III}_U$ with $6 < q < 18$, and then with $q>6$ and $p=6^{-}$ coming from region $\mathbb{II}$ versus $p=6^{+}$ coming from region $\mathbb{III}_L$ with $q > 6$.