Exercise 4.1: Triangular (x, y) Area

Triangular two-dimensional area

A two-dimensional random variable is defined by the adjacent sketch:

For $(x, \ y)$ only values within the triangular-shaped region defined by the three vertices $(0,\ 1)$,$ $(4,\ 3)$,$ and $(4,\ 5)$ can occur.
Within the triangle, all the random variables $(x,\ y)$ are equally probable.
For the 2D–PDF, in this domain:

$$f_{xy}(x,y) = A .$$

In addition, the straight line $x = y$ ⇒ "angle bisector" is drawn in the graphic ⇒ see subtask (2).

Hint:

The exercise belongs to the chapter Two-Dimensional Random Variables.

Questions

$A \ = \ $

${\rm Pr}(x > y) \ = \ $

${\rm Pr}(x ≥ 2)\ = \ $

${\rm Pr}(y ≥ 3)\ = \ $

${\rm Pr}\big[(x ≥ 2) ∩ (y ≥ 3)\big]\ = \ $

${\rm Pr}\big[x ≥ 2\hspace{0.05cm} | \hspace{0.05cm} y ≥ 3\big]\ = \ $

${\rm Pr}\big[y ≥ 3\hspace{0.05cm} | \hspace{0.05cm} x ≥ 2\big]\ = \ $

Solution

(1) The volume under the two dimensional PDF is by definition equal to $1$:

Triangular 2D–PDF

$$\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}f_{xy}(x,y)\, {\rm d}x\, {\rm d}y=1.$$

The triangle area is $D = 0.5 \cdot 2 \cdot 4 = 4$.
Since in this definition area the PDF is constantly equal to $A$, we get.

$$A= 1/D\hspace{0.15cm}\underline{= 0.25}.$$

(2) For the solution we start from the adjacent sketch.

The area $x>y$ lies to the right of the angle bisector $x=y$ and is marked in green.
This green triangular area is $D_{\rm (2)} = 0.5 \cdot 1 \cdot 2 = 1 $, i.e. exactly one quarter of the total area $D$ of the definition area.
From this follows ${\rm Pr}(x > y)\hspace{0.15cm}\underline{= 0.25}$.

(3) For the wanted marginal PDF holds in this case:

Marginal PDF with respect to $x$

$$f_x(x)=\int_{-\infty}^{+\infty}f_{xy}(x,y)\, {\rm d}y=A\cdot B_y (x).$$

Here denotes $B_y(x)$ the width of the area $f_{xy} \ne 0$ in $y$–direction at the considered $x$–value.
It holds: $B_y(x) = x/2$. With $A = 0.25$ it follows $f_{x}(x) = x/8$ for the range $ 0 \le x \le 4$.

The wanted probability corresponds to the shaded area in the accompanying sketch. One obtains:

$$\rm Pr(\it x\ge \rm 2) = \rm 1-\rm Pr(\it x < \rm 2) = \rm 1-\frac{1}{2}\cdot2\cdot 0.25\hspace{0.15cm}\underline{ =0.75}. $$

The same result is obtained using 2D–PDF: To the right of $x = 2$ lies $3/4$ of the total definition area.

Marginal PDF with respect to $y$

(4) Analogous to the solution of the subtask (3) holds:

$$f_y(y)=\int_{-\infty}^{+\infty}f_{xy}(x,y)\, {\rm d}x=A\cdot B_x (y).$$

The width of PDF area in $x$–direction is zero for $y \le 1$ and $y \ge 5$ respectively.
The maximum is at $y=3$ and gives $B_x(y=3) = 2$.
In between, the increase and decrease of $B_x(y)$ is linear, yielding a triangular-shaped PDF.
The probability that $y=3$ corresponds to the green shaded area in the adjacent sketch.
Because of the symmetry, one obtains:

$${\rm Pr}(y ≥ 3)\hspace{0.15cm}\underline{ =0.5}. $$

The same result is obtained using 2D–PDF: Above the horizontal $y= 3$ lies half of the total definition area.

On subtask (5)

(5) If $y \ge 3$ $($red highlighted triangle $D)$, it is always true $x \ge 2$ $($green outlined trapezoid $T)$.

This means: In this example $D$ is a subset of $T$, and it holds:

$${\rm Pr}[(x ≥ 2) ∩ (y ≥ 3)] = {\rm Pr}(y ≥ 3) \hspace{0.15cm}\underline{= 0.50}.$$

(6) According to the solution to the subtask (5), it follows from $y \ge 3$ with certainty also $x \ge 2$.

So the conditional probability we are looking for is:

$${\rm Pr}[x ≥ 2\hspace{0.05cm} | \hspace{0.05cm} y ≥ 3]\hspace{0.15cm}\underline{= 1}.$$

(7) This subtask can be solved using Bayes' theorem and the results from (2) and (5):

$$\rm Pr(\it y \ge \rm 3\hspace{0.1cm}|\hspace{0.1cm} \it x \ge \rm 2) = \frac{ \rm Pr((\it x \ge \rm 2)\cap(\it y \ge \rm 3))} {\rm Pr(\it x \ge \rm 2)}=2/3\hspace{0.15cm}\underline{=0.667}.$$

Or expressed differently: The area $D$ of the triangle with red background makes $2/3$ of the area of the trapezoid with green border $T$.