Let (X, Y) have joint density f(x,y) = A(2 − x − y) for 0 ≤ x ≤ 1 and 0 ≤ y ≤ x, and 0 otherwise. (1) Find A. (2) Find P(2X − Y < 1). (3) Find F(1/2, 1/4) and F(1/2, −1).

Step-by-step solution

1. 1

   (1) Impose the normalisation condition on the correct region. The support is the triangle $0\le y\le x\le 1$, so integrating $y$ from $0$ to $x$ and then $x$ from $0$ to $1$:

   $\int_{0}^{x}(2-x-y)\,dy=\left[(2-x)y-\frac{y^{2}}{2}\right]_{0}^{x}=2x-x^{2}-\frac{x^{2}}{2}=2x-\frac32x^{2}$

2. 2

   Finish the normalisation.

   $A\int_{0}^{1}\left(2x-\tfrac32x^{2}\right)dx=A\left[x^{2}-\tfrac12x^{3}\right]_{0}^{1}=A\left(1-\tfrac12\right)=\frac{A}{2}=1$

   $\boxed{A=2}$

3. 3

   (2) Split the region where the event holds. The event $2X-Y<1$ means $y>2x-1$. For $x\le\tfrac12$ the bound $2x-1$ is $\le 0$, so the whole strip $0\le y\le x$ qualifies; for $x>\tfrac12$ only $2x-1<y\le x$ does. Hence

   $P=\int_{0}^{1/2}\!\!\int_{0}^{x}f\,dy\,dx+\int_{1/2}^{1}\!\!\int_{2x-1}^{x}f\,dy\,dx$

   Splitting at $x=\tfrac12$ is the whole difficulty of this part.

4. 4

   Evaluate the two pieces. With $f=2(2-x-y)$,

   $\int_{0}^{1/2}\!\!\int_{0}^{x}f\,dy\,dx=\frac38,\qquad \int_{1/2}^{1}\!\!\int_{2x-1}^{x}f\,dy\,dx=\frac{5}{24}$

   $P\{2X-Y<1\}=\frac38+\frac{5}{24}=\frac{9+5}{24}=\frac{14}{24}=\frac{7}{12}\approx 0.583$

5. 5

   (3) Compute F(1/2, 1/4). By definition $F(u,v)=P\{X\le u,\ Y\le v\}$, so integrate the density over the part of the triangle with $x\le\tfrac12$ and $y\le\tfrac14$. That region again splits, at $x=\tfrac14$:

   $F=\int_{0}^{1/4}\!\!\int_{0}^{x}f\,dy\,dx+\int_{1/4}^{1/2}\!\!\int_{0}^{1/4}f\,dy\,dx=\frac{7}{64}+\frac{3}{16}=\frac{7+12}{64}=\frac{19}{64}$

6. 6

   Compute F(1/2, −1) with no integration at all. The support requires $y\ge 0$, so the event $\{Y\le -1\}$ is impossible:

   $F\left(\tfrac12,-1\right)=0$

   $\boxed{A=2,\quad P\{2X-Y<1\}=\tfrac{7}{12},\quad F\left(\tfrac12,\tfrac14\right)=\tfrac{19}{64},\quad F\left(\tfrac12,-1\right)=0}$

7. 7

   Sanity-check the three numbers. All are in $[0,1]$ ✓. Since $F(u,v)$ is non-decreasing in each argument, $F\left(\tfrac12,\tfrac14\right)=0.297$ must be at most $P\{X\le\tfrac12\}$; integrating gives $P\{X\le\tfrac12\}=\tfrac38=0.375$ ✓. And $P\{2X-Y<1\}=0.583$ exceeds $0.375$, as it must, since the event contains $\{X\le\tfrac12\}$ entirely.