Section 2.7 (Part 1)

Joint Distributions

Question 2.7.1

Let \(X\) and \(Y\) be jointly absolutely continuous random variables, with joint density function \[ f_{X,Y}(x,y) = \frac{3}{2}x^2I_{(0,1)}(x)I_{(0,2)}(y) \]

Compute \(\mathbb{P}(0.5 < X < 1.5, \ 0 < Y < 1)\).

\[ \begin{aligned} \mathbb{P}(0.5 < X < 1.5, \ 0 < Y < 1) &= \mathbb{P}(0.5 < X < 1, \ 0 < Y < 1) \\ &= \int_0^1 \int_{0.5}^{1} \frac{3}{2}x^2 y \, dx\, dy \\ &= 0.21875 \end{aligned} \]

Question 2.7.2

Suppose \[ p_{X,Y}(x,y) = \begin{cases} 1/5 & (x,y) = (2,3) \\ 1/5 & (x,y) = (3,2) \\ 1/5 & (x,y) = (-3,-2) \\ 1/5 & (x,y) = (-2,-3) \\ 1/5 & (x,y) = (17,19) \\ 0 & \text{otherwise} \end{cases} \]

  1. Compute \(\mathbb{P}(Y > X)\).

  2. Compute \(\mathbb{P}(Y = X)\).

\[ \mathbb{P}(Y > X) = p_{X,Y}(2,3) + p_{X,Y}(-3,-2) + p_{X,Y}(17,19) = 3/5 \]

\[ \mathbb{P}(Y = X) = 0 \] since this never occurs (none of the five pairs has equal coordinates).

Question 2.7.3

Let \(f_{X,Y}(x,y) = 2x^2y + Cy^5I_{(0,1)}(x)I_{(0,1)}(y)\). Find the value of \(C\) that makes the joint PDF valid, and calculate \(\mathbb{P}(X < 0.8, Y < 0.6)\).

Setting the double integral equal to 1: \[ \int_0^1\int_0^1 \left(2x^2y + Cy^5\right)dx\,dy = \int_0^1 \left(\frac{2}{3}y + Cy^5\right)dy = \frac{1}{3} + \frac{C}{6} = 1 \] so \(C = 4\).

\[ \mathbb{P}(X<0.8, Y<0.6) = \int_0^{0.6}\int_0^{0.8}(2x^2y+4y^5)\,dx\,dy \approx 0.0863 \]

Question 2.7.4

The provincial avalanche centre monitors two adjacent weather stations in the Cascade Mountains. Let \(X\) be the snow depth (cm) at the lower station and \(Y\) be the snow depth (cm) at the upper station on a randomly selected day in January. Their joint PDF is modelled as:

\[ f_{X,Y}(x,y)=\frac{1}{3000}(x+2y), \hspace{1cm} 0≤x≤10,0≤y≤15 \]

  1. On any given day, what is the probability that the snow depth at the lower station is greater than 5cm and the snow depth of the upper station is less than 9cm?

\[ \begin{aligned} P(X > 5,\ Y < 9) &= \frac{1}{3000}\int_5^{10}\int_0^{9} \frac{1}{3000}(x+2y)\ dy\ dx\\ &= \frac{1}{3000}\int_5^{10}\left[xy + y^2\right]_0^9\ dx\\ &= \frac{1}{3000}\int_5^{10}(9x + 81)\ dx\\ &= \frac{1}{3000}\left[\frac{9x^2}{2}+81x\right]_5^{10}\\ &= \frac{1}{3000}\left[(450+810)-\left(\frac{225}{2}+405\right)\right]\\ &= \frac{1}{3000}\cdot\frac{1485}{2}\\ &= 0.2475 \end{aligned} \]

  1. Derive the joint CDF for \(0≤x≤10,0≤y≤15\)

\[ \begin{aligned} F_{X,Y}(x,y) &= \int_0^x\int_0^y \frac{1}{3000}(s+2t)\ dt\ ds\\ &= \frac{1}{3000}\int_0^x\left[st + t^2\right]_0^y\ ds\\ &= \frac{1}{3000}\int_0^x[sy + y^2]\ ds\\ &= \frac{1}{3000}\left[\frac{s^2 y}{2}+sy^2\right]_0^x\\ &= \frac{x^2y + 2xy^2}{6000} & \text{when } 0≤x≤10,0≤y≤15 \end{aligned} \]

  1. On any given day, what is the probability that the snow depth at the lower station is more than 3cm greater than snow depth at the upper station?

We need the region \(x > y + 3\), while also considering the support \(0 \leq x \leq 10\), \(0 \leq y \leq 15\). Since \(x \leq 10\).

We can consider \(y < x - 3 \le 7\), so \(y\) ranges from \(0\) to \(7\) and \(x\) from \(y+3\) to \(10\).

\[ \begin{aligned} P(X > Y+3) &= \int_0^{7}\int_{y+3}^{10} \frac{1}{3000}(x+2y)\ dx\ dy\\ &= \frac{1}{3000}\int_0^{7}\left[\frac{x^2}{2}+2xy\right]_{y+3}^{10}\ dy\\ &= \frac{1}{3000}\int_0^{7}\left[(50+20y) - \left(\frac{(y+3)^2}{2}+2y(y+3)\right)\right]\ dy\\ &= \frac{1}{3000}\int_0^{7}\left[\frac{91}{2} + 11y - \frac{5y^2}{2}\right]\ dy\\ &= \frac{1}{3000}\left[\frac{91y}{2}+\frac{11y^2}{2}-\frac{5y^3}{6}\right]_0^7\\ &= \frac{1}{3000}\cdot\frac{1813}{6}\\ &= 0.1007 \end{aligned} \]

You can also compute \(P(Y < X - 3)\).

We need to be super careful about the integration region. For \(y \ge 0\), we need \(x - 3 > 0\), so \(x > 3\). Thus \(x\) runs from \(3\) to \(10\), and for each \(x\), \(y\) runs from \(0\) to \(x - 3\):

\[ \begin{aligned} P(X > Y+3) &= \int_3^{10}\int_{0}^{x-3} \frac{1}{3000}(x+2y)\ dy\ dx\\ &= \frac{1}{3000}\int_3^{10}\left[xy + y^2\right]_0^{x-3}\ dx\\ &= \frac{1}{3000}\int_3^{10}(x(x-3) + (x-3)^2)\ dx\\ &= \frac{1}{3000}\int_3^{10}(x^2-3x +x^2 - 6x +9)\ dx\\ &= \frac{1}{3000}\int_3^{10}(2x^2-9x +9)\ dx\\ &= \frac{1}{3000}\left[(\frac{2}{3}x^3-\frac{9}{2}x^2 +9x)\right]_{x = 3}^{x=10}\\ &= \frac{1}{3000}\left[(\frac{2}{3}(10)^3 -\frac{9}{2}(10^2) + 9(10)) - (\frac{2}{3}(3)^3 -\frac{9}{2}(3^2) + 9(3))\right]\\ &=0.1007 \end{aligned} \]