Question 2.7.1
Let \(X\) and \(Y\) be jointly absolutely continuous random variables, with joint density function \[
f_{X,Y}(x,y) = \frac{3}{2}x^2yI_{(0,1)}(x)I_{(0,2)}(y)
\]
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 \[
p_{X,Y}(x,y) =
\begin{cases}
1/5 & (x,y) = (2,3) \\
1/5 & (x,y) = (3,2) \\
1/5 & (x,y) = (-3,-2) \\
1/5 & (x,y) = (-2,-3) \\
1/5 & (x,y) = (17,19) \\
0 & \text{otherwise}
\end{cases}
\]
Compute \(\mathbb{P}(Y > X)\).
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 \(f_{X,Y}(x,y) = (2x^2y + Cy^5)I_{(0,1)}(x)I_{(0,1)}(y)\). 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:
\[
f_{X,Y}(x,y)=\frac{1}{3000}(x+2y), \hspace{1cm} 0≤x≤10,0≤y≤15
\]
- 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?
\[
\begin{aligned}
P(X > 5,\ Y < 9) &= \int_5^{10}\int_0^{9} \frac{1}{3000}(x+2y)\ dy\ dx\\
&= \frac{1}{3000}\int_5^{10}\left[xy + y^2\right]_0^9\ dx\\
&= \frac{1}{3000}\int_5^{10}(9x + 81)\ dx\\
&= \frac{1}{3000}\left[\frac{9x^2}{2}+81x\right]_5^{10}\\
&= \frac{1}{3000}\left[(450+810)-\left(\frac{225}{2}+405\right)\right]\\
&= \frac{1}{3000}\cdot\frac{1485}{2}\\
&= 0.2475
\end{aligned}
\]
- Derive the joint CDF for \(0≤x≤10,0≤y≤15\)
\[
\begin{aligned}
F_{X,Y}(x,y) &= \int_0^x\int_0^y \frac{1}{3000}(s+2t)\ dt\ ds\\
&= \frac{1}{3000}\int_0^x\left[st + t^2\right]_0^y\ ds\\
&= \frac{1}{3000}\int_0^x[sy + y^2]\ ds\\
&= \frac{1}{3000}\left[\frac{s^2 y}{2}+sy^2\right]_0^x\\
&= \frac{x^2y + 2xy^2}{6000} & \text{when } 0≤x≤10,0≤y≤15
\end{aligned}
\]
- 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 \(x - 3\):
\[
\begin{aligned}
P(X > Y+3) &= \int_3^{10}\int_{0}^{x-3} \frac{1}{3000}(x+2y)\ dy\ dx\\
&= \frac{1}{3000}\int_3^{10}\left[xy + y^2\right]_0^{x-3}\ dx\\
&= \frac{1}{3000}\int_3^{10}(x(x-3) + (x-3)^2)\ dx\\
&= \frac{1}{3000}\int_3^{10}(x^2-3x +x^2 - 6x +9)\ dx\\
&= \frac{1}{3000}\int_3^{10}(2x^2-9x +9)\ dx\\
&= \frac{1}{3000}\left[(\frac{2}{3}x^3-\frac{9}{2}x^2 +9x)\right]_{x = 3}^{x=10}\\
&= \frac{1}{3000}\left[(\frac{2}{3}(10)^3 -\frac{9}{2}(10^2) + 9(10)) - (\frac{2}{3}(3)^3 -\frac{9}{2}(3^2) + 9(3))\right]\\
&=0.1007
\end{aligned}
\]
Question 2.7.5 (Challenge)
In any given 15 minute block of time, the R4 shows up with the “SORRY, BUS FULL” sign 5 times. Assume the arrival times of full buses are independent. Let \(X\) be the time in minutes between full buses.
What is the probability that the total time for 2 consecutive full buses to arrive is less than 12 minutes?
We can find this by finding the joint PDF of \(X_1\) and \(X_2\), which is
\[
\begin{aligned}
f_(X_1,X_2)(x_1, x_2) &= \frac{1}{3} e^{-\frac{1}{3} x_1}\frac{1}{3} e^{-\frac{1}{3} x_2}I_{[0,\infty)}(x_1)I_{[0,\infty)}(x_2)\\
&= \frac{1}{9} e^{-\frac{1}{3} (x_1 + x_2)}I_{[0,\infty)}(x_1)I_{[0,\infty)}(x_2)
\end{aligned}
\]
Now, we are interested in \(\mathbb{P}(X_1 + X_2 < 12)\). We can think of this as \(\mathbb{P}(X_1 < 12 - X_2)\) (or \(\mathbb{P}(X_2 < 12 - X_1)\). WLOG we will show the solution using \(\mathbb{P}(X_1 < 12 - X_2)\).
We can integrate this, but we do need to be careful about the bounds of the integral. If \(X_2 > 12\), this then \(12 - X_2 < 0\), and our support is only for \(X_1 \ge 0\). Therefore, this part of the integral will just be 0. If \(0 \le X_2 \le 12\), then \(12 - X_2\) will be between 0 and 1, which is allowed for our support of \(X_1\). So, we really only care about the case where \(0 \le X_2 \le 12\).
\[
\begin{aligned}
\mathbb{P}(X_1 + X_2 < 12) &= \mathbb{P}(X_1 < 12 - X_2)\\
&= \int_0^{12}\int_{0}^{12 - x_2}\frac{1}{9} e^{-\frac{1}{3} (x_1 + x_2)}dx_1dx_2\\
&=\frac{1}{9}\int_0^{12} e^{-\frac{1}{3}x_2}\left[\int_{0}^{12 - x_2} e^{-\frac{1}{3}x_1}dx_1\right]dx_2\\
&=\frac{1}{9}\int_0^{12} e^{-\frac{1}{3}x_2}\left[(-3)e^{-\frac{1}{3}x_1}\Big|_{x_1 = (0)}^{x_1 =12 - x_2}\right] dx_2\\
&=\frac{1}{3}\int_0^{12} e^{-\frac{1}{3}x_2}\left(1 - e^{-\frac{1}{3}(12 - x_2)}\right) dx_2\\
&= \frac{1}{3}\int_{0}^{12}\left(e^{-\frac{1}{3}x_2} - e^{-\frac{1}{3}(12)}\right)dx_2\\
&= \frac{1}{3}\left((-3)e^{-\frac{1}{3}x_2} - e^{-\frac{1}{3}(12)}x_2\right)\Big|_{x_2 = 0}^{x_2 = 12}\\
&= \left(-e^{-\frac{1}{3}x_2} - \frac{1}{3}e^{-\frac{1}{3}(12)}x_2\right)\Big|_{x_2 = 0}^{x_2 = 12}\\
&= \left(-e^{-\frac{1}{3}(12)} - \frac{1}{3}e^{-\frac{1}{3}(12)}(12)\right) - \left(-e^{-\frac{1}{3}(0)} - \frac{1}{3}e^{-\frac{1}{3}(12)}(0)\right) \\
&= 1 - e^{-4} - 4e^{-4}\\
&= 1 - 5e^{-4}\\
&= \approx 0.9084
\end{aligned}
\]
Aside: You could also look at this as a Poisson distribution. With \(\lambda = \frac{1}{3}\) buses/min, the expected number of full buses in 12 minutes is: \(\mu = \lambda \cdot t = \frac{1}{3} \times 12 = 4\). So we can define \(Y\) as the number of busses full within 12 minutes, where \(Y ∼ Poisson(4)\). Then,
\(P(Y \ge 1)=1−\mathbb{P}(Y=0)−\mathbb{P}(Y=1) = 1 - \frac{e^{-4} \cdot 4^0}{0!} - \frac{e^{-4} \cdot 4^1}{1!}= 0.9084\).