Question 2.7.5
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:
\[
f_{X,Y}(x,y)=\frac{1}{3000}(x+2y), \hspace{1cm} 0≤x≤10,0≤y≤15
\]
- Calculate the marginal pdf \(f_{X}(x)\).
\[
f_X(x) = \int_0^{15} \frac{1}{3000}(x+2y)\ dy
= \frac{1}{3000}\left[xy+y^2\right]_0^{15}
= \frac{1}{3000}(15x + 225)I_{[0, 10](x)}
\]
- What is the probability that the snow depth at the lower station is more than 3cm?
\[
\begin{aligned}
\mathbb{P}(X > 3) &= \int_{3}^{10}\frac{1}{3000}(15x + 225)dx \\
&=\frac{1}{3000}(15x^2/2 + 225x)\Big|_{x = 3}^{x=10} \\
&= \frac{1}{3000}(15(10)^2/2 + 225(2) - 15(3)^2/2-225(3))\\
&=0.1525
\end{aligned}
\]
Question 2.7.6
Suppose \[
p_{X,Y}(x,y) =
\begin{cases}
1/5 & (x,y) = (2,3) \\
1/5 & (x,y) = (3,2) \\
1/5 & (x,y) = (-3,-2) \\
1/5 & (x,y) = (-2,-3) \\
1/5 & (x,y) = (17,19) \\
0 & \text{otherwise}
\end{cases}
\]
- Calculate \(p_X(x)\)
Here, \(X\) has a non-zero probability at \(\{-3, -2, 2, 3, 17\}\). For each of these values of \(X\), there is a non-zero probability for one unique value of \(Y\). So the marginal distribution looks similar to the joint:
\[
p_{X}(x) =
\begin{cases}
1/5 & (x = 2 \\
1/5 & x = 3 \\
1/5 & x = -3 \\
1/5 & x = -2 \\
1/5 & x = 17 \\
0 & \text{otherwise}
\end{cases}
\]
- What is the marginal CDF \(F_X(x)?\)
\[
F_{X}(x) =
\begin{cases}
0 & x< -3\\
1/5 & -3 \le x <-2 \\
2/5 & -2 \le x <2 \\
3/5 & 2 \le x < 3 \\
4/5 & 3 \le x < 17 \\
1 & x \ge 17 \\
\end{cases}
\]
Question 2.7.7
Let \(X\sim {\mathrm{Exp}}(\lambda)\) and \(Y\sim{\mathrm{Exp}}(\mu)\) be independent random variables. Find the distribution of \(U = \min\{X, Y\}\).
Hint: If \(Z\sim{\mathrm{Exp}}(\theta)\), then \(F_Z(z) = \mathbb{P}(Z \le z) = 1 - e^{-\theta z}\) for \(z > 0\).
We have that for \(u > 0\), \[\begin{aligned}
\mathbb{P}(U < u) &= \mathbb{P}(\min\{X, Y\} < u) = 1 - \mathbb{P}(\min\{X, Y\} \geq u)\\
&= 1 - \mathbb{P}(X \geq u, Y \geq u)\\
&= 1 - \mathbb{P}(X \geq u) \mathbb{P}(Y \geq u) \\
&= 1 - e^{-\lambda u} e^{-\mu u} \\
&= 1 - e^{-(\lambda + \mu) u}.
\end{aligned}\] Therefore, \(U\) has CDF \(F_U(u) = 1 - e^{-(\lambda + \mu) u}\) for \(u > 0\), so \(U
\sim {\mathrm{Exp}}(\lambda + \mu)\).
Question 2.7.8
The joint PMF of discrete random variables \(X\) and \(Y\) is given by:
| \(X=0\) |
0.30 |
0.05 |
0.35 |
| \(X=1\) |
0.20 |
0.25 |
0.45 |
| \(X=2\) |
0.10 |
0.10 |
0.20 |
| \(P(Y=y)\) |
0.60 |
0.40 |
1.00 |
Are \(X\) and \(Y\) independent? Answer by using marginal PMFs.
The marginal PMF of X is:
\[
p_{X}(x) =
\begin{cases}
0.3 + 0.05 = 0.35 & x = 0\\
0.20 +0.25 = 0.45 & x = 1\\
0.10 + 0.10 = 0.20 & x = 2\\
0 & \text{otherwise}
\end{cases}
\]
The marginal PMF of X is:
\[
p_{Y}(y) =
\begin{cases}
0.3 + 0.2 + 0.1 = 0.60 & y = 0\\
0.25 + 0.05 + 0.10 = 0.4 & y = 1\\
0 & \text{otherwise}
\end{cases}
\]
Pick any combination of \(X\) and \(Y\).
From the table, we have \(\mathbb{P}(X = 1, Y = 0) = p_{X,Y}(1,0) = 0.20\). We also have \(\mathbb{P}(X = 1)\times \mathbb{P}(Y = 0) = 0.45 \times 0.60 = 0.27\). These are not equal. Therefore, \(X\) and \(Y\) are not independent.
Note: If any combination of \(X\) and \(Y\) has \(\mathbb{P}(X = x, Y = y) \ne \mathbb{P}(X=x)\mathbb{P}(Y=y)\) then they are not independent.