Question 2.8.1
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:
\[
f_{X,Y}(x,y)=\frac{1}{3000}(x+2y), \hspace{1cm} 0≤x≤10,0≤y≤15
\]
What is the conditional distribution \(f_{X|Y{(x|y)}}\)?
First we can derive the marginal \(f_Y(y)\).
\[
\begin{aligned}
f_Y(y) &= \int_0^{10}\frac{1}{3000}(x+2y)\ dx\\
&= \frac{1}{3000}\left[\frac{x^2}{2}+2xy\right]_0^{10}\\
&= \frac{50+20y}{3000}\\
&= \frac{5+2y}{300}I_{[0, 15]}(y)
\end{aligned}
\]
Then,
\[
\begin{aligned}
f_{X|Y}(x \ \vert\ y) &= \frac{f_{X,Y}(x,y)}{f_Y(y)}I_{[0, 10]}(x) \\
&= \frac{\dfrac{x+2y}{3000}}{\dfrac{5+2y}{300}}I_{[0, 10]}(x) \\
&= \frac{x+2y}{10(5+2y)}I_{[0, 10]}(x)
\end{aligned}
\]
which ONLY defined for \(0 \le y \le 15\)
Question 2.8.2
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 |
- Compute the conditional distribution \(\mathbb{P}_{Y | X}(y \ \vert\ X = 1)\) for all \(y \in {\mathbb{R}}\).
By definition, \(P_{Y \ \vert\ X}(y \ \vert\ X = x) = \dfrac{P(X = x, Y = y)}{P(X = x)}\). For \(X = 1\), we have \(P(X = 1) = 0.45\):
\[
\begin{aligned}
P_{Y \ \vert\ X}(0 \ \vert\ X = 1) &= \frac{P(X = 1, Y = 0)}{P(X = 1)} = \frac{0.20}{0.45} = \frac{4}{9}\\[6pt]
P_{Y \ \vert\ X}(1 \ \vert\ X = 1) &= \frac{P(X = 1, Y = 1)}{P(X = 1)} = \frac{0.25}{0.45} = \frac{5}{9}
\end{aligned}
\]
Check: \(\dfrac{4}{9} + \dfrac{5}{9} = 1\).
- Calculate \(\mathbb{P}(Y = 1 | X \geq 1)\). Show your work.
We need \(\mathbb{P}(Y = 1 \ \vert\ X \geq 1)\). By the definition of conditional probability,
\[
\mathbb{P}(Y = 1 \ \vert\ X \geq 1) = \frac{\mathbb{P}(Y = 1 \ \text{and}\ X \geq 1)}{\mathbb{P}(X \geq 1)}
\]
For the numerator, use the joint table: \(X \geq 1\) and \(Y = 1\) means \((X=1, Y=1)\) or \((X=2, Y=1)\), so
\[
\begin{aligned}
\mathbb{P}(Y = 1 \ \text{and}\ X \geq 1) &= P(X=1, Y=1) + P(X=2, Y=1) \\
&= 0.25 + 0.10 = 0.35
\end{aligned}
\]
For the denominator: \(\mathbb{P}(X \geq 1) = 1 - \mathbb{P}(X = 0) = 1 - 0.35 = 0.65\). Therefore,
\[
\mathbb{P}(Y = 1 \ \vert\ X \geq 1) = \frac{0.35}{0.65} = \frac{7}{13} \approx 0.5385
\]
- Compute the conditional distribution \(P_{X | Y}(x \ \vert\ Y = 0)\) for all \(x \in {\mathbb{R}}\).
By definition, \(P_{X \ \vert\ Y}(x \ \vert\ Y = 0) = \dfrac{P(X = x, Y = 0)}{P(Y = 0)}\). For \(Y = 0\), we have \(P(Y = 0) = 0.60\):
\[
\begin{aligned}
P_{X \ \vert\ Y}(0 \ \vert\ Y = 0) &= \frac{P(X = 0, Y = 0)}{P(Y = 0)} = \frac{0.30}{0.60} = \frac{1}{2}\\[6pt]
P_{X \ \vert\ Y}(1 \ \vert\ Y = 0) &= \frac{P(X = 1, Y = 0)}{P(Y = 0)} = \frac{0.20}{0.60} = \frac{1}{3}\\[6pt]
P_{X \ \vert\ Y}(2 \ \vert\ Y = 0) &= \frac{P(X = 2, Y = 0)}{P(Y = 0)} = \frac{0.10}{0.60} = \frac{1}{6}
\end{aligned}
\]
Check: \(\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} = 1\).
- Are \(X\) and \(Y\) independent? Justify your answer mathematically.
Two equivalent definitions of independence:
1. Checking by factorization: \(X\) and \(Y\) are independent if \(P(X = x, Y = y) = P(X = x)\,P(Y = y)\) for all \((x, y)\) in the support. We can provide any single counter example where this fails:
\[
\begin{aligned}
P(0, 0) &= 0.30 \ne P(X=0)\,P(Y=0) = (0.35)(0.60) = 0.21 \\
P(0, 1) &= 0.05 \ne P(X=0)\,P(Y=1) = (0.35)(0.40) = 0.14 \\
P(1, 0) &= 0.20 \ne P(X=1)\,P(Y=0) = (0.45)(0.60) = 0.27 \\
P(1, 1) &= 0.25 \ne P(X=1)\,P(Y=1) = (0.45)(0.40) = 0.18 \\
P(2, 0) &= 0.10 \ne P(X=2)\,P(Y=0) = (0.20)(0.60) = 0.12 \\
P(2, 1) &= 0.10 \ne P(X=2)\,P(Y=1) = (0.20)(0.40) = 0.08 \\
\end{aligned}
\]
Therefore, \(X\) and \(Y\) are not independent. (A single counterexample is sufficient to conclude \(X\) and \(Y\) are not independent).
2. Checking if conditional distribution = marginal distribution: \(X\) and \(Y\) are independent if \(P_{X \ \vert\ Y}(x \ \vert\ y) = P(X = x)\) for all \((x, y)\). From part (c):
\[
P_{X \ \vert\ Y}(0 \ \vert\ Y=0) = \frac{1}{2} = 0.50 \neq P(X = 0) = 0.35
\]
(A single counterexample is sufficient to conclude \(X\) and \(Y\) are not independent).
Question 2.8.3
Suppose \(X\) and \(Y\) have joint probability:
\[
p_{X,Y}(x,y) = \begin{cases}
1/9 & x = -4, y = -2\\
2/9 & x = 5, y = -2\\
3/9 & x = 9, y = -2\\
2/9 & x = 9, y = 0\\
1/9 & x = 9, y = 4\\
0&\text{otherwise}
\end{cases}
\]
- Compute \(\mathbb{P}(Y = 4 \ \vert\ X = 9)\)
\[
\begin{aligned}
\mathbb{P}(Y = 4 \ \vert\ X = 9) &= \frac{\mathbb{P}(Y = 4 \cap X = 9)}{X = 9}\\
&= \frac{1/9}{3/9 + 2/9 + 1/9}\\
&= \frac{1}{6}
\end{aligned}
\]
- \(\mathbb{P}(Y = -2 \ \vert\ X = 9)\)
\[
\begin{aligned}
\mathbb{P}(Y = -2 \ \vert\ X = 9) &= \frac{\mathbb{P}(Y = -2 \cap X = 9)}{X = 9}\\
&= \frac{3/9}{3/9 + 2/9 + 1/9}\\
&= \frac{1}{2}
\end{aligned}
\]
- \(\mathbb{P}(Y = 0 \ \vert\ X = -4)\)
\[
\begin{aligned}
\mathbb{P}(Y = 0 \ \vert\ X =-4) &= \frac{\mathbb{P}(Y = 0 \cap X = -4)}{X = -4}\\
&= \frac{0}{1/9}\\
&= 0
\end{aligned}
\]
- \(\mathbb{P}(Y = -2 \ \vert\ X = 5)\)
\[
\begin{aligned}
\mathbb{P}(Y = -2\ \vert\ X = 5) &= \frac{\mathbb{P}(Y = -2 \cap X = 5)}{X = 5}\\
&= \frac{2/9}{2/9}\\
&= 1
\end{aligned}
\]
- \(\mathbb{P}(X = 5 \ \vert\ Y = -2)\)
\[
\begin{aligned}
\mathbb{P}(X = 5 \ \vert\ Y = -2) &= \frac{\mathbb{P}(X = 5 \cap Y = -2)}{Y=-2}\\
&= \frac{2/9}{1/2 + 2/9 + 3/9}\\
&= 1/3
\end{aligned}
\]
Question 2.8.4
Let \(X \sim {\mathrm{Binom}}(1, 1/3)\) and \(Y \sim {\mathrm{Poiss}}(\lambda)\). Let \(X\) and \(Y\) be independent and \(\lambda > 0\). Compute \(\mathbb{P}(X = 1, Y = 1)\).
Since \(X\) and \(Y\) are independent, \(\mathbb{P}(X = 1 \ \vert\ Y = 5) = \mathbb{P}(X = 1) = 1/3\)
Question 2.8.5
Let \(X\) and \(Y\) have joint density \(f_{X,Y}(x,y) = \frac{1}{4}(x^2 + y)I_{(0, 2)}(x)I_{(x, 2)}(y)\).
Compute the conditional density \(f_{X \ \vert\ Y}(x \ \vert\ y)\) for \(f_X(x) > 0\).
Recall \(f_{Y \ \vert\ X}(y \ \vert\ x) = \frac{f_{X,Y}(x,y)}{f_X(x)}\).
We first find the marginal pdf \(f_X(x)\):
\[
\begin{aligned}
f_X(x) &= \int_{x}^2\frac{1}{4}(x^2+y)dy\\
&= \frac{1}{4}(x^2y + y^2/2 \Big|_{y = x}^{y=2})\\
&= \frac{4 + 3x^2 - 2x^3}{8}I_{(0, 2)}(x)
\end{aligned}
\] Then, the conditional distribution \(f_{Y \ \vert\ X}(y \ \vert\ x) = \frac{f_{X,Y}(x,y)}{f_X(x)}\) is
\[
\begin{aligned}
f_{Y \ \vert\ X}(y \ \vert\ x) &= \frac{f_{X,Y}(x,y)}{f_X(x)} \\
&= \frac{\frac{1}{4}(x^2 + y)I_{(0, 2)}(x)I_{(x, 2)}(y)}{\frac{4 + 3x^2 - 2x^3}{8}I_{(0, 2)}(x)} \\
&= \frac{2(x^2y)}{4 + 3x^2 - 2x^3}I_{(x, 2)}(y)
\end{aligned}
\] which is defined only for \(x \in (0,2)\).