We earlier defined a random variable as a real-valued function over a sample space with a probability measure. Many different random variables can be defined over one and the same sample space.
In this section we concern ourselves first with the bivariate case — situations where we are interested simultaneously in a pair of random variables defined over a joint sample space. We then extend the discussion to the multivariate case, covering any finite number of random variables.
If \(X\) and \(Y\) are discrete random variables, we write the probability that \(X\) takes on value \(x\) and \(Y\) takes on value \(y\) as:
\[\Pr(X = x,\; Y = y)\]
This is the probability of the intersection of the events \(\{X = x\}\) and \(\{Y = y\}\).
Example 1
Two caplets are selected at random from a bottle containing three aspirins, two sedatives, and four laxative caplets. Let \(X\) = number of aspirin caplets and \(Y\) = number of sedative caplets among the two drawn. Find the probabilities for all possible pairs \((x, y)\).
Solution
The possible pairs are \((0,0),\,(0,1),\,(1,0),\,(1,1),\,(0,2),\,(2,0)\).
For \((1,0)\): we need 1 aspirin, 0 sedatives, 1 laxative:
\[\binom{3}{1}\binom{2}{0}\binom{4}{1} = 3 \times 1 \times 4 = 12\]
Total ways to choose 2 from 9: \(\binom{9}{2} = 36\).
\[\Pr(X=1, Y=0) = \frac{12}{36} = \frac{1}{3}\]
The complete joint probability table is:
| \(y \backslash x\) | 0 | 1 | 2 |
|---|---|---|---|
| 0 | \(\frac{1}{6}\) | \(\frac{1}{3}\) | \(\frac{1}{12}\) |
| 1 | \(\frac{2}{9}\) | \(\frac{1}{6}\) | \(0\) |
| 2 | \(\frac{1}{36}\) | \(0\) | \(0\) |
A closed-form expression for this distribution is:
\[f(x,y) = \Pr(X=x,\,Y=y) = \frac{\binom{3}{x}\binom{2}{y}\binom{4}{2-x-y}}{\binom{9}{2}}, \quad x=0,1,2;\; y=0,1,2;\; x+y \leq 2\]
Definition 1 – Joint Probability Distribution Function (Discrete)
If \(X\) and \(Y\) are discrete random variables, the function
\[f(x,y) = \Pr(X=x,\,Y=y)\]
for each pair \((x,y)\) within the range of \(X\) and \(Y\) is called the joint probability distribution function of \(X\) and \(Y\).
Theorem 1
A bivariate function \(f(x,y)\) can serve as a joint probability distribution function of a pair of discrete random variables \(X\) and \(Y\) if and only if:
Example 2
Determine the value of \(k\) for which \(f(x,y) = kxy\), for \(x = 1,2,3\) and \(y = 1,2,3\), can serve as a joint probability distribution function.
Solution
Summing all values:
\[\sum_x \sum_y kxy = k(1+2+3)(1+2+3) = k \cdot 6 \cdot 6 = 36k = 1\]
\[\boxed{k = \frac{1}{36}}\]
Definition 2 – Joint Cumulative Distribution Function (Discrete)
For discrete random variables \(X\) and \(Y\):
\[F(x,y) = \Pr(X \leq x,\, Y \leq y) = \sum_{s \leq x}\sum_{t \leq y} f(s,t), \quad -\infty < x < \infty,\; -\infty < y < \infty\]
Example 3
With reference to Example 1, find \(F(1,1)\).
Solution
\[F(1,1) = \Pr(X \leq 1,\, Y \leq 1) = f(0,0)+f(0,1)+f(1,0)+f(1,1)\] \[= \frac{1}{6}+\frac{2}{9}+\frac{1}{3}+\frac{1}{6} = \frac{8}{9}\]
Definition 3 – Joint Probability Density Function (Continuous)
A bivariate function \(f(x,y)\) defined over the \(XY\)-plane is called a joint probability density function of continuous random variables \(X\) and \(Y\) if and only if:
\[\Pr\bigl[(X,Y) \in A\bigr] = \iint_A f(x,y)\,dx\,dy\]
for any region \(A\) in the \(XY\)-plane.
Theorem 2
A bivariate function \(f(x,y)\) can serve as a joint probability density function of continuous random variables \(X\) and \(Y\) if and only if:
Example 4
Given the joint PDF
\[f(x,y) = \begin{cases} \dfrac{3}{5}(x^2 + y), & 0 \leq x \leq 1,\; 0 \leq y \leq 2 \\ 0, & \text{elsewhere} \end{cases}\]
find \(\Pr[(X,Y) \in A]\) where \(A = \{(x,y): 0 \leq x \leq \tfrac{1}{2},\; 1 \leq y \leq 2\}\).
Solution
\[\Pr[(X,Y) \in A] = \int_1^2\int_0^{1/2} \frac{3}{5}(x^2+y)\,dx\,dy\]
\[= \int_1^2 \left[\frac{3}{5}\left(\frac{x^3}{3}+xy\right)\right]_0^{1/2} dy = \int_1^2 \frac{3}{5}\left(\frac{1}{24}+\frac{y}{2}\right) dy = \frac{11}{80}\]
Definition 4 – Joint CDF (Continuous)
For continuous random variables \(X\) and \(Y\):
\[F(x,y) = \Pr(X \leq x,\, Y \leq y) = \int_{-\infty}^{x}\int_{-\infty}^{y} f(s,t)\,dt\,ds\]
and we can recover the density via:
\[f(x,y) = \frac{\partial^2 F(x,y)}{\partial x\,\partial y}\]
wherever the partial derivatives exist.
Example 5
Given \(f(x,y) = 1\) for \(0 \leq x \leq 1\), \(0 \leq y \leq 1\) (zero elsewhere), find \(F(x,y)\).
Solution
\[F(x,y) = \int_0^x\int_0^y ds\,dt = xy - \frac{x^2 y}{2} - \frac{x y^2}{2} + \frac{x^2 y^2}{4}\]
\[= \frac{(2x - x^2)(2y - y^2)}{4}\]
Example 6
The joint CDF of \(X\) and \(Y\) is:
\[F(x,y) = \begin{cases} (1-e^{-x})(1-e^{-y}), & x > 0,\; y > 0 \\ 0, & \text{elsewhere} \end{cases}\]
(a) Find the joint PDF.
\[f(x,y) = \frac{\partial^2 F}{\partial x\,\partial y} = e^{-x}e^{-y} = e^{-(x+y)}, \quad x>0,\; y>0\]
(b) Find \(\Pr(1 < X < 3,\; 1 < Y < 2)\).
\[= \int_1^2\int_1^3 e^{-(x+y)}\,dx\,dy = (e^{-1}-e^{-3})(e^{-1}-e^{-2}) \approx 0.074\]
All definitions generalise to \(n\) random variables \(X_1, X_2, \ldots, X_n\).
Joint probability distribution (discrete):
\[f(x_1, x_2, \ldots, x_n) = \Pr(X_1=x_1,\, X_2=x_2,\, \ldots,\, X_n=x_n)\]
Joint CDF (continuous):
\[F(x_1,\ldots,x_n) = \int_{-\infty}^{x_1}\cdots\int_{-\infty}^{x_n} f(t_1,\ldots,t_n)\,dt_n\cdots dt_1\]
Recovery of density:
\[f(x_1,\ldots,x_n) = \frac{\partial^n F}{\partial x_1 \cdots \partial x_n}\]
Example 7
The joint distribution of \(X\), \(Y\), \(Z\) is:
\[f(x,y,z) = \begin{cases} \dfrac{x+y+z}{63}, & x=1,2;\; y=1,2,3;\; z=1,2 \\ 0, & \text{elsewhere} \end{cases}\]
Find \(\Pr(X=2,\; Y+Z=3)\).
Solution
The pairs \((y,z)\) with \(y+z=3\) are: \((1,2),\,(2,1)\).
\[\Pr = f(2,1,2)+f(2,2,1) = \frac{5}{63}+\frac{5}{63}+\frac{6}{63} = \frac{13}{63}\]
Example 8
The trivariate PDF is:
\[f(x_1,x_2,x_3) = \begin{cases} x_1 x_2 e^{-x_3}, & 0<x_1<1,\; 0<x_2<1,\; x_3>0 \\ 0, & \text{elsewhere} \end{cases}\]
Find \(\Pr[(X_1,X_2,X_3)\in A]\) where \(A = \{0<x_1<\tfrac{1}{2},\; x_1<x_2<1,\; x_3>x_1+x_2\}\).
Solution
\[\Pr = \int_0^{1/2}\int_{x_1}^{1}\int_{x_1+x_2}^{\infty} x_1 x_2 e^{-x_3}\,dx_3\,dx_2\,dx_1 \approx 0.158\]
Exercise 1. Given the joint distribution table below:
| \(y \backslash x\) | 0 | 1 | 2 |
|---|---|---|---|
| 0 | \(\frac{1}{12}\) | \(\frac{1}{6}\) | \(\frac{1}{24}\) |
| 1 | \(\frac{1}{4}\) | \(\frac{1}{4}\) | \(\frac{1}{40}\) |
| 2 | \(\frac{1}{8}\) | \(\frac{1}{20}\) | \(0\) |
| 3 | \(\frac{1}{120}\) | \(0\) | \(0\) |
Find: (a) \(\Pr(X\leq1, Y=2)=\frac{1}{20}\); (b) \(\Pr(X>0, 1\leq Y<3)=\frac{3}{8}\); (c) \(\Pr(X+Y\leq1)=\frac{1}{2}\); (d) \(F(1.2, 0.9)=\frac{1}{4}\); (e) \(F(3, 1.5)=0\); (f) \(F(2, 0)=\frac{7}{24}\); (g) \(F(4, 2.7)=\frac{119}{120}\).
Exercise 2. If \(f(x,y)=c(x^2+y^2)\) for \(x=1,0,1,3\); \(y=1,2,3\), find \(c\). Answer: \(c=\frac{1}{89}\).
Exercise 3. Determine \(k\) so that \(f(x,y)=kx(x+y)\) for \(0<x<1\), \(x<y<x\) can serve as a joint PDF. Answer: \(k=2\).
Exercise 4. Find \(k\) if the joint distribution of \(X,Y,Z\) is \(f(x,y,z)=kxyz\) for \(x=1,2\); \(y=1,2,3\); \(z=1,2\). Answer: \(k=\frac{1}{54}\).
Exercise 5. Find \(k\) if \(f(x,y,z)=kxy^2z\) for \(0\leq x,y,z\leq1\), \(x+y+z\leq1\). Answer: \(k=144\).
Definition 5 – Marginal Distribution (Discrete)
If \(X\) and \(Y\) are discrete with joint distribution \(f(x,y)\), the marginal distribution of \(X\) is:
\[g(x) = \sum_y f(x,y) \quad \text{for each } x \text{ in the range of } X\]
The marginal distribution of \(Y\) is:
\[h(y) = \sum_x f(x,y) \quad \text{for each } y \text{ in the range of } Y\]
Example 9
Given the joint distribution:
| \(y \backslash x\) | 1 | 2 | 3 |
|---|---|---|---|
| 1 | \(0\) | \(\frac{1}{6}\) | \(\frac{1}{6}\) |
| 2 | \(\frac{1}{6}\) | \(0\) | \(\frac{1}{6}\) |
| 3 | \(\frac{1}{6}\) | \(\frac{1}{6}\) | \(0\) |
Marginal distribution of \(X\):
| \(x\) | 1 | 2 | 3 |
|---|---|---|---|
| \(g(x)\) | \(\frac{1}{3}\) | \(\frac{1}{3}\) | \(\frac{1}{3}\) |
Marginal distribution of \(Y\):
| \(y\) | 1 | 2 | 3 |
|---|---|---|---|
| \(h(y)\) | \(\frac{1}{3}\) | \(\frac{1}{3}\) | \(\frac{1}{3}\) |
Definition 6 – Marginal Density (Continuous)
For continuous random variables \(X\) and \(Y\) with joint density \(f(x,y)\):
\[g(x) = \int_{-\infty}^{\infty} f(x,y)\,dy \qquad \text{(marginal density of } X\text{)}\]
\[h(y) = \int_{-\infty}^{\infty} f(x,y)\,dx \qquad \text{(marginal density of } Y\text{)}\]
Example 10
Given
\[f(x,y) = \begin{cases} \frac{2}{3}(x+2y), & 0\leq x\leq1,\; 0\leq y\leq1 \\ 0, & \text{elsewhere} \end{cases}\]
Marginal density of \(X\):
\[g(x) = \int_0^1 \frac{2}{3}(x+2y)\,dy = \frac{2}{3}\left[xy+y^2\right]_0^1 = \frac{2}{3}(x+1), \quad 0\leq x\leq1\]
Marginal density of \(Y\):
\[h(y) = \int_0^1 \frac{2}{3}(x+2y)\,dx = \frac{2}{3}\left[\frac{x^2}{2}+2xy\right]_0^1 = \frac{1}{3}(1+4y), \quad 0\leq y\leq1\]
For \(n\) random variables \(X_1, X_2, \ldots, X_n\) with joint density \(f(x_1,\ldots,x_n)\):
Marginal of \(X_1\) alone: \(g(x_1) = \sum_{x_2}\cdots\sum_{x_n} f(x_1,\ldots,x_n)\)
Joint marginal of \(X_1, X_2, X_3\): \(m(x_1,x_2,x_3) = \sum_{x_4}\cdots\sum_{x_n} f(x_1,\ldots,x_n)\)
(Replace sums with integrals for the continuous case.)
Example 11
For the trivariate density \(f(x_1,x_2,x_3) = x_1 x_2 e^{-x_3}\) (\(0<x_1<1\), \(0<x_2<1\), \(x_3>0\)):
(i) Joint marginal density of \(X_1\) and \(X_3\):
\[m(x_1,x_3) = \int_0^1 x_1 x_2 e^{-x_3}\,dx_2 = x_1 e^{-x_3}\left[\frac{x_2^2}{2}\right]_0^1 = \frac{x_1}{2}e^{-x_3}\]
\[\Rightarrow\quad m(x_1,x_3) = \frac{x_1}{2}e^{-x_3}, \quad 0<x_1<1,\; x_3>0\]
(ii) Marginal density of \(X_1\) alone:
\[g(x_1) = \int_0^{\infty} \frac{x_1}{2}e^{-x_3}\,dx_3 = \frac{x_1}{2}, \quad 0<x_1<1\]
Recall the definition of conditional probability:
\[\Pr(A \mid B) = \frac{\Pr(A \cap B)}{\Pr(B)}, \quad \Pr(B) \neq 0\]
Definition 7 – Conditional Distribution (Discrete)
If \(f(x,y)\) is the joint distribution and \(h(y)\) is the marginal distribution of \(Y\), the conditional distribution of \(X\) given \(Y=y\) is:
\[f(x \mid y) = \frac{f(x,y)}{h(y)}, \quad h(y) \neq 0\]
Similarly, the conditional distribution of \(Y\) given \(X=x\) is:
\[w(y \mid x) = \frac{f(x,y)}{g(x)}, \quad g(x) \neq 0\]
Definition 8 – Conditional Density (Continuous)
For continuous random variables:
\[f(x \mid y) = \frac{f(x,y)}{h(y)}, \quad h(y) \neq 0, \quad -\infty < x < \infty\]
\[w(y \mid x) = \frac{f(x,y)}{g(x)}, \quad g(x) \neq 0, \quad -\infty < y < \infty\]
Example 12
Given \(f(x,y) = 4xy\) for \(0\leq x\leq1\), \(0\leq y\leq1\):
\[g(x) = \int_0^1 4xy\,dy = 2x, \quad h(y) = \int_0^1 4xy\,dx = 2y\]
Conditional density of \(X\) given \(Y=y\):
\[f(x \mid y) = \frac{4xy}{2y} = 2x, \quad 0\leq x\leq1\]
Note this does not depend on \(y\) — a sign of independence.
Example 13
Using Example 1 results:
Marginal of \(X\): \(g(0)=\frac{5}{12}\), \(g(1)=\frac{1}{2}\), \(g(2)=\frac{1}{12}\)
Marginal of \(Y\): \(h(0)=\frac{7}{12}\), \(h(1)=\frac{7}{18}\), \(h(2)=\frac{1}{36}\)
(i) Conditional distribution of \(X\) given \(Y=1\):
| \(x\) | 0 | 1 | 2 |
|---|---|---|---|
| \(f(x\mid Y=1)\) | \(\frac{4}{7}\) | \(\frac{3}{7}\) | \(0\) |
(ii) Conditional distribution of \(Y\) given \(X=0\):
| \(y\) | 0 | 1 | 2 |
|---|---|---|---|
| \(f(y\mid X=0)\) | \(\frac{2}{5}\) | \(\frac{8}{15}\) | \(\frac{1}{15}\) |
Definition 9 – Independence (Discrete)
The \(n\) discrete random variables \(X_1, X_2, \ldots, X_n\) with joint distribution \(f(x_1,\ldots,x_n)\) and marginals \(f_i(x_i)\) are independent if and only if:
\[f(x_1, x_2, \ldots, x_n) = f_1(x_1)\cdot f_2(x_2)\cdots f_n(x_n)\]
for all \((x_1,\ldots,x_n)\) within their range.
Definition 10 – Independence (Continuous)
The same condition holds for continuous random variables, replacing distributions with densities.
Example 14
Given \(f(x,y) = 24xy(1-x-y)\) for \(x\geq0\), \(y\geq0\), \(x+y\leq1\):
\[g(x) = 4x(1-x)^3, \quad h(y) = 12y^2(1-y)\]
Since \(g(x)\cdot h(y) = 48x(1-x)^3 y^2(1-y) \neq f(x,y)\), \(X\) and \(Y\) are not independent.
Example 15
From Example 1:
\[g(0)\cdot h(0) = \frac{5}{12}\cdot\frac{7}{12} = \frac{35}{144} \neq \frac{1}{6} = f(0,0)\]
Hence \(X\) and \(Y\) are not independent.
Example 16
\(n\) independent flips of a fair coin; \(X_i \in \{0,1\}\) (heads on flip \(i\)):
\[f_i(x_i) = \frac{1}{2} \quad\Rightarrow\quad f(x_1,\ldots,x_n) = \left(\frac{1}{2}\right)^n\]
Example 17
Independent random variables with exponential densities:
\[f_1(x_1)=e^{-x_1}, \quad f_2(x_2)=2e^{-2x_2}, \quad f_3(x_3)=3e^{-3x_3}, \quad x_i>0\]
Joint density: \(f(x_1,x_2,x_3) = 6e^{-(x_1+2x_2+3x_3)}\)
\[\Pr(X_1+X_2+X_3 \leq 1) \approx 0.020\]
Exercise 1. For the joint distribution below, find the marginal distributions, and the conditional distributions of \(X\) given \(Y=1\) and \(Y\) given \(X=0\).
| \(y \backslash x\) | \(-1\) | \(1\) |
|---|---|---|
| \(-1\) | \(\frac{1}{8}\) | \(\frac{1}{2}\) |
| \(0\) | \(0\) | \(\frac{1}{4}\) |
| \(1\) | \(\frac{1}{8}\) | \(0\) |
Exercise 2. Given \(f(x,y,z) = \frac{xyz}{108}\) for \(x=1,2,3\); \(y=1,2,3\); \(z=1,2\), find:
Exercise 3. Check independence: (a) \(f(x,y) = \frac{1}{4}\) — independent; (b) \(f(x,y) = \frac{1}{3}\) — not independent.
Exercise 4. If \(f(x,y) = \frac{1}{4}(x+y)\) for \(0<x<1\), \(0<y<2\), find marginal of \(Y\) and conditional density of \(X\) given \(Y=1\).
Exercise 5. For \(f(x,y) = \frac{2}{5}(4+xy)\), find the probability that at least 30% respond to the first solicitation: \(0.742\).
Definition 1 – Bivariate Normal Distribution
Random variables \(X\) and \(Y\) have a bivariate normal distribution if their joint density is:
\[f(x,y) = \frac{1}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}} \exp\!\left\{-\frac{1}{2(1-\rho^2)}\left[\left(\frac{x-\mu_1}{\sigma_1}\right)^2 - 2\rho\left(\frac{x-\mu_1}{\sigma_1}\right)\!\left(\frac{y-\mu_2}{\sigma_2}\right) + \left(\frac{y-\mu_2}{\sigma_2}\right)^2\right]\right\}\]
for \(-\infty < x < \infty\), \(-\infty < y < \infty\), where \(\sigma_1 > 0\), \(\sigma_2 > 0\), \(-1 < \rho < 1\).
By integrating over \(y\), the marginal density of \(X\) is:
\[g(x) = \frac{1}{\sigma_1\sqrt{2\pi}}\exp\!\left\{-\frac{(x-\mu_1)^2}{2\sigma_1^2}\right\}\]
i.e., \(X \sim N(\mu_1, \sigma_1^2)\). By symmetry, \(Y \sim N(\mu_2, \sigma_2^2)\).
The parameter \(\rho\) is the correlation coefficient: \(\text{Cov}(X,Y) = \rho\sigma_1\sigma_2\).
Theorem 1 – Conditional Distributions
If \(X\) and \(Y\) have a bivariate normal distribution, then:
\[\mu_{Y|X} = \mu_2 + \rho\frac{\sigma_2}{\sigma_1}(x - \mu_1), \qquad \sigma_{Y|X}^2 = \sigma_2^2(1-\rho^2)\]
\[\mu_{X|Y} = \mu_1 + \rho\frac{\sigma_1}{\sigma_2}(y - \mu_2), \qquad \sigma_{X|Y}^2 = \sigma_1^2(1-\rho^2)\]
Theorem 2 – Independence Condition
If \(X\) and \(Y\) have a bivariate normal distribution, they are independent if and only if \(\rho = 0\).
When \(\rho = 0\), the variables are called uncorrelated.
Note: Two random variables can each be marginally normal without their joint distribution being bivariate normal.
Example 1
Husband height \(X\) and wife height \(Y\) (in metres) have a bivariate normal distribution with:
\[Q(x,y) = 25(x-1.6)^2 - 40(x-1.6)(y-1.5) + 20(y-1.5)^2\]
and \(f(x,y) = ce^{-Q(x,y)/2}\).
Part (i): Find \(\mu_1, \mu_2, \sigma_1, \sigma_2, \rho\).
Comparing with the standard form \(\frac{1}{1-\rho^2}\left[\frac{(x-\mu_1)^2}{\sigma_1^2} - \frac{2\rho(x-\mu_1)(y-\mu_2)}{\sigma_1\sigma_2} + \frac{(y-\mu_2)^2}{\sigma_2^2}\right]\):
\[\mu_1 = 1.6\text{ m}, \quad \mu_2 = 1.5\text{ m}\]
From the equations:
\[\frac{1}{\sigma_1^2(1-\rho^2)} = 25, \quad \frac{1}{\sigma_2^2(1-\rho^2)} = 20, \quad \frac{\rho}{\sigma_1\sigma_2(1-\rho^2)} = 20\]
Solving: \(\sigma_1 \approx 0.4472\) m, \(\sigma_2 \approx 0.2236\) m, \(\rho = 0.5\).
Part (ii): Expected height of husband whose wife is 1.55 m tall.
\[E(X \mid Y = 1.55) = 1.6 + 0.5\cdot\frac{0.4472}{0.2236}(1.55-1.5) = 1.6 + 0.04 = 1.64 \text{ m}\]
Conditional variance:
\[\sigma_{X|Y}^2 = (0.4472)^2(1-0.25) = 0.2\times0.75 = 0.15\]
Example 2
\(X\) and \(Y\) have a bivariate normal distribution with \(\mu_1=70\), \(\mu_2=50\), \(\sigma_1^2=100\), \(\sigma_2^2=64\), \(\rho=\frac{4}{5}\).
(i) \(\Pr(42 < Y < 46)\):
\[\Pr\!\left(\frac{42-50}{8} < Z < \frac{46-50}{8}\right) = \Pr(-1 < Z < -0.5) \approx 0.1498\]
(ii) \(\Pr(42 < Y < 46 \mid X = 74)\):
The conditional distribution of \(Y\) given \(X=74\) is normal with:
\[\mu_{Y|X} = 50 + \frac{4}{5}\cdot\frac{8}{10}(74-70) = 50 + 2.56 = 52.56\]
\[\sigma_{Y|X}^2 = 64\left(1-\frac{16}{25}\right) = 64\cdot\frac{9}{25} = 23.04 \;\Rightarrow\; \sigma_{Y|X} = 4.8\]
\[\Pr(42 < Y < 46 \mid X=74) = \Pr\!\left(\frac{42-52.56}{4.8} < Z < \frac{46-52.56}{4.8}\right)\] \[= \Pr(-2.2 < Z < -1.367) \approx 0.0714\]
Exercise 1. The exponent of \(e\) in a bivariate normal density is:
\[-\frac{1}{2}\left[2(x-2)^2 - 2.8(x-2)(y-1) + 4(y-1)^2\right] \cdot \frac{1}{10}\]
Find \(\mu_1, \mu_2, \sigma_1, \sigma_2, \rho\). Answer: \(\mu_1=2,\;\mu_2=1,\;\sigma_1=\sqrt{10},\;\sigma_2=\sqrt{5},\;\rho=0.7\).
Exercise 2. Exponent: \(-\frac{1}{54}[4x^2-2xy+2y^2-8x-4y+4]\).
Find \(\mu_1, \mu_2, \sigma_1, \sigma_2, \rho\). Answer: \(\sigma_1=\sqrt{6},\;\sigma_2=\sqrt{3},\;\rho=-\frac{1}{2\sqrt{2}}\).
Exercise 3. \(X\) and \(Y\) have bivariate normal with \(\mu_1=2,\;\mu_2=5,\;\sigma_1=3,\;\sigma_2=6,\;\rho=\frac{2}{3}\).
Find \(\mu_{X|Y=\frac{1}{2}}\) and \(\sigma^2_{X|Y}\). Answer: \(\mu_{X|Y=1/2}=\frac{11}{2},\;\sigma^2_{X|Y}=\frac{3}{20}\).
Exercise 4. If \(U=X+Y\) and \(V=X-Y\) where \(X,Y\) are jointly normal, find \(\rho_{UV}\).
\[\rho_{UV} = \frac{\sigma_1^2-\sigma_2^2}{\sqrt{(\sigma_1^2+\sigma_2^2)^2-4\rho^2\sigma_1^2\sigma_2^2}}\]
Exercise 6. \(X\) (height, inches) and \(Y\) (weight, pounds) have bivariate normal with \(\mu_1=18,\;\mu_2=15,\;\sigma_1=3,\;\sigma_2=2,\;\rho=0.75\).
| Concept | Discrete | Continuous |
|---|---|---|
| Joint distribution | \(f(x,y) = \Pr(X=x, Y=y)\) | \(\Pr[(X,Y)\in A] = \iint_A f\,dx\,dy\) |
| Marginal of \(X\) | \(g(x) = \sum_y f(x,y)\) | \(g(x) = \int f(x,y)\,dy\) |
| Marginal of \(Y\) | \(h(y) = \sum_x f(x,y)\) | \(h(y) = \int f(x,y)\,dx\) |
| Conditional \(X\mid Y=y\) | \(f(x\mid y) = f(x,y)/h(y)\) | \(f(x\mid y) = f(x,y)/h(y)\) |
| Independence | \(f(x,y) = g(x)h(y)\) | \(f(x,y) = g(x)h(y)\) |
| Joint CDF | \(F(x,y)=\sum_{s\leq x}\sum_{t\leq y}f(s,t)\) | \(F(x,y)=\int_{-\infty}^x\int_{-\infty}^y f\,dt\,ds\) |
For the bivariate normal, the key parameters are \((\mu_1, \mu_2, \sigma_1, \sigma_2, \rho)\), where \(\rho\) controls the dependence. Independence holds if and only if \(\rho = 0\).